WBCHSE Class 12 Maths Solutions For Differential Equations

Differential Equations – Chapter 1 Differential Equations

Differential Equation An equation containing an independent variable, a dependent variable and the derivatives of the dependent variable is called a differential equation.

Examples Each of the following equations is a differential equation:

(1) \(\frac{d y}{d x}+5 y=e^x\)

(2) \(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+3 y=\sin x\)

(3) \(\frac{d y}{d x}=\frac{x^3-y^3}{x y^2-x^2 y}\)

(4) x2dx + y2dy = 0

Order Of A Differential Equation The order of the highest-order derivative occurring in a differential equation, is called the order of the differential equation.

Degree Of A Differential Equation The power of the highest-order derivative occurring in a differential equation, after it is made free from radicals and fractions, is called the degree of the differential equation.

Examples (1) Consider the equation \(\left(\frac{d y}{d x}\right)^2+5 y=\sin x .\)

In this equation, the order of the highest-order derivative is 1.

So, its order is 1.

The power of the highest-order derivative is 2.

So, its degree is 2.

Hence, the above equation is of order 1 and degree 2.

(2) Consider the equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^3+2 y=0 .\)

In this equation, the order of the highest-order derivative is 2.

So, its order is 2.

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The power of the highest-order derivative is 1.

So, its degree is 1.

Hence, the above equation is of order 2 and degree 1.

(3) The equation \(x\left(\frac{d^2 y}{d x^2}\right)^3+y\left(\frac{d y}{d x}\right)^4+y^2=0\) is an equation of order 2 and degree 3.

(4) The equation y dx = x dy may be written as \(\frac{d y}{d x}\) = \(\frac{y}{x}\).

So, it is differential equation of order 1 and degree 1.

Solved Examples

Example 1 Determine the order and degree of the differential equation

\(\left(\frac{d^2 y}{d x^2}\right)=\sqrt{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}} .\)

Solution

Given

\(\left(\frac{d^2 y}{d x^2}\right)=\sqrt{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}} .\)

The above equation when free from radicals, takes the form

\(\left(\frac{d^2 y}{d x^2}\right)^2=1+\left(\frac{d y}{d x}\right)^2 .\)

Clearly, it is an equation of order 2 and degree 2.

Example 2 Determine the order and degree of the differential equation \(y=p x+\sqrt{a^2 p^2+b^2}, \text { where } p=\frac{d y}{d x} \text {. }\)

Solution

Given

\(y=p x+\sqrt{a^2 p^2+b^2}, \text { where } p=\frac{d y}{d x} \text {. }\) \(y=p x+\sqrt{a^2 p^2+b^2}\)

⇔ \(y-p x=\sqrt{a^2 p^2+b^2}\)

⇔ (y – px)2 = a2p2 + b2

⇔ y2 + x2p2 – 2xyp = a2p2 + b2

⇔ (x2-a2)p2 – 2xyp + (y2 – b2) = 0

⇔ \(\left(x^2-a^2\right)\left(\frac{d y}{d x}\right)^2-2 x y \cdot\left(\frac{d y}{d x}\right)+\left(y^2-b^2\right)=0\).

Clearly, it is a differential equation of order 1 and degree 2.

Solution Of A Differential Equation A function of the form y = f(x) + C which satisfies a given differential equation is called its solution.

General Solution Of A Differential Equation Suppose a differential equation of order n is being given. If its solution contains n arbitrary constants then it is called a general solution.

Particular Solution Of A Differential Equation Giving particular values to arbitrary constants in the general solution of a differential equation, we get its particular solutions.

Solved Examples

Example 1 Verify that y = A cos x – B sin x is a solution of the differential equation \(\frac{d^2 y}{d x^2}+y=0\).

Solution

Given: y = A cos x – B sin x …(1)

⇒ \(\frac{d y}{d x}\) = -A sin x – B cos x

⇒ \(\frac{d^2 y}{d x^2}\) = -A cos x + B sin x

= -(A cos x – B sin x) = -y [from (1)]

⇒ \(\frac{d^2 y}{d x^2}+y=0\).

Hence, y = A cos x – B sin x is a solution of the differential equation \(\frac{d^2 y}{d x^2}+y=0\).

Example 2 Verify that y = a e2x + b e-x is a solution of the differential equation \(\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y=0 .\)

Solution

Given: y = a e2x + b e-x …(1)

⇒ \(\frac{d y}{d x}=2 a e^{2 x}-b e^{-x}\) …(2)

⇒ \(\frac{d^2 y}{d x^2}=4 a e^{2 x}+b e^{-x}\) …(3)

∴ \(\left(\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y\right)=\left(4 a e^{2 x}+b e^{-x}\right)-\left(2 a e^{2 x}-b e^{-x}\right)-2\left(a e^{2 x}+b e^{-x}\right)\) [using (1),(2) and (3)]

= 0

Hence, y = a e2x + b e-x is a solution of the differential equation \(\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y=0 .\)

Example 3 Verify that \(y=A x+\frac{B}{x}\) is a solution of the differential equation \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=0 .\)

Solution

Given: y = Ax + \(\frac{B}{x}\) …(1)

⇒ \(\frac{d y}{d x}=A-\frac{B}{x^2}\) …(2)

⇒ \(\frac{d^2 y}{d x^2}=\frac{2 B}{x^3}\) …(3)

Substituting the values of y, \(\frac{d y}{d x}\) and \(\frac{d^2 y}{d x^2}\) from (1), (2) and (3), we get

\(\left(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y\right)=x^2 \cdot \frac{2 B}{x^3}+x\left(A-\frac{B}{x^2}\right)-\left(A x+\frac{B}{x}\right)\)

= \(\left(\frac{2 B}{x}+A x-\frac{B}{x}-A x-\frac{B}{x}\right)=0 \text {. }\)

Thus, y = Ax + \(\frac{B}{x}\) satisfies \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=0 .\)

Hence, y = Ax + \(\frac{B}{x}\) is a solution of \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=0 .\)

Example 4 Verify that y = a cos(log x) + b sin (log x) is a solution of the differential equation \(x^2+\frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0 .\)

Solution

Given: y = a cos(log x) + b sin (log x) …(1)

⇒ \(\frac{d y}{d x}=\frac{-a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\) [on differentiating (1) w.r.t. x]

⇒ \(x \frac{d y}{d x}=-a \sin (\log x)+b \cos (\log x)\) …(2)

⇒ \(x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{-a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\) [on differentiating (2) w.r.t. x]

⇒ \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=-a \cos (\log x)-b \sin (\log x)\)

⇒ \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0\) [using (1)].

Hence, y = a cos (log x) + b sin (log x) is a solution of \(x^2+\frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0 .\)

Example 5 Verify that y = emsin-1x is a solution of the differential equation \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0\)

Solution

Given:

y = emsin-1x …(1)

⇒ \(\frac{d y}{d x}=\frac{e^{m \sin ^{-1} x}}{\sqrt{1-x^2}} \cdot m\) [on differentiating (1)]

⇒ \(\sqrt{1-x^2}\left(\frac{d y}{d x}\right)=m y\) …(2) [∵ em sin-1x = y]

⇒ \(\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=m^2 y^2\) …(3) [on squaring both sides of (2)]

⇒ \(\left(1-x^2\right) 2 \frac{d y}{d x} \cdot\left(\frac{d^2 y}{d x^2}\right)-2 x\left(\frac{d y}{d x}\right)^2=2 m^2 y \frac{d y}{d x}\)

[on differentiating both sides of (3)]

⇒ \(2\left(\frac{d y}{d x}\right) \cdot\left\{\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y\right\}=0\)

⇒ \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0\)

Hence, y = emsin-1x is a solution of the differential equation \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0\)

Example 6 Verify that v = \(\frac{A}{r}\) + B is a solution of the differential equation \(\frac{d^2 v}{d r^2}+\frac{2}{r} \cdot \frac{d v}{d r}=0\)

Solution

Since the given relation contains two arbitrary constants, we differentiate it two times w.r.t. r, and eliminate A and B.

∴ v = \(\frac{A}{r}\) + B ⇒ \(\frac{d v}{d r}=\frac{-A}{r^2}\) …(1)

⇒ \(\frac{d^2 v}{d r^2}=\frac{2 A}{r^3}\) …(2)

On dividing (2) by (1), we get

\(\frac{\left(d^2 v / d r^2\right)}{(d v / d r)}=\left\{\frac{2 A}{r^3} \times \frac{r^2}{(-A)}\right\}=\frac{-2}{r}\)

⇒ \(\frac{d^2 v}{d r^2}=\frac{-2}{r} \cdot \frac{d v}{d r}\)

⇒ \(\frac{d^2 v}{d r^2}+\frac{2}{r} \cdot \frac{d v}{d r}=0 .\)

Hence, v = \(\frac{A}{r}\) + B is a solution of the differential equation \(\frac{d^2 v}{d r^2}+\frac{2}{r} \cdot \frac{d v}{d r}=0 .\)

Formation of a Differential Equation whose General Solution is Given

Method Suppose an equation of a family of curves contains n arbitrary constants (called parameters).

Then, we obtain its differential equation, as given below.

Step 1. Differentiate the equation of the given family of curves n times to get n more equations.

Step 2. Eliminate n constants, using these (n+1) equations.

This gives us the required differential equation of order n.

Solved Examples

Example 1 Find the differential equation of the family of curves y = Aex + Be-x, where A and B are arbitrary constants.

Solution

The equation of the given family of curves is y = Aex + Be-x …(1)

Since the given equation contains two arbitrary constants, we differentiate it two times w.r.t. x.

Now, \(\frac{d y}{d x}=A e^x-B e^{-x}\)

⇒ \(\frac{d^2 y}{d x^2}=A e^x+B e^{-x}\) ⇒ \(\frac{d^2 y}{d x^2}=y\)

⇒ \(\frac{d^2 y}{d x^2}-y=0\), which is the required differential equation.

Example 2 Find the differential equation of the family of curves y = ex(A cos x + B sin x), where A and B are arbitrary constants.

Solution

The equation of the given family of curves is y = ex(A cos x + B sin x) …(1)

Since the given equation contains two arbitrary constants, we differentiate it two times w.r.t x.

Now, \(\frac{d y}{d x}=e^x(-A \sin x+B \cos x)+e^x(A \cos x+B \sin x)\)

⇒ \(\frac{d y}{d x}-y=e^x(-A \sin x+B \cos x)\) …(2)

⇒ \(\frac{d^2 y}{d x^2}-\frac{d y}{d x}=e^x(-A \cos x-B \sin x)+e^x(-A \sin x+B \cos x)\)

⇒ \(\frac{d^2 y}{d x^2}-\frac{d y}{d x}=-y+\left(\frac{d y}{d x}-y\right)\) [using (1) and (2)]

⇒ \(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0\), which is the required differential equation of the given family of curves.

Example 3 Find the differential equation for the family of all concentric circles centred at the origin and having different radii.

Solution

The equation of the given family of circles is x2 + y2 = r2, r being a parameter …(1)

On differentiating (1) w.r.t. x, we get

Class 12 Maths Differential Equations Example 3

\(2 x+2 y \cdot \frac{d y}{d x}=0\)

⇒ \(x+y \frac{d y}{d x}=0\), which is the required differential equation.

Example 4 Find the differential equation of the family of all straight lines passing through the origin.

Solution

The general equation of the family of all straight lines passing through the origin is y = mx …(1)

Differentiating (1) w.r.t. x, we get \(\frac{d y}{d x}\) = m …(2)

Class 12 Maths Differential Equations Example 4

Substituting the value of m from (2) in (1), we get

y = \(x\left(\frac{d y}{d x}\right)\), which is the required differential equation.

Example 5 Find the differential equation of the family of all straight lines.

Solution

The general equation of the family of all straight lines is given by y = mx + c, where m and c are parameters.

Class 12 Maths Differential Equations Example 5

Now, y = mx + c ⇒ \(\frac{d y}{d x}=m\)

⇒ \(\frac{d^2 y}{d x^2}=0 \text {. }\)

So, the required differential equation is \(\frac{d^2 y}{d x^2}=0 \text {. }\)

Example 6 Find the differential equation of the family of all circles touching the x-axis at the origin.

Solution

We know that the equation of a circle with centre (0, a) and radius a is given by x2 + (y-a)2 = a2 or x2 + y2 = 2ay

Class 12 Maths Differential Equations Example 6

So, the general equation of the family of all circles touching the x-axis at the origin is given by

x2 + y2 = 2ay …(1)

where a is a parameter.

On differentiating (1) w.r.t. x, we get

\(2 x+2 y \frac{d y}{d x}=2 a \frac{d y}{d x} \Rightarrow a \frac{d y}{d x}=x+y \frac{d y}{d x}\)

⇒ \(a=\frac{x}{\left(\frac{d y}{d x}\right)}+y\) …(2)

Substituting the value of a from (2) in (1), we get

\(x^2+y^2=2 y\left\{\frac{x}{\left(\frac{d y}{d x}\right)}+y\right\} \Rightarrow\left(x^2-y^2\right)=\frac{2 x y}{\left(\frac{d y}{d x}\right)}\)

⇒ \(\frac{d y}{d x}=\frac{2 x y}{\left(x^2-y^2\right)}\)

Hence, \(\frac{d y}{d x}=\frac{2 x y}{\left(x^2-y^2\right)}\) is the required differential equation.

Example 7 Find the differential equation of the family of all parabolas having vertex at the origin and axis along the positive direction of the x-axis.

Solution

Let C be the family of all parabolas having vertex at origin and axis along positive direction of x-axis. Let F(a,0) be the focus of a member of this family, where a is an arbitrary constant.

Class 12 Maths Differential Equations Example 7

Then, the equation of this family is

y2 = 4ax …(1)

where a is the parameter.

Now, y2 = 4ax ⇒ \(2 y \frac{d y}{d x}=4 a\) …(2)

Substituting the value of 4a from (2) in (1), we get

y2 = \(2 x y \frac{d y}{d x}\) ⇒ \(\left(y^2-2 x y \frac{d y}{d x}\right)=0\)

Hence, \(y^2-2 x y \frac{d y}{d x}=0\) is the required differential equation.

Example 8 Find the differential equation of the family of all ellipses having foci on the x-axis and centre at the origin.

Solution

Let C be the family of all ellipses having foci on the x-axis and centre at the origin.

Class 12 Maths Differential Equations Example 8

The general equation of such a family is

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) …(1)

where a and b are parameters.

Differentiating (1) w.r.t. x, we get

\(\frac{2 x}{a^2}+\frac{2 y}{b^2} \cdot \frac{d y}{d x}=0 \Rightarrow \frac{y}{b^2} \cdot \frac{d y}{d x}=\frac{-x}{a^2}\)

⇒ \(\frac{y}{x} \cdot \frac{d y}{d x}=\frac{-b^2}{a^2}\) …(2)

Differentiating (2) w.r.t. x, we get

\(\frac{y}{x} \cdot\left(\frac{d^2 y}{d x^2}\right)+\frac{d y}{d x} \cdot \frac{\left(x \frac{d y}{d x}-y\right)}{x^2}=0\)

⇒ \((x y)\left(\frac{d^2 y}{d x^2}\right)+x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)=0 .\)

Hence, \((x y)\left(\frac{d^2 y}{d x^2}\right)+x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)=0 .\) is the required differential equation.

Example 9 Find the differential equation of the family of all circles in second quadrant and touching the coordinate axes.

Solution

Let C be the family of all circles in second quadrant and touching the coordinate axes.

The coordinates of centre of an arbitrary member of this family is (-a, a) and its radius is a.

Class 12 Maths Differential Equations Example 9

Thus, the equation of such a family is

(x+a)2 + (y-a)2 = a2 …(1)

where a is an arbitrary constant.

This equation may be written as

x2 + y2 + 2ax – 2ay + a2 = 0 …(2)

⇒ \(2 x+2 y \frac{d y}{d x}+2 a-2 a \frac{d y}{d x}=0\) [on differentiating (1) w.r.t. x]

⇒ \(x+y \frac{d y}{d x}=a\left(\frac{d y}{d x}-1\right)\) ⇒ \(a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}-1}\)

⇒ \(a=\frac{x+y y^{\prime}}{\left(y^{\prime}-1\right)}\) …(3)

where \(\frac{d y}{d x}\) = y’.

Putting the values of a from (3) in (1), we get

\(\left(x+\frac{x+y y^{\prime}}{y^{\prime}-1}\right)^2+\left(y-\frac{x+y y^{\prime}}{y^{\prime}-1}\right)^2=\left(\frac{x+y y^{\prime}}{y^{\prime}-1}\right)^2\)

⇒ (xy’ – x + x + yy’)2 + (yy’ – y – x – yy’)2 = (x + yy’)2

⇒ (xy’ + yy’)2 + (x + y)2 = (x + yy’)2

⇒ (x + y)2 . [y’2 + 1] = (x + yy’)2. which is the required differential equation.

Example 10 Find the differential equation of the family of all circles of radius r.

Solution

The equation of the family of all circles of radius r is (x-a)2 + (y-b)2 = r2 …(1)

where a and b are arbitrary constants.

Differentiating (1) w.r.t. x, we get

2(x-a)+2(y-b) \(\frac{d y}{d x}=0\)

⇒ (x-a)+(y-b) \(\frac{d y}{d x}=0\) …(2)

⇒ \(1+(y-b) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0\) [on differentiating (2) w.r.t.x]

⇒ \((y-b)=-\left\{\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)}\right\}\) …(3)

Substituting the value of (y-b) in (2), we get

\((x-a)=-\left\{\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)}\right\} \cdot\left(\frac{d y}{d x}\right)\) …(4)

Putting the values of (y – b) and (x – a) from (3) and (4) in (1), we get

\(\frac{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^2}{\left(\frac{d^2 y}{d x^2}\right)^2} \cdot\left(\frac{d y}{d x}\right)^2+\frac{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^2}{\left(\frac{d^2 y}{d x^2}\right)^2}=r^2\)

⇒ \(\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^3=r^2\left(\frac{d^2 y}{d x^2}\right)^2\), which is the required differential equation.

Solution of Differential Equations

Solving Differential Equations With Variables Separable If the given differential equation can be expressed in the form f(x)dx = g(y)dy, then \(\int f(x) d x=\int g(y) d y+C\) is the solution of such a differential equation.

Solved Examples

Example 1 Solve the differential equation \(\frac{d y}{d x}=\frac{x-1}{y+2}(y \neq-2) .\)

Solution

We have \(\frac{d y}{d x}=\frac{x-1}{y+2}(y \neq-2) .\)

⇒ (y+2)dy = (x-1)dx [separating the variables]

⇒ \(\int(y+2) d y=\int(x-1) d x\)

⇒ \(\frac{y^2}{2}+2 y=\frac{x^2}{2}-x+C_1\)

⇒ y2 + 4y = x2 – 2x + 2C1

⇒ y2 + 4y – x2 + 2x = C, where C = 2C1.

Hence, y2 + 4y – x2 + 2x = C is the general solution of the given differential equation.

Example 2 Solve the differential equation \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2} .\)

Solution

We have \(\frac{d y}{d x}=\frac{\left(1+y^2\right)}{\left(1+x^2\right)}\)

⇒ \(\frac{1}{\left(1+y^2\right)} d y=\frac{1}{\left(1+x^2\right)} d x\) [separating the variables]

⇒ \(\int \frac{1}{\left(1+y^2\right)} d y=\int \frac{1}{\left(1+x^2\right)} d x\)

⇒ tan-1y = tan-1x + C1

⇒ tan-1y – tan-1x = C1

⇒ \(\tan ^{-1}\left(\frac{y-x}{1+y x}\right)=C_1\)

⇒ \(\frac{y-x}{1+y x}=\tan C_1\)

⇒ \(\frac{y-x}{1+y x}=C\), where C = tan C1.

Hence, \(\frac{y-x}{1+y x}=C\) is the general solution of the given differential equation.

Example 3 Solve the differential equation \(\frac{d y}{d x}=\log (x+1) \text {. }\)

Solution

\(\frac{d y}{d x}=\log (x+1)\)

⇒ dy = log(x+1)dx [seperating the variables]

⇒ \(\int d y=\int \log (x+1) d x\)

⇒ \(y=\int 1 \cdot \log (x+1)+C\)

= \(x \log (x+1)-\int \frac{1}{(x+1)} \cdot x d x+C\) [integrating by parts]

= \(x \log (x+1)-\int \frac{(x+1)-1}{(x+1)} d x+C\)

= \(x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) d x+C\)

= x log(x+1) – x + log(x+1) + C

= (x+1) log(x+1) – x + C.

Hence, y = (x+1) log(x+1) – x + C is the required solution.

Example 4 Solve the differential equation \(\frac{d y}{d x}=\sin ^{-1} x .\)

Solution

\(\frac{d y}{d x}=\sin ^{-1} x\)

⇒ dy = sin-1x dx [separating the variables]

⇒ \(\int d y=\int \sin ^{-1} x d x\)

⇒ \(y=\int 1 \cdot \sin ^{-1} x d x\)

= \(\left(\sin ^{-1} x\right) x-\int \frac{1}{\sqrt{1-x^2}} \cdot x d x+C\)

= \(\left(\sin ^{-1} x\right) x+\frac{1}{2} \cdot \int \frac{-2 x}{\sqrt{1-x^2}} d x+C\)

= \(\left(\sin ^{-1} x\right) x+\frac{1}{2} \int \frac{1}{\sqrt{t}} d t+C\), where (1 – x2) = t

= \(\left(\sin ^{-1} x\right) x+\frac{1}{2} \times 2 \sqrt{t}+C\)

= \(\left(\sin ^{-1} x\right) x+\sqrt{1-x^2}+\text { C. }\)

Hence, y = \(\left(\sin ^{-1} x\right) x+\sqrt{1-x^2}+\text { C. }\)

Example 5 Solve the differential equation \(\log \left(\frac{d y}{d x}\right)=(a x+b y)\)

Solution

We have

\(\log \left(\frac{d y}{d x}\right)=(a x+b y)\)

⇒ \(\frac{d y}{d x}=e^{a x+b y}=e^{a x} \cdot e^{b y}\)

⇒ \(\frac{1}{e^{b y}} d y=e^{a x} d x\) [on separating the variables]

⇒ \(\int e^{-b y} d y=\int e^{a x} d x\) [integrating both sides]

⇒ \(\frac{e^{-b y}}{-b}=\frac{e^{a x}}{a}+C\)

⇒ ae-by + beax = C’, where C’ = -abC.

Thus, ae-by + beax = C’ is the required solution.

Example 6 Solve the differential equation \(\frac{d y}{d x}=\sqrt{4-y^2}\)

Solution

We have \(\frac{d y}{d x}=\sqrt{4-y^2}\)

⇒ \(\frac{d y}{\sqrt{4-y^2}}=d x\) [on separating the variables]

⇒ \(\int \frac{d y}{\sqrt{2^2-y^2}}=\int d x\) [integrating both sides]

⇒ \(\sin ^{-1}\left(\frac{y}{2}\right)=x+C \text {. }\)

Hence, \(\sin ^{-1}\left(\frac{y}{2}\right)=x+C\) is the required solution.

Example 7 Solve the differential equation \(\frac{d y}{d x}=1-x+y-x y\)

Solution

We have

\(\frac{d y}{d x}=1-x+y-x y\)

⇒ \(\frac{d y}{d x}=(1-x)+y(1-x)\)

⇒ \(\frac{d y}{d x}=(1-x)(1+y)\)

⇒ \(\frac{d y}{(1+y)}=(1-x) d x\) [on separating the variables]

⇒ \(\int \frac{d y}{(1+y)}=\int(1-x) d x\) [integrating both sides]

⇒ \(\log |1+y|=x-\frac{x^2}{2}+C\)

Hence, \(\log |1+y|=x-\frac{x^2}{2}+C\) is the required solution.

Example 8 Solve the differential equation x(1 + y2)dx – y(1 + x2)dy = 0, given that y = 0 when x = 1.

Solution

The given differential equation is

x(1 + y2)dx – y(1 + x2)dy = 0

⇒ x(1+y2)dx = y(1+x2)dy

⇒ \(\frac{y}{\left(1+y^2\right)} d y=\frac{x}{\left(1+x^2\right)} d x\)

⇒ \(\int \frac{y}{\left(1+y^2\right)} d y=\int \frac{x}{\left(1+x^2\right)} d x\)

⇒ \(\frac{1}{2} \int \frac{2 y}{\left(1+y^2\right)} d y=\frac{1}{2} \int \frac{2 x}{\left(1+x^2\right)} d x\)

⇒ \(\frac{1}{2} \log \left(1+y^2\right)=\frac{1}{2} \log \left(1+x^2\right)+C\) …(1)

Putting x= 1 and y = 0 in (1), we get C = –\(\frac{1}{2}\) log 2.

∴ \(\frac{1}{2} \log \left(1+y^2\right)=\frac{1}{2} \log \left(1+x^2\right)-\frac{1}{2} \log 2\)

⇒ log(1+y2) = log(1+x2) – log 2

⇒ \(\log \left(1+y^2\right)=\log \left(\frac{1+x^2}{2}\right)\)

⇒ \(\left(1+y^2\right)=\left(\frac{1+x^2}{2}\right)\)

⇒ (x2 – 2y2) = 1.

Hence, the required solution is (x2 – 2y2) = 1.

Example 9 Solve the differential equation \(\frac{d y}{d x}=e^{x+y}+x^2 \cdot e^y \text {. }\)

Solution

We have

\(\frac{d y}{d x}=e^{x+y}+x^2 \cdot e^y\)

= ex . ey + x2 . ey

= (ex + x2).ey

⇒ e-ydy = (ex+x2)dx [separating the variables]

⇒ \(\int e^{-y} d y=\int\left(e^x+x^2\right) d x\)

⇒ \(-e^{-y}=e^x+\frac{x^3}{3}+C_1\)

⇒ \(e^x+e^{-y}+\frac{x^3}{3}=C\), where C = -C1.

Hence, \(e^x+e^{-y}+\frac{x^3}{3}=C\) is the required solution.

Example 10 Solve the differential equation \(\frac{d y}{d x}=y \sin 2 x\), given that y(0) = 1.

Solution

We have

\(\frac{d y}{d x}=y \sin 2 x\)

⇒ \(\frac{1}{y} d y=\sin 2 x d x\) [separating the variables]

⇒ \(\int \frac{1}{y} d y=\int \sin 2 x d x+C\)

⇒ \(\log y=-\frac{1}{2} \cos 2 x+C\)

Putting x = 0 and y = 1 in (1), we get C = \(\frac{1}{2}\).

∴ \(\log y=-\frac{1}{2} \cos 2 x+\frac{1}{2}\)

= \(\frac{1}{2}(1-\cos 2 x)=\left(\frac{1}{2} \times 2 \sin ^2 x\right)=\sin ^2 x\)

⇒ log y = sin2x ⇒ y = esin2x.

Hence, y = esin2x is the required solution.

Example 11 Solve the differential equation (1+e2x)dy + ex(1+y2)dx = 0, it being given that y = 1 when x = 0.

Solution

We have

(1+e2x)dy + ex(1+y2)dx = 0

⇒ \(\frac{1}{\left(1+y^2\right)} d y+\frac{e^x}{\left(1+e^{2 x}\right)} d x=0\) [separating the variables]

⇒ \(\int \frac{1}{\left(1+y^2\right)} d y+\int \frac{e^x}{\left(1+e^{2 x}\right)} d x=\mathrm{C}\)

⇒ \(\tan ^{-1} y+\int \frac{d t}{\left(1+t^2\right)}=C\), where ex = t

⇒ tan-1y + tan-1t = C

⇒ tan-1y + tan-1ex = C …(1) [∵ t = ex]

Putting x = 0 and y = 1 in (1), we get

C = \(\tan ^{-1} 1+\tan ^{-1} e^0=\left(\tan ^{-1} 1+\tan ^{-1} 1\right)=\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\frac{\pi}{2} .\)

∴ tan-1y + tan-1ex = \(\frac{\pi}{2}\) is the required solution.

Example 12 Solve the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0 .\)

Solution

\(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)

⇒ \(\frac{1}{\sqrt{1-y^2}} d y+\frac{1}{\sqrt{1-x^2}} d x=0\) [on separating the variables]

⇒ \(\int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=C\) [integrating both sides]

⇒ sin-1y + sin-1x = C .

Hence, sin-1y + sin-1x = C is the required solution.

Example 13 Solve the differential equation \(x \sqrt{1-y^2} d x+y \sqrt{1-x^2} d y=0\).

Solution

We have

\(x \sqrt{1-y^2} d x+y \sqrt{1-x^2} d y=0\)

⇒ \(\frac{x}{\sqrt{1-x^2}} d x+\frac{y}{\sqrt{1-y^2}} d y=0\) [on separating the variables]

⇒ \(\int \frac{x}{\sqrt{1-x^2}} d x+\int \frac{y}{\sqrt{1-y^2}} d y=C\) [integrating both sides]

⇒ \(-\frac{1}{2} \cdot \int \frac{(-2 x)}{\sqrt{1-x^2}} d x-\frac{1}{2} \cdot \int \frac{(-2 y)}{\sqrt{1-y^2}} d y=C\)

⇒ \(-\frac{1}{2} \cdot \int \frac{1}{\sqrt{t}} d t-\frac{1}{2} \cdot \int \frac{1}{\sqrt{s}} d s=\mathrm{C}\), where (1-x2) = t and (1-y2) = s

⇒ \(-\frac{1}{2} \int t^{-1 / 2} d t-\frac{1}{2} \int s^{-1 / 2} d s=C\)

⇒ -√t -√s = C

⇒ √t + √s = k [where k = -C]

⇒ \(\sqrt{1-x^2}+\sqrt{1-y^2}=k\)

Hence, \(\sqrt{1-x^2}+\sqrt{1-y^2}=k\) is the required solution.

Example 14 Solve the differential equation \(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x} .\)

Solution

We have

\(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}\)

= \(\frac{2 \sin ^2(x / 2)}{2 \cos ^2(x / 2)}=\tan ^2\left(\frac{x}{2}\right)\)

⇒ \(d y=\tan ^2\left(\frac{x}{2}\right) d x\) [separating the variables]

⇒ \(\int d y=\int \tan ^2 \frac{x}{2} d x+C\)

= \(\int\left(\sec ^2 \frac{x}{2}-1\right) d x+C\)

⇒ \(y=2 \tan \frac{x}{2}-x+C \text {. }\)

Hence, \(y=2 \tan \frac{x}{2}-x+C\) is the required solution.

Example 15 Solve the differential equation (1+y2)(1+log x)dx + x dy = 0, it being given that y = 1 when x = 1.

Solution

We have

(1+y2)(1+log x)dx + x dy = 0

⇒ \(\frac{(1+\log x)}{x} d x+\frac{1}{\left(1+y^2\right)} d y=0\) [on separating the variables]

⇒ \(\int \frac{(1+\log x)}{x} d x+\int \frac{1}{\left(1+y^2\right)} d y=C\) [integrating both sides]

⇒ \(\int t d t+\tan ^{-1} y=C, where (1+log x) = t ⇒ \frac{1}{2} t^2+\tan ^{-1} y=C\)

⇒ \(\frac{1}{2}(1+\log x)^2+\tan ^{-1} y=C\) …(1)

Putting x = 1 and y = 1 in (1), we get:

\(C=\frac{1}{2}+\tan ^{-1} 1 \Rightarrow C=\left(\frac{1}{2}+\frac{\pi}{4}\right)\) …(2)

∴ \(\frac{1}{2}(1+\log x)^2+\tan ^{-1} y=\left(\frac{1}{2}+\frac{\pi}{4}\right)\) [using (2) in (1)]

⇒ \(\frac{1}{2}(\log x)^2+\log x+\tan ^{-1} y=\frac{\pi}{4}\), where is the required solution.

Example 16 Find the equation of the curve that passes through the point (1,2) and satisfies the differential equation \(\frac{d y}{d x}=\frac{-2 x y}{\left(x^2+1\right)} \text {. }\)

Solution

We have

\(\frac{d y}{d x}=\frac{-2 x y}{\left(x^2+1\right)}\)

⇒ \(\frac{d y}{y}=\frac{-2 x}{\left(x^2+1\right)} d x\) [on separating the variables]

⇒ \(\int \frac{d y}{y}=\int \frac{-2 x}{\left(x^2+1\right)} d x\) [integrating both sides]

⇒ log y = -log(x2+1) + log C

⇒ log y + log(x2+1) = log C

⇒ log{y(x2+1)} = log C

⇒ y(x2+1) = C

Now, it is given that the curve passes through (1,2).

So, putting x = 1 and y = 2 in (1), we get C = 4.

∴ y(x2+1) = 4 is the required equation of the curve.

Example 17 Solve the differential equation \((x-1) \frac{d y}{d x}=2 x^3 y .\)

Solution

We have

\((x-1) \frac{d y}{d x}=2 x^3 y\)

⇒ \(\frac{1}{y} d y=\frac{2 x^3}{(x-1)} d x\) [on separating the variables]

⇒ \(\int \frac{d y}{y}=\int \frac{2 x^3}{(x-1)} d x\) [integrating both sides]

⇒ \(\log y=2 \int\left\{x^2+x+1+\frac{1}{(x-1)}\right\} d x\)

⇒ \(\log y=2\left(\frac{x^3}{3}+\frac{x^2}{2}+x\right)+2 \log |x-1|+C\)

Hence, \(\log y=2\left(\frac{x^3}{3}+\frac{x^2}{2}+x\right)+2 \log |x-1|+C\) is the required solution.

Example 18 Solve the differential equation cos x(1+cos y)dx – sin y(1+sin x)dy = 0.

Solution

We have

cos x(1+cos y)dx – sin y(1+sin x)dy = 0 …(1)

⇒ \(\frac{\cos x}{(1+\sin x)} d x-\frac{\sin y}{(1+\cos y)} d y=0\)

⇒ \(\int \frac{\cos x}{(1+\sin x)} d x-\int \frac{\sin y}{(1+\cos y)} d y=\log C\), where C is a constant

⇒ log|1 + sin x| + log|1 + cos y| = log C

⇒ log|(1 + sin x)(1 + cos y)| = log C

⇒ (1 + sin x)(1 + cos y) = C.

Hence, (1+ sin x)(1 + cos y) = C is the required solution.

Example 19 Solve the differential equation \(y-x \frac{d y}{d x}=a\left(y^2+\frac{d y}{d x}\right)\).

Solution

We have

\(y-x \frac{d y}{d x}=a\left(y^2+\frac{d y}{d x}\right)\)

⇒ \(\left(y-a y^2\right)=(a+x) \frac{d y}{d x}\)

⇒ \(\frac{d y}{y(1-a y)}=\frac{d x}{(a+x)}\) [on separating the variables]

⇒ \(\int \frac{d y}{y(1-a y)}=\int \frac{d x}{(a+x)}\)

⇒ \(\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y=\int \frac{d x}{(a+x)}\) [by partial fractions]

⇒ log|y| – log|1 – ay| = log|a + x| + C

⇒ \(\log \left|\frac{y}{(1-a y)(a+x)}\right|=C\)

⇒ \(\frac{y}{(1-a y)(a+x)}=e^c=k\) (say)

⇒ y = k(1 – ay)(a + x), where k is a constant.

Hence, y = k(1 – ay)(a + x) is the required solution.

Example 20 Solve the differential equation \((\sqrt{a+x}) \frac{d y}{d x}+x=0 \text {. }\)

Solution

We have

\(\frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\)

⇒ \(d y=\frac{-x}{\sqrt{a+x}} d x\) [on separating the variables]

⇒ \(\int d y=\int \frac{-x}{\sqrt{a+x}} d x\) [integrating both sides]

⇒ \(y=-\int \frac{[(a+x)-a]}{\sqrt{a+x}} d x\)

⇒ \(y=-\int\left(\sqrt{a+x}-\frac{a}{\sqrt{a+x}}\right) d x\)

⇒ \(y=-\int \sqrt{a+x} d x+a \int(a+x)^{-1 / 2} d x\)

⇒ \(y=-\frac{2}{3}(a+x)^{3 / 2}+2 a \sqrt{a+x}+C\) is the required solution.

Example 21 Solve the differential equation x cos y dy = (x ex log x + ex)dx.

Solution

We have

x cos y dy = (x ex log x + ex)dx

⇒ \(\cos y d y=e^x\left(\log x+\frac{1}{x}\right) d x\) [on separating the variables]

⇒ \(\int \cos y d y=\int e^x\left(\log x+\frac{1}{x}\right) d x\) [integrating both sides]

⇒ sin y = ex(log x) + C [∵ \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)\)].

Hence, sin y = ex(log x) + C is the required solution.

Example 22 Solve the differential equation \(\frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)} .\)

Solution

We have

\(\frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)}\)

⇒ y(2 log y + 1)dy = ex(sin2x + sin 2x)dx

⇒ \(2 \int y \log y d y+\int y d y=\int e^x\left(\sin ^2 x+\sin 2 x\right) d x\)

⇒ \(2\left[(\log y) \cdot \frac{y^2}{2}-\int \frac{1}{y} \cdot \frac{y^2}{2} d y\right]+\frac{1}{2} y^2=\int e^x\left(\sin ^2 x+\sin 2 x\right) d x\) [integrating by parts]

⇒ \(y^2(\log y)-\int y d y+\frac{1}{2} y^2=\int e^x\left(\sin ^2 x+\sin 2 x\right) d x\)

⇒ y2(log y) = ex sin2x + C [∵ \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)\)].

∴ y2(log y) = ex sin2x + C is the required solution.

Example 23 Solve the differential equation (1+x)(1+y2)dx + (1+y)(1+x2)dy = 0.

Solution

(1+x)(1+y2)dx + (1+y)(1+x2)dy = 0

⇒ \(\frac{(1+x)}{\left(1+x^2\right)} d x+\frac{(1+y)}{\left(1+y^2\right)} d y=0\) [on separating the variables]

⇒ \(\int \frac{(1+x)}{\left(1+x^2\right)} d x+\int \frac{(1+y)}{\left(1+y^2\right)} d y=C\) [integrating both sides]

⇒ \(\int\left\{\frac{1}{\left(1+x^2\right)}+\frac{x}{\left(1+x^2\right)}\right\} d x+\int\left\{\frac{1}{\left(1+y^2\right)}+\frac{y}{\left(1+y^2\right)}\right\} d y=C\)

⇒ \(\int \frac{1}{\left(1+x^2\right)} d x+\frac{1}{2} \cdot \int \frac{2 x}{\left(1+x^2\right)} d x+\int \frac{1}{\left(1+y^2\right)} d y+\frac{1}{2} \cdot \int \frac{2 y}{\left(1+y^2\right)} d y=C\)

⇒ \(\tan ^{-1} x+\frac{1}{2} \log \left(1+x^2\right)+\tan ^{-1} y+\frac{1}{2} \log \left(1+y^2\right)=C\)

⇒ \(\tan ^{-1} x+\tan ^{-1} y+\frac{1}{2}\left\{\log \left(1+x^2\right)+\log \left(1+y^2\right)\right\}=C\), which is the required solution.

Example 24 Find the equation of a curve which passes through the point (-2,3) and the slope of whose tangent at any point (x,y) is \(\frac{2 x}{y^2}\).

Solution

We know that the slope of a curve at a point (x,y) is \(\frac{d x}{d y}\).

∴ \(\frac{d y}{d x}=\frac{2 x}{y^2}\) …(1)

⇒ y2dy = 2xdx [separating the variables]

⇒ \(\int y^2 d y=\int 2 x d x\)

⇒ \(\frac{1}{3} y^3=x^2+C\) …(2)

where C is a constant.

Thus, (2) is the equation of the curve whose differential equation is given by (1).

Since the given curve passes through the point (-2,3), we have

C = \(\left(\frac{1}{3} \times 27\right)-(-2)^2=(9-4)=5\)

Hence, the required equation of the curve is

\(\frac{1}{3}\)y3 = x2 + 5 ⇒ y3 = 3x2 + 15.

Homogeneous Equation

Homogeneous Function A function f(x,y) in x and y is said to be a homogeneous function of degree n, if the degree of each term is n.

Examples

(1) f(x,y) = (x2 + y2 – xy) is a homogeneous function of degree 2.

(2) g(x,y) = (x3 – 3xy3 + 3x2y + y3) is a homogeneous function of degree 3.

In general, a homogeneous function f(x,y) of degree n is expressible as

\(f(x, y)=x^n f\left(\frac{y}{x}\right)\)

Homogeneous Differential Equation An equation of the form \(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)}\), where both f(x,y) and g(x,y) are homogeneous functions of degree n, is called a homogeneous differential equation.

Example \(\frac{d y}{d x}=\frac{x^2-y^2}{x y}\) is a homogeneous differential equation.

Method Of Solving A Homogeneous Differential Equation

Let \(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)}\) be a homogeneous differential equation.

Putting y = vx and \(\frac{d y}{d x}=\left(v+x \frac{d v}{d x}\right)\) in the given equation, we get

\(v+x \frac{d v}{d x}=F(v)\)

⇒ \(\frac{d v}{\{F(v)-v\}}=\frac{d x}{x}\)

⇒ \(\int \frac{d v}{\{F(v)-v\}}=\int \frac{d x}{x}\)

⇒ \(\int \frac{d v}{\{F(v)-v\}}=\log |x|+C .\)

Now, replace v by (y/x) to obtain the required solution.

Solved Examples

Example 1 Solve \(\frac{d y}{d x}=\frac{y^2-x^2}{2 x y}\).

Solution

Clearly, since each of the functions (y2-x2) and 2xy is a homogeneous function of degree 2, the given equation is homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\), the given equation becomes

\(v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 v x^2}\)

⇒ \(v+x \frac{d v}{d x}=\frac{v^2-1}{2 v}\)

⇒ \(x \frac{d v}{d x}=\left(\frac{v^2-1}{2 v}-v\right)\)

⇒ \(x \frac{d v}{d x}=\frac{-\left(1+v^2\right)}{2 v}\)

⇒ \(\frac{2 v}{\left(1+v^2\right)} d v=-\frac{1}{x} d x\)

⇒ \(\int \frac{2 v}{\left(1+v^2\right)} d v=-\int \frac{1}{x} d x\)

⇒ log|1 + v2| = -log|x| + log C

⇒ log|1 + v2| + log|x| = log C

⇒ log|x(1+v2)| = log C

⇒ x(1+v2) = ±C

⇒ x(1+v2) = C1

⇒ \(x\left(1+\frac{y^2}{x^2}\right)=C_1\)

⇒ (x2 + y2) = C1x, which is the required solution.

Example 2 Solve (x3+y3)dy – x2ydx = 0.

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{x^2 y}{x^3+y^3}\) …(1)

Clearly, each of the functions (x2y) and (x3 + y3) is a homogeneous function of degree 3.

So, the given differential equation is homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in (1), we get

\(v+x \frac{d v}{d x}=\frac{v x^3}{x^3+v^3 x^3}\)

⇒ \(v+x \frac{d v}{d x}=\frac{v}{1+v^3}\)

⇒ \(x \frac{d v}{d x}=\left(\frac{v}{1+v^3}-v\right)=\frac{-v^4}{\left(1+v^3\right)}\)

⇒ \(\frac{\left(1+v^3\right)}{v^4} d v=\frac{-1}{x} d x\)

⇒ \(\int\left(\frac{1}{v^4}+\frac{1}{v}\right) d v=-\int \frac{1}{x} d x\)

⇒ \(\int\left(\frac{1}{v^4}+\frac{1}{v}\right) d v+\int \frac{1}{x} d x=\mathrm{C}\)

⇒ \(\frac{-1}{3 v^3}+\log |v|+\log |x|=C\)

⇒ \(\frac{-1}{3 v^3}+\log |v x|=C\)

⇒ \(\frac{-x^3}{3 y^3}+\log |y|=C\), which is the required solution.

Example 3 Solve (3xy + y2)dx + (x2 + xy)dy = 0.

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{-\left(3 x y+y^2\right)}{\left(x^2+x y\right)}\) …(1)

Clearly, each of the functions (3xy + y2) and (x2 + xy) is a homogeneous function of degree 2.

Therefore, the given equation is homogeneous.

Putting y = vx and \(\frac{d y}{d x}=\left(v+x \frac{d v}{d x}\right)\) in (1), it becomes

\(v+x \frac{d v}{d x}=\frac{-\left(3 v x^2+v^2 x^2\right)}{\left(x^2+v x^2\right)}\)

⇒ \(v+x \frac{d v}{d x}=\frac{-\left(3 v+v^2\right)}{(1+v)}\)

⇒ \(x \frac{d v}{d x}=\left[\frac{-\left(3 v+v^2\right)}{(1+v)}-v\right]\)

⇒ \(x \frac{d v}{d x}=\frac{-2\left(2 v+v^2\right)}{(1+v)}\)

⇒ \(\frac{(1+v)}{\left(2 v+v^2\right)} d v=\frac{-2}{x} d x\)

⇒ \(\int \frac{(1+v)}{\left(2 v+v^2\right)} d v+\int \frac{2}{x} d x=\log C\)

⇒ \(\frac{1}{2} \log \left|2 v+v^2\right|+2 \log |x|=\log C\)

⇒ \(\log \left|x^2 \sqrt{2 v+v^2}\right|=\log C\)

⇒ \(\log \left|x \sqrt{2 x y+y^2}\right|=\log C\) [putting v = \(\frac{x}{y}\)]

⇒ \(x \sqrt{2 x y+y^2}= \pm C\)

⇒ x2(2xy + y2) = C2, which is the required solution.

Example 4 Solve (x3 – 3xy2)dx = (y3 – 3x2y)dy.

Solution

The given equation may be written as \(\frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y}\), which is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{d x}=\frac{x^3-3 v^2 x^3}{v^3 x^3-3 v x^3}\)

⇒ \(v+x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}\)

⇒ \(x \frac{d v}{d x}=\left(\frac{1-3 v^2}{v^3-3 v}-v\right)\)

⇒ \(x \frac{d v}{d x}=\frac{\left(1-v^4\right)}{\left(v^3-3 v\right)}\)

⇒ \(\frac{\left(3 v-v^3\right)}{\left(v^4-1\right)} d v=\frac{d x}{x}\)

⇒ \(\int \frac{\left(3 v-v^3\right)}{\left(v^4-1\right)} d v=\int \frac{d x}{x}\)

⇒ \(\int\left[\frac{1}{2(v+1)}+\frac{1}{2(v-1)}-\frac{2 v}{\left(v^2+1\right)}\right] d v=\int \frac{d x}{x}\) [by partial fractions]

⇒ \(\frac{1}{2} \log |v+1|+\frac{1}{2} \log |v-1|-\log \left|v^2+1\right|=\log |x|+\log C\)

⇒ \(\log \left|\frac{(\sqrt{v+1})(\sqrt{v-1})}{x\left(v^2+1\right)}\right|=\log C\)

⇒ \(\frac{\sqrt{v^2-1}}{x\left(v^2+1\right)}= \pm C\)

⇒ (v2-1) = C12x2(v2+1)2, where C1 = ±C

⇒ (y2 – x2) = C12(y2+x2)2, which is the required solution.

Example 5 Solve y2dx + (x2-xy+y2)dy = 0.

Solution

The given equation may be written as

\(\frac{d y}{d x}=-\frac{y^2}{\left(x^2-x y+y^2\right)}\), which is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{d x}=\frac{-v^2}{\left(1-v+v^2\right)}\)

⇒ \(x \frac{d v}{d x}=-\left[\frac{v^2}{\left(1-v+v^2\right)}+v\right]\)

⇒ \(x \frac{d v}{d x}=\frac{-\left(v+v^3\right)}{\left(1-v+v^2\right)}\)

⇒ \(\frac{\left(1-v+v^2\right)}{v\left(1+v^2\right)} d v=-\frac{1}{x} d x\)

⇒ \(\int \frac{\left(1+v^2\right)-v}{v\left(1+v^2\right)} d v=-\int \frac{d x}{x}\)

⇒ \(\int \frac{d v}{v}-\int \frac{d v}{\left(1+v^2\right)}+\int \frac{d x}{x}=\log C\)

⇒ log|v| – tan-1v + log|x| = log C

⇒ \(\tan ^{-1} v=\log \frac{|v x|}{C}\)

⇒ \(\tan ^{-1} \frac{y}{x}=\log \left(\frac{|y|}{C}\right)\) [putting v = \(\frac{y}{x}\)]

⇒ \(\frac{|y|}{C}=e^{\tan ^{-1}(y / x)}\)

⇒ \(|y|=\dot{C} e^{\tan ^{-1}(y / x)}\), which is the required solution.

Example 6 Solve \(x \frac{d y}{d x}-y=\sqrt{x^2+y^2} .\)

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}\), which is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^2+v^2 x^2}}{x}=v+\sqrt{1+v^2}\)

⇒ \(x \frac{d v}{d x}=\sqrt{1+v^2}\)

⇒ \(\frac{d v}{\sqrt{1+v^2}}=\frac{1}{x} d x\)

⇒ \(\int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x}\)

⇒ \(\log \left|v+\sqrt{1+v^2}\right|=\log |x|+\log C\)

⇒ \(\log \left|\frac{v+\sqrt{1+v^2}}{x}\right|=\log C\)

⇒ \(\frac{v+\sqrt{1+v^2}}{x}= \pm C\)

⇒ \(v+\sqrt{1+v^2}=C_1 x\), where C1 = ±C

⇒ \(y+\sqrt{x^2+y^2}=C_1 x^2\), which is the required solution.

Example 7 Solve \(\frac{d y}{d x}=\frac{y-x}{y+x}\)

Solution

The given equation is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{d x}=\frac{v x-x}{v x+x}\)

⇒ \(v+x \frac{d v}{d x}=\frac{v-1}{v+1}\)

⇒ \(x \frac{d v}{d x}=\left(\frac{v-1}{v+1}-v\right)\)

⇒ \(x \frac{d v}{d x}=\frac{-\left(1+v^2\right)}{(1+v)}\)

⇒ \(\frac{(1+v)}{\left(1+v^2\right)} d v=\frac{-1}{x} d x\)

⇒ \(\int \frac{(1+v)}{\left(1+v^2\right)} d v=-\int \frac{d x}{x}\)

⇒ \(\int \frac{1}{\left(1+v^2\right)} d v+\frac{1}{2} \int \frac{2 v}{\left(1+v^2\right)} d v=-\int \frac{d x}{x}\)

⇒ \(\tan ^{-1} v+\frac{1}{2} \log \left|1+v^2\right|=-\log |x|+C\)

⇒ \(\tan ^{-1} v+\log \left|x \sqrt{1+v^2}\right|=C\)

⇒ \(\tan ^{-1} \frac{y}{x}+\log \left|\sqrt{x^2+y^2}\right|=C\) [putting v = \(\frac{y}{x}\)]

⇒ \(\tan ^{-1} \frac{y}{x}+\frac{1}{2} \log \left(x^2+y^2\right)=C\), which is the required solution.

Example 8 Solve x2dy + y(x+y)dx = 0.

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{-y(x+y)}{x^2}\), which is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{x}=\frac{-v x(x+v x)}{x^2}\)

⇒ \(v+x \frac{d v}{d x}=-\left(v+v^2\right)\)

⇒ \(x \frac{d v}{d x}=-\left(v^2+2 v\right)=-v(v+2)\)

⇒ \(\frac{d v}{v(v+2)}=-\frac{1}{x} d x\)

⇒ \(\int \frac{d v}{v(v+2)}=-\int \frac{1}{x} d x\)

⇒ \(\frac{1}{2} \int\left\{\frac{1}{v}-\frac{1}{v+2}\right\} d v=-\int \frac{d x}{x}\) [by partial fractions]

⇒ \(\frac{1}{2} \log |v|-\frac{1}{2} \log |v+2|=-\log |x|+\log C\)

⇒ \(\frac{1}{2} \log |v|+\log |x|-\frac{1}{2} \log |v+2|=\log C\)

⇒ \(\log \left|\frac{x \sqrt{v}}{\sqrt{v+2}}\right|=\log C\)

⇒ \(\frac{x \sqrt{v}}{\sqrt{v+2}}= \pm C\)

⇒ \(\frac{x^2 v}{v+2}=C^2\)

⇒ x2y = C2(y + 2x) is the required solution.

Example 9 Solve \(x^2\left(\frac{d y}{d x}\right)=\left(x^2-2 y^2+x y\right)\)

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{\left(x^2-2 y^2+x y\right)}{x^2}\), which is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{d x}=\frac{x^2-2 v^2 x^2+v x^2}{x^2}\)

⇒ \(v+x \frac{d v}{d x}=1-2 v^2+v\)

⇒ \(x \frac{d v}{d x}=1-2 v^2\)

⇒ \(\frac{d v}{\left(1-2 v^2\right)}=\frac{d x}{x}\)

⇒ \(\int \frac{d v}{\left(1-2 v^2\right)}=\int \frac{d x}{x}\)

⇒ \(\frac{1}{2} \int \frac{d v}{\left(\frac{1}{2}-v^2\right)}=\int \frac{d x}{x}\)

⇒ \(\frac{1}{2} \int \frac{d v}{\left\{\left(\frac{1}{\sqrt{2}}\right)^2-v^2\right\}}=\int \frac{d x}{x}\)

⇒ \(\frac{1}{2} \cdot \frac{1}{\left(2 \times \frac{1}{\sqrt{2}}\right)} \log \left|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}\right|=\log |x|+C\)

⇒ \(\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\sqrt{2} y}{x-\sqrt{2} y}\right|-\log |x|=C\) [∵ v = \(\frac{y}{x}\)].

This is the required solution.

Example 10 Solve \(\left(x \sqrt{x^2+y^2}-y^2\right) d x+x y d y=0 .\)

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{y^2-x \sqrt{x^2+y^2}}{x y}\), which is clearly homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in it, we get

\(v+x \frac{d v}{d x}=\frac{v^2 x^2-x \sqrt{x^2+v^2 x^2}}{v x^2}\)

⇒ \(x \frac{d v}{d x}=\left(\frac{v^2-\sqrt{1+v^2}}{v}-v\right)\)

⇒ \(x \frac{d v}{d x}=\frac{-\sqrt{1+v^2}}{v}\)

⇒ \(\int \frac{v}{\sqrt{1+v^2}} d v=-\int \frac{d x}{x}\)

⇒ \(\sqrt{1+v^2}=-\log |x|+C\)

⇒ \(\sqrt{x^2+y^2}+x \log |x|=C x\), which is the required solution.

Example 11 Solve \(x \frac{d y}{d x}=y-x \tan \frac{y}{x} .\)

Solution

The given equation may be written as

\(\frac{d y}{d x}=\frac{y}{x}-\tan \frac{y}{x}\) …(1)

This is of the form \(\frac{d y}{d x}=x^0 f\left(\frac{y}{x}\right) .\)

So, the given equation is homogeneous.

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) in (1), we get

\(v+x \frac{d v}{d x}=v-\tan v\)

⇒ \(x \frac{d v}{d x}=-\tan v\)

⇒ \(\frac{d v}{\tan v}=-\frac{d x}{x}\)

⇒ \(\int \cot v d v=-\int \frac{d x}{x}\)

⇒ \(\int \frac{\cos v}{\sin v} d v+\int \frac{d x}{x}=\log C\)

⇒ log|sin v| + log |x| = log C

⇒ log |x sin v| = log C

⇒ x sin v = ±C = C1 (say)

⇒ x sin \(\frac{y}{x}\) = C1, which is the required solution.

Example 12 Solve \(\left(x \cos \frac{y}{x}\right)(y d x+x d y)=\left(y \sin \frac{y}{x}\right)(x d y-y d x)\).

Solution

The given equation may be written as

\(\left(x \cos \frac{y}{x}+y \sin \frac{y}{x}\right) y-\left(y \sin \frac{y}{x}-x \cos \frac{y}{x}\right) x \cdot \frac{d y}{d x}=0\)

⇒ \(\frac{d y}{d x}=\frac{\{x \cos (y / x)+y \sin (y / x)\} y}{\{y \sin (y / x)-x \cos (y / x)] x}\)

⇒ \(\frac{d y}{d x}=\frac{\{\cos (y / x)+(y / x) \sin (y / x)\}(y / x)}{\{(y / x) \sin (y / x)-\cos (y / x)\}}\)

[dividing num. and denom. by x2], which is clearly homogeneous, being a function of (y/x).

Putting y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

\(v+x \frac{d v}{d x}=\frac{v(\cos v+v \sin v)}{(v \sin v-\cos v)}\)

⇒ \(x \frac{d v}{d x}=\left[\frac{v(\cos v+v \sin v)}{(v \sin v-\cos v)}-v\right]\)

⇒ \(x \frac{d v}{d x}=\frac{2 v \cos v}{(v \sin v-\cos v)}\)

⇒ \(\int \frac{(v \sin v-\cos v)}{v \cos v} d v=\int \frac{2}{x} d x\)

⇒ \(\int \tan v d v-\int \frac{d v}{v}=\int \frac{2}{x} d x\)

⇒ -log |cos v| – log |v| + log C = 2 log |x|

⇒ log |cos v| + log |v| + 2log |x| = log C

⇒ log |x2v cos v| = log C

⇒ |x2v cos v| = C

⇒ x2v cos v = ±C = C1 (say)

⇒ xycos(\(\frac{y}{x}\)) = C1, which is the required solution.

Linear Differential Equations

The most general form of linear differential equations is \(\frac{d y}{d x}+P y=Q\), where P is a constant and Q is a constant or a function of x. The other common form of linear differential equations is \(\frac{d x}{d y}+P x=Q\), where P is a constant and Q is a constant or a function of y.

Solution Of \(\frac{d y}{d x}+P y=Q\)

First, we find \(e^{\int P d x}\), which is known as the integrating factor, i.e., IF.

Now, \(\frac{d y}{d x}+P y=Q ⇒ e^{\int P d x} \frac{d y}{d x}+P y e^{\int P d x}=Q \cdot e^{\int P d x}\)

⇒ \(e^{\int P d x} d y+P y \cdot e^{\int P d x} d x=Q \cdot e^{\int P d x} d x \Rightarrow d\left(y \cdot e^{\int P d x}\right)=Q \cdot e^{\int P d x} d x\) …(1)

⇒ \(y \cdot e^{\int P d x}=\int Q \cdot e^{\int P d x} d x+C\), which is the required solution

[integrating both sides of (1)].

Working Rule for Solving \(\frac{d y}{d x}+P y=Q\)

(1) Find IF = \(e^{\int P d x}\)

(2) The solution is given by

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\)

Solved Examples

Example 1 Solve \(x \frac{d y}{d x}-y=x^2 .\)

Solution

The given differential equation may be written as \(\frac{d y}{d x}-\frac{1}{x} \cdot y=x\).

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = –\(\frac{1}{x}\) and Q = x.

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=e^{\log \left(x^{-1}\right)}=x^{-1}=\frac{1}{x} .\)

So, the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times \frac{1}{x}=\int\left(x \times \frac{1}{x}\right) d x+C\)

= \(\int d x+C=x+C \text {. }\)

∴ \(\frac{y}{x}=x+C \Rightarrow y=x^2+C x \text {. }\)

Hence, y = x2 + Cx is the required solution.

Example 2 Solve \(\left(1+x^2\right) \frac{d y}{d x}+y=\tan ^{-1} x\)

Solution

The given differential equation may be written as

\(\frac{d y}{d x}+\frac{1}{\left(1+x^2\right)} \cdot y=\frac{\tan ^{-1} x}{\left(1+x^2\right)}\) …(1)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = \(\frac{1}{\left(1+x^2\right)}\) and Q = \(\frac{\tan ^{-1} x}{\left(1+x^2\right)}\).

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x} .\)

∴ the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times e^{\tan ^{-1} x}=\int\left\{\frac{\tan ^{-1} x}{\left(1+x^2\right)} \cdot e^{\tan ^{-1} x}\right\} d x+C\)

= \(\int\left(t e^t\right) d t+C\), where tan-1x = t

= \(t e^t-\int 1 \cdot e^t d t+C\) [integrating by parts]

= t et – et + C = et(t-1) + C

= etan-1x(tan-1x – 1) + C.

Hence, y = tan-1x – 1 + Ce-tan-1x is the required solution.

Example 3 Solve \(x \frac{d y}{d x}+2 y=x \cos x\)

Solution

The given differential equation may be written as

\(\frac{d y}{d x}+\frac{2}{x} \cdot y=\cos x\)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = \(\frac{2}{x}\) and Q = cos x.

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log \left(x^2\right)}=x^2 .\)

So, the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y x^2=\int x^2 \cos x d x+C\)

= \(x^2 \sin x-\int 2 x \sin x d x+C\) [integrating by parts]

= \(x^2 \sin x-2\left[x(-\cos x)-\int 1 \cdot(-\cos x) d x\right]+C\) [integrating by parts]

= x2sin x + 2 x cos x – 2 sin x + C.

Hence, \(y=\sin x+\frac{2}{x} \cos x-\frac{2}{x^2} \sin x+\frac{C}{x^2}\) is the required solution.

Example 4 Solve \(\left(x^2-1\right) \frac{d y}{d x}+2 x y=\frac{2}{\left(x^2-1\right)}\).

Solution

The given differential equation may be written as

\(\frac{d y}{d x}+\frac{2 x}{\left(x^2-1\right)} \cdot y=\frac{2}{\left(x^2-1\right)^2}\)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = \(\frac{2 x}{\left(x^2-1\right)}\) and Q = \(\frac{2}{\left(x^2-1\right)^2}\).

Thus, the given equation is linear.

IF = \(e^{\int \frac{2 x}{\left(x^2-1\right)} d x}=e^{\log \left(x^2-1\right)}=x^2-1\)

So, the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y\left(x^2-1\right)=\int\left\{\frac{2}{\left(x^2-1\right)^2} \times\left(x^2-1\right)\right\} d x+C\)

= \(2 \int \frac{d x}{\left(x^2-1\right)}+C\)

= \(2 \int \frac{1}{2}\left\{\frac{1}{(x-1)}-\frac{1}{(x+1)}\right\} d x+C\) [by partial fraction]

= \(\log \left|\frac{x-1}{x+1}\right|+C\)

Hence, \(y\left(x^2-1\right)=\log \left|\frac{x-1}{x+1}\right|+C\) is the required solution.

Example 5 Solve \(\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}\)

Solution

The given differential equation may be written as

\(\frac{d y}{d x}+\frac{1}{\left(1+x^2\right)} \cdot y=\frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)} .\)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = \(\frac{1}{\left(1+x^2\right)}\) and Q = \(\frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}\)

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int \frac{1}{\left(1+x^2\right)} d x}=e^{\tan ^{-1} x}\)

So, the required solution is given by

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times e^{\tan ^{-1} x}=\int\left\{\frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)} \times e^{\tan ^{-1} x}\right\} d x+C\)

= \(\int \frac{e^{2 \tan ^{-1} x}}{\left(1+x^2\right)} d x+C\)

= \(\int e^{2 t} d t+C\), where tan-1x = t

= \(\frac{1}{2} e^{2 t}+C=\frac{1}{2} e^{2 \tan ^{-1} x}+C .\)

Hence, y = \(\frac{1}{2}\)etan-1x + Ce-tan-1x is the required solution.

Example 6 Solve \(\frac{d y}{d x}-3 y \cot x=\sin 2 x\), it being given tath y = 2 when x = \(\frac{\pi}{2}\).

Solution

The given differential equation is

\(\frac{d y}{d x}-3 y \cot x=\sin 2 x\) …(1)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = -3 cot x and Q = sin2x.

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int-3 \cot x d x}=e^{\log (\sin x)^{-3}}=(\sin x)^{-3}=\frac{1}{\sin ^3 x}\)

So, the solution of (1) is given by

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times \frac{1}{\sin ^3 x}=\int\left(\sin 2 x \times \frac{1}{\sin ^3 x}\right) d x+C\)

= \(2 \int \frac{\cos x}{\sin ^2 x} d x+C\)

= \(2 \int \frac{1}{t^2} d t+C\), where sin x = t

= \(\frac{-2}{t}+C=\frac{-2}{\sin x}+C .\)

Thus, y = -2 sin2x + C sin3x …(2)

It is given that y = 2 when x = (π/2).

Putting x = \(\frac{\pi}{2}\) and y = 2 in (2), we get C = 4.

Hence, y = -2 sin2x + 4 sin3x is the required solution.

Example 7 Solve \(\frac{d y}{d x}+(\sec x) y=\tan x\)

Solution

The given equation is of the form

\(\frac{d y}{d x}+P y=Q\), where P = sec x and Q = tan x.

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=(\sec x+\tan x) .\)

So, the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+C\)

= \(\int \sec x \tan x d x+\int \tan ^2 x d x+C\)

= \(\sec x+\int\left(\sec ^2 x-1\right) d x+C\)

= sec x + tan x – x + C.

So, y(sec x + tan x) = sec x + tan x – x + C is the required solution.

Example 8 Solve \(x \frac{d y}{d x}-y=\log x .\)

Solution

The given equation may be written as

\(\frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x} .\)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = –\(\frac{1}{x}\) and Q = \(\frac{\log x}{x}\).

Thus, the given equation is linear.

IF = \(e^{\int-\frac{1}{x} d x}=e^{-\log x}=e^{\log \left(x^{-1}\right)}=x^{-1}=\frac{1}{x} .\)

So the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times \frac{1}{x}=\int\left(\frac{\log x}{x} \times \frac{1}{x}\right) d x+C\)

= \(\int(\log x) \cdot \frac{1}{x^2} d x+C\)

= \((\log x)\left(-\frac{1}{x}\right)-\int \frac{1}{x} \cdot\left(-\frac{1}{x}\right) d x+C\) [integrating by parts]

= \(-\frac{\log x}{x}+\int \frac{1}{x^2} d x+C\)

= \(\frac{-\log x}{x}-\frac{1}{x}+C .\)

Hence, y = Cx – (log x + 1) is the required solution.

Example 9 Solve \(\frac{d y}{d x}+y \cot x=2 \cos x \text {. }\)

Solution

The given equation is of the form

\(\frac{d y}{d x}+P y=Q\), where P = cot x and Q = 2 cos x.

Thus, the given differential equation is linear.

IF = \(e^{\int P d x}=e^{\int \cot x d x}=e^{\log (\sin x)}=\sin x .\)

So, the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

= \(\int \sin 2 x d x+C=\frac{-\cos 2 x}{2}+C\)

Thus, 2y sin x + cos 2x = C’ is the required solution, where C’ = 2C.

Example 10 Solve \(\frac{d y}{d x}-2 y=\cos 3 x\)

Solution

The given equation is of the form

\(\frac{d y}{d x}+P y=Q\), where P = -2 and Q = cos 3x.

Thus, it is a linear equation.

IF = \(e^{\int P d x}=e^{\int-2 d x}=e^{-2 x} \text {. }\)

So, the required solution is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y \times e^{-2 x}=\int e^{-2 x} \cos 3 x d x+C\) …(1)

Now, we have

I = \(\int e^{-2 x} \cos 3 x d x\)

= \(e^{-2 x} \cdot \frac{\sin 3 x}{3}+2 \int e^{-2 x} \cdot \frac{\sin 3 x}{3} d x\) [integrating by parts]

= \(e^{-2 x} \cdot \frac{\sin 3 x}{3}+2 \int e^{-2 x} \cdot \frac{\sin 3 x}{3} d x\)

= \(\frac{1}{3} e^{-2 x} \sin 3 x+\frac{2}{3} \cdot\left[e^{-2 x} \cdot \frac{(-\cos 3 x)}{3}+2 \int e^{-2 x} \frac{(-\cos 3 x)}{3} d x\right]\) [integrating by parts]

= \(\frac{1}{3} e^{-2 x} \sin 3 x-\frac{2}{9} e^{-2 x} \cos 3 x-\frac{4}{9} \int e^{-2 x} \cos 3 x d x\)

= \(\frac{1}{3} e^{-2 x} \sin 3 x-\frac{2}{9} e^{-2 x} \cos 3 x-\frac{4}{9} I\)

⇒ \(\left(\frac{13}{9}\right) I=\frac{1}{3} e^{-2 x} \sin 3 x-\frac{2}{9} e^{-2 x} \cos 3 x\)

⇒ \(I=\frac{3\left(e^{-2 x} \sin 3 x\right)}{13}-\frac{2\left(e^{-2 x} \cos 3 x\right)}{13} .\)

Putting this value in (1), we get

\(y e^{-2 x}=\frac{3\left(e^{-2 x} \sin 3 x\right)}{13}-\frac{2\left(e^{-2 x} \cos 3 x\right)}{13}+C\)

⇒ \(y=\frac{(3 \sin 3 x-2 \cos 3 x)}{13}+C e^{2 x}\), which is the required solution.

Example 11 Solve \((x \log x) \frac{d y}{d x}+y=\frac{2}{x} \log x .\)

Solution

The given equation may be written as

\(\frac{d y}{d x}+\frac{1}{(x \log x)} \cdot y=\frac{2}{x^2}\)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = \(\frac{1}{(x \log x)}\) and Q = \(\frac{2}{x^2}\).

Thus, the given equation is linear.

IF = \(e^{\int P d x}=e^{\int \frac{1}{x \log x} d x}=e^{\int \frac{1}{t} d t}\), where log x = t

= elogt = t = log x.

So, the solution of the given differential equation is

\(y \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d x+C\),

i.e., \(y(\log x)=\int\left(\frac{2}{x^2} \log x\right) d x+C\)

= \(2 \int(\log x) \cdot \frac{1}{x^2} d x+C\)

= \(2\left[(\log x)\left(-\frac{1}{x}\right)-\int \frac{1}{x} \cdot\left(-\frac{1}{x}\right) d x\right]+C\) [integrating by parts]

= \(2\left[(\log x)\left(-\frac{1}{x}\right)-\int \frac{1}{x} \cdot\left(-\frac{1}{x}\right) d x\right]+C\)

Hence, \(\frac{-2 \log x}{x}-\frac{2}{x}+C\) is the required solution.

Example 12 An equation related to the stability of an aeroplane is given by \(\frac{d v}{d t}=g \cos \alpha-k v\), where v is the velocity and g, α, k are constants. Find an expression for the velocity if v = 0 when t = 0.

Solution

The given differential equation is

\(\frac{d v}{d t}+k v=g \cos \alpha\) …(1)

This is of the form \(\frac{d y}{d x}+P y=Q\), where P = k and Q = g cos α.

Thus, the given equation is linear.

IF = \(e^{\int P d t}=e^{\int k d t}=e^{k t}\)

So, the solution of the given differential equation is

\(v \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d t+C\),

i.e., \(v e^{k t}=\int(g \cos \alpha) e^{k t} d t+C\)

= \(\frac{(g \cos \alpha) e^{k t}}{k}+C\) …(2)

Now, it is given that v = 0 when t = 0.

Putting t = 0 and v = 0 in (2), we get C = \(\frac{-g \cos \alpha}{k}\)

∴ \(v e^{k t}=\frac{(g \cos \alpha) e^{k t}}{k}-\frac{g \cos \alpha}{k}\)

⇒ \(v=\frac{1}{k}(g \cos \alpha)\left(1-e^{-k t}\right)\), which is the required expression.

Solution Of \(\frac{d x}{d y}+P x=Q\)

Working Rule for Solving \(\frac{d x}{d y}+P x=Q\)

(1) Find IF = \(e^{\int P d y}\)

(2) The solution is given by \(x \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d y+C\)

Example 13 Solve \(\left(x+2 y^3\right) \frac{d y}{d x}=y .\)

Solution

The given differential equation may be written as

\(y \frac{d x}{d y}=x+2 y^3\)

⇒ \(\frac{d x}{d y}-\frac{1}{y} \cdot x=2 y^2\) …(1)

This is of the form \(\frac{d x}{d y}+P x=Q\), where P = –\(\frac{1}{y}\) and Q = 2y2.

Thus, the given equation is linear.

IF = \(e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log \left(y^{-1}\right)}=y^{-1}=\frac{1}{y} .\)

So, the required solution is

\(x \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d y+C\),

i.e., \(x \times \frac{1}{y}=\int\left(2 y^2 \times \frac{1}{y}\right) d y+C\)

= \(\int 2 y d y+C=y^2+C\).

Hence, x = y3 + Cy is the required solution.

Example 14 Solve \(\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y\)

Solution

The given differential equation may be written as

\(\frac{d x}{d y}=\frac{\tan ^{-1} y-x}{\left(1+y^2\right)}\)

⇒ \(\frac{d x}{d y}+\frac{1}{\left(1+y^2\right)} \cdot x=\frac{\tan ^{-1} y}{\left(1+y^2\right)}\) …(1)

This is of the form \(\frac{d x}{d y}+P x=Q\), where P = \(\frac{1}{\left(1+y^2\right)}\) and Q = \(\frac{\tan ^{-1} y}{\left(1+y^2\right)}\).

Clearly, the given equation is linear.

IF = \(e^{\int P d y}=e^{\int \frac{1}{\left(1+y^2\right)} d y}=e^{\tan ^{-1} y} .\)

So the required solution is

\(x \times \mathrm{IF}=\int\{Q \times \mathrm{IF}\} d y+C\),

i.e., \(x \times e^{\tan ^{-1} y}=\int\left\{\frac{\tan ^{-1} y}{\left(1+y^2\right)} \times e^{\tan ^{-1} y}\right\} d y+C\)

= \(\int\left(t e^t\right) d t+C\), where tan-1y = t

= \(t e^t-\int 1 \cdot e^t d t+C\)

= t et – et + C = (t – 1)et + C

= \(\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+C\)

⇒ \(x=\left(\tan ^{-1} y-1\right)+C e^{-\tan ^{-1} y}\), which is the required solution.

Application Of Differential Equations

Solved Examples

Example 1 Obtain the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin.

Solution

Given

y + 2x

The differential equation of the curve is

\(\frac{d y}{d x}=y+2 x \Rightarrow \frac{d y}{d x}-y=2 x\) …(1) which is linear.

IF = \(e^{-\int d x}=e^{-x}\)

∴ the solution of (1) is given by

\(y e^{-x}=\int 2 x e^{-x} d x\)

= \(2\left[\left(-x e^{-x}\right)-\int-e^{-x} d x\right]+C=-2 x e^{-x}-2 e^{-x}+C\)

⇒ y + 2x + 2 = Cex …(2) which is the equation of the curve.

Since the curve passes through (0,0), by putting x = 0 and y = 0 in (2), we get C = 2.

∴ y + 2x + 2 = 2ex is the required equation of the curve.

Example 2 The line normal to a given curve at each point (x,y) on the curves passes through the point (3,0). If the curve contains the point (3,4), find its equation.

Solution

Given

The line normal to a given curve at each point (x,y) on the curves passes through the point (3,0). If the curve contains the point (3,4),

The equation of the normal to a curve at a point (x,y) is

\((Y-y) \frac{d y}{d x}+(X-x)=0\).

Since it passes through the point (3,0), we have

\((0-y) \frac{d y}{d x}+(3-x)=0 \Rightarrow y \frac{d y}{d x}=(3-x)\)

⇒ \(\int y d y=\int(3-x) d x\)

⇒ \(\int y d y=\int(3-x) d x\)

⇒ x2 + y2 – 6x – 2C = 0 …(1)

which is the equation of the curve.

Since the curve passes through (3,4), we have

9 + 16 – 18 – 2C = 0 ⇒ c = (7/2).

∴ x2 + y2 – 6x – 7 = 0 is the required equation of the curve.

Example 3 A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is 3 mm and 1 hour later it reduces to 2 mm, find an expression for the radius of the raindrop at any time.

Solution

Given

A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is 3 mm and 1 hour later it reduces to 2 mm

Let R be the radius of the raindrop at time t. Then,

\(\frac{d R}{d t}=-k\left(4 \pi R^2\right)\)

⇒ \(-\int \frac{d R}{R^2}=\int 4 \pi k d t\)

⇒ \(\frac{1}{R}=4 \pi k t+C\) …(1)

When t = 0, R = 3.

Putting these values in (1), we get C = \(\frac{1}{3}\).

∴ \(\frac{1}{R}=4 \pi k t+\frac{1}{3}\) …(2)

Also, when t = 1, R = 2.

Putting these values in (2), we get k = \(\frac{1}{24 \pi}\)

∴ \(\frac{1}{R}=\frac{4 \pi t}{24 \pi}+\frac{1}{3}\)

⇒ \(\frac{1}{R}=\frac{t}{6}+\frac{1}{3}\)

⇒ R(t + 2) = 6, which is the required expression.

Example 4 A population grows at the rate of 5% per year. How long does it take for the population to double?

Solution

Given

A population grows at the rate of 5% per year.

Let the population be P at any time t. Then,

\(\frac{d P}{d t}=5 \% \text { of } P\)

⇒ \(\frac{d P}{d t}=\frac{P}{20}\)

⇒ \(\int \frac{d P}{P}=\frac{1}{20} \int d t\)

⇒ \(\log P=\frac{t}{20}+C\) …(1)

Let the population be P0 at t = 0.

Putting P = P0 and t = 0 in (1), we get C = log P0.

∴ \(\log P=\frac{t}{20}+\log P_0\) …(2)

Let P = 2P0 at t = t1. Then,

\(\log \left(2 P_0\right)=\frac{t_1}{20}+\log P_0\)

⇒ \(\log 2+\log P_0=\frac{t_1}{20}+\log P_0\)

⇒ \(\log 2=\frac{t_1}{20}\)

⇒ t1 = 20(log 2), which is the required time.

Example 5 The population of a country doubles in 40 years. Assuming that the rate of increase is proportional to the number of inhabitants, find the number of years in which it would treble itself.

Solution

Given

The population of a country doubles in 40 years. Assuming that the rate of increase is proportional to the number of inhabitants,

Let the population be P at time t.

Then, \(\frac{d P}{d t}=k P\)

⇒ \(\int \frac{d P}{P}=k \int d t\)

⇒ log P = kt + C …(1)

Let the population be P0 at t = 0. Then, c = log P0.

∴ log P = kt + log P0 …(2)

Now, P = 2P0 when t = 40.

Putting these values in (2), we get

log(2P0) = 40k + log P0

⇒ log 2 + log P0 = 40k + log P0

⇒ 40k = log 2

⇒ \(k=\frac{1}{40}(\log 2)\)

Putting this value of k in (2), we get

\(\log P=\frac{t}{40}(\log 2)+\log P_0\) …(3)

Let P = 3P0 at t = t1. Then,

\(\log P=\frac{t}{40}(\log 2)+\log P_0\)

⇒ \(\log 3+\log P_0=\frac{t_1(\log 2)}{40}+\log P_0\)

⇒ \(\log 3+\log P_0=\frac{t_1(\log 2)}{40}+\log P_0\) years, which is the required time.

The number of years in which it would treble itself ⇒ \(\log 3+\log P_0=\frac{t_1(\log 2)}{40}+\log P_0\)

Example 6 The rate of increase of bacteria in a culture is proportional to the number of bacteria present, and it is found that the number doubles in 6 hours. Calculate how many times the bacteria may be expected to grow at the end of 18 hours.

Solution

Given

The rate of increase of bacteria in a culture is proportional to the number of bacteria present, and it is found that the number doubles in 6 hours.

Let the number of bacteria be N at time t. Then,

\(\frac{d N}{d t}=k N\)

⇒ \(\int \frac{d N}{N}=\int k d t\)

⇒ log N = kt + C …(1)

Let the number of bacteria be N0 at t = 0.

Putting t = 0 and N = N0 in (1), we get C = log N0.

∴ log N = kt + log N0 …(2)

Now, when t = 6, then N = 2N0.

Substituting these values in (2), we get

log(2N0) = 6k + log N0

⇒ log 2 + log N0 = 6k + log N0

⇒ \(k=\frac{1}{6}(\log 2) \text {. }\)

∴ \(\log N=\frac{t}{6}(\log 2)+\log N_0\) …(3)

Putting t = 18 in (3), we get

log N = 3(log 2) + log N0 = log (8N0)

⇒ N = 8N0, when t = 18.

∴ the number of bacteria becomes 8 times at the end of 18 hours.

Example 7 A wet porous substance in the open air loses its moisture at a rate proportional to themoisture content. If a sheet hung in the wind loses half its moisture during the first hour, when will it have lost 90%, weather conditions remaining the same?

Solution

Given

A wet porous substance in the open air loses its moisture at a rate proportional to themoisture content.

Let the moisture content be M at time t. Then,

\(\frac{d M}{d t}=-k M\)

⇒ \(\int \frac{d M}{M}=\int-k d t\)

⇒ log M = -kt + C …(1)

Let M0 be the moisture content at t = 0.

Putting M = M0 and t = 0 in (1), we get C = log M0.

∴ log M = -kt + log M0 …(2)

Now, it is being given that when t = 1, then M = \(\frac{1}{2}\)M0.

Putting these values in (2), we get

\(\log \left(\frac{1}{2} M_0\right)=-k+\log M_0\)

⇒ \(\log \frac{1}{2}+\log M_0=-k+\log M_0\)

⇒ -log 2 = -k i.e., k = log 2.

∴ log M = -t(log 2) + log M0 …(3)

When 90% of M0 is lost, we are left with 10% of M0, i.e., \(\frac{1}{10}\)M0.

Let M = \(\frac{1}{10}\)M0 when t = t1. Then,

\(\log \left(\frac{1}{10} M_0\right)=-t_1(\log 2)+\log M_0\)

⇒ \(\log \frac{1}{10}+\log M_0=-t_1(\log 2)+\log M_0\)

⇒ -log 10 = -t1(log 2)

⇒ \(\log \frac{1}{10}+\log M_0=-t_1(\log 2)+\log M_0\)

Thus, 90% moisture content is lost in \(\frac{1}{(\log 2)}\) hours.

Example 8 Experiments show that the rate of inversion of cane sugar in dilute solution is proportional to the concentration y(t) of the unaltered sugar. Suppose that the concentration is (1/100) at t = 0 and (1/300) at t = 10 hours. Find y(t).

Solution

Given

Experiments show that the rate of inversion of cane sugar in dilute solution is proportional to the concentration y(t) of the unaltered sugar. Suppose that the concentration is (1/100) at t = 0 and (1/300) at t = 10 hours.

Let the concentration be y at time t. Then,

\(\frac{d y}{d t}=-k y\)

⇒ \(\int \frac{d y}{y}=\int-k d t\)

⇒ log y = -kt + C …(1)

Putting t = 0 and y = \(\frac{1}{100}\) in (1), we get C = -2.

∴ log y = -kt – 2 …(2)

Putting t = 10 and y = \(\frac{1}{300}\) in (2), we get

\(\log \frac{1}{300}=-10 k-2 \Rightarrow \log 1-\log 3-\log 100=-10 k-2\)

⇒ -10k = -log 3

⇒ \(k=\frac{1}{10} \log 3 .\)

∴ \(\log y=\left(-\frac{1}{10} \log 3\right) t-2\)

or \(\log y+\left(\frac{1}{10} \log 3\right) t+2=0\)

Example 9 A radioactive substance has a half-life of h days. Find a formula for its mass m in terms of t, the time , if the initial mass is m0. What is its initial decay rate?

Solution

Given

A radioactive substance has a half-life of h days.

Having a half-life of h days means that half of the mass decays in h days.

Let M be the mass of the substance at t days. Then,

\(\frac{d M}{d t}=-k M\)

⇒ \(\int \frac{d M}{M}=\int-k d t\)

⇒ log M = -kt + C …(1)

When t = 0, we have M = m0.

∴ log M = -kt + log m0 …(2)

Again, when t = h, we have M = \(\frac{1}{2}\)m0.

∴ \(\log \left(\frac{1}{2} m_0\right)=-k h+\log m_0\)

⇒ \(\log \frac{1}{2}+\log m_0=-k h+\log m_0\)

⇒ \(k h=\log 2 \text {, i.e., } k=\frac{1}{h}(\log 2) \text {. }\)

∴ \(\log M=\frac{-t}{h}(\log 2)+\log m_0\)

⇒ \(\log M=\log \left(2^{-t / h} m_0\right)\)

⇒ M = 2-t/hm0, which is the required formula.

The initial decay rate is given by \(\frac{d M}{d t}\), when M = m0 and \(k=\frac{1}{h}(\log 2)\).

∴ initial decay rate is given by \(\frac{d M}{d t}=-\frac{1}{h}(\log 2) m_0\).

Example 10 A bank pays interest by treating the interest rate as the instantaneous rate of change of the principal. Suppose in an account, interest accures at 8% per annum, compounded continuosly. Calculate the percentage increase in such an account over one year. [Take e0.08 = 1.0833].

Solution

Given:

A bank pays interest by treating the interest rate as the instantaneous rate of change of the principal. Suppose in an account, interest accures at 8% per annum, compounded continuosly.

Let the principal be P at t years. Then,

\(\frac{d P}{d t}=\frac{8}{100} P\)

⇒ \(\int \frac{d P}{P}=\frac{2}{25} \int d t\)

⇒ log P = \(\frac{2}{25} t+C\) …(1)

Let the original principal be P0 at t = 0.

Putting t = 0 and P = P0 in (1), we get C = log P0.

∴ \(\log P=\frac{2}{25} t+\log P_0\) …(2)

When t = 1, we have

\(\log P=\frac{2}{25}+\log P_0 \Rightarrow \log \left(\frac{P}{P_0}\right)=\frac{2}{25}=0.08\)

⇒ \(\frac{P}{P_0}=e^{0.08}=1.0833\)

⇒ P = 1.0833P0.

∴ percentage increase in the principal in 1 year = \(\frac{0.0833 P_0}{P_0} \times 100\)

= 8.33%.

The percentage increase in such an account over one year. = 8.33%.

Example 11 The acceleration after t seconds of a particle which starts from rest and moves in a straight line is (8 – t/5)cm/s2. Find the velocity of the particle at the instant when the acceleration is zero.

Solution

Given

The acceleration after t seconds of a particle which starts from rest and moves in a straight line is (8 – t/5)cm/s2.

Let v be the velocity of the particle at time t. Then,

\(\frac{d v}{d t}=\left(8-\frac{t}{5}\right)\)

⇒ \(\int d v=\int\left(8-\frac{t}{5}\right) d t\)

⇒ \(v=8 t-\frac{t^2}{10}+C\) …(1)

When t = 0, we have v = 0. This gives, C = 0.

∴ \(v=8 t-\frac{t^2}{10}\) …(2)

Now, acceleration is zero ⇒ \(\frac{d v}{d t}\) = 0

⇒ 8 – \(\frac{t}{5}\) – 0, i.e., t = 40.

Putting t = 40 in (2), we get

v = 320 – 160 = 160.

Hence, the required velocity is 160 cm/s.

Example 12 A steamboat is moving at velocity v0 when steam is shut off. It is given that the retardation at any subsequent time is equal to the magnitude of the velocity at that time. Find the velocity and distance travelled in time t after the steam is shut off.

Solution

Given

A steamboat is moving at velocity v0 when steam is shut off. It is given that the retardation at any subsequent time is equal to the magnitude of the velocity at that time.

Let v be the velocity at time t after the steam is shut off. Then,

\(-\frac{d v}{d t}=v\)

⇒ \(\int \frac{d v}{v}=-\int d t\)

⇒ log v = -t + C …(1)

When t = 0, we are given that v = v0.

Putting t = 0 and v = v0 in (1), we get C = log v0.

∴ log v = -t + log v0

⇒ \(\log \left(\frac{v}{v_0}\right)=-t \text {, i.e., } \frac{v}{v_0}=e^{-t}\)

⇒ v = v0e-t

⇒ \(\frac{d x}{d t}=v_0 e^{-t}\), where x is the distance covered

⇒ \(\int d x=v_0 \cdot \int e^{-t} d t\)

⇒ x = -v0e-t + C’ …(2)

Clearly, when t = 0, we have x = 0. This gives C’ = v0.

∴ x = -v0e-t + v0 ⇒ x = v0(1 – e-t).

Hence, the required velocity = v0e-t, and the required distance = v0(1 – e-t).

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