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NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations Exercise 9.4

Last updated date: 02nd Aug 2024
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NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.4 - FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations Exercise 9.4 provides comprehensive answers and explanations for problems related to solving differential equations. This exercise focuses on advanced techniques such as separating variables, integrating factors, and solving homogeneous and non-homogeneous differential equations. Designed to enhance problem-solving skills, these solutions help students grasp complex concepts and apply them effectively. They are an essential resource for mastering the subject and performing well in exams, offering clear, step-by-step guidance that aligns with the NCERT curriculum.

Table of Content
1. NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.4 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 9 Exercise 9.4 Class 12 | Vedantu
3. Access NCERT Solutions for Maths Class 12 Chapter 9 - Differential Equations
4. Class 12 Maths Chapter 9: Exercises Breakdown
5. CBSE Class 12 Maths Chapter 9 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Glance on NCERT Solutions Maths Chapter 9 Exercise 9.4 Class 12 | Vedantu

• This exercise focuses exclusively on homogeneous differential equations.

• Learn to identify these equations based on their specific characteristics - where the degree of each term (considering x and y) is the same.

• A special technique for solving homogeneous equations will be explored. This technique involves separating the variables and then integrating both of the equations.

• There are 17 questions in Exercise 9.4 Class 12 Maths which experts at Vedantu fully solved.

Competitive Exams after 12th Science

Access NCERT Solutions for Maths Class 12 Chapter 9 - Differential Equations

EXERCISE 9.4

Refer for exercise 9.4 in the PDF

1. Solve the differential equation $\left( {{x^2} + xy} \right)dy = \left( {{x^2} + {y^2}} \right)dy$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + {y^2}}}{{{x^2} + xy}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{{x^2} + {{\left( {vx} \right)}^2}}}{{{x^2} + x\left( {vx} \right)}}$

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{{x^2}}}{{{x^2}}}\left( {\dfrac{{1 + {v^2}}}{{1 + v}}} \right)$

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 + v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 + v}} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2} - v - {v^2}}}{{1 + v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{1 - v}}{{1 + v}}$

$dfrac{1}{x}dx = \dfrac{{1 + v}}{{1 - v}}dv$

Taking integration on both side,

$\int {\dfrac{1}{x}dx} = \int {\dfrac{{1 + v}}{{1 - v}}dv}$

$\log x + C = \int {\dfrac{{1 + v + 1 - 1}}{{1 - v}}dv}$

$\log x + C = \int {\dfrac{{2 + 1 - v}}{{1 - v}}dv}$

$\log x + C = \int {\left( {\dfrac{2}{{1 - v}} + 1} \right)dv}$

$\log x + C = - 2\log \left( {1 - v} \right) + v$

Substituting the value of $v = \dfrac{y}{x}$.

$\log x + C = - 2\log \left( {1 - \dfrac{y}{x}} \right) + \dfrac{y}{x}$

$\log x + \log {\left( {\dfrac{{x - y}}{x}} \right)^2} = \dfrac{y}{x} + C$

$\log \left[ {{{\left( {\dfrac{{x - y}}{x}} \right)}^2}x} \right] = \dfrac{y}{x} - C$

${\left( {\dfrac{{x - y}}{x}} \right)^2}x = {e^{\dfrac{y}{x} - C}}$

$\dfrac{{{{\left( {x - y} \right)}^2}}}{x} = {e^{\dfrac{y}{x} - C}}$

${\left( {x - y} \right)^2} = x{e^{\dfrac{y}{x}}}{e^{ - C}}$

${\left( {x - y} \right)^2} = x{e^{\dfrac{y}{x}}}C$

This is the required differential equation.

Where $C = {e^{ - C}}$.

2. Solve the differential equation $y' = \dfrac{{x + y}}{x}$

Ans:$\dfrac{{dy}}{{dx}} = \dfrac{{x + y}}{x}$

On rearranging the equation, we get

$\dfrac{{dy}}{{dx}} = 1 + \dfrac{y}{x}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = 1 + v$

$dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {1.dv} = \int {\dfrac{1}{x}dx}$

$v = \log x + C$

Substituting the value of $v = \dfrac{y}{x}$.

$\dfrac{y}{x} = \log x + C$

$y = x\log x + Cx$

This is the required differential equation.

3. Solve the differential equation $\left( {x - y} \right)dy = \left( {x + y} \right)dx$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{x + y}}{{x - y}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{x + vx}}{{x - vx}}$

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v}}{{1 - v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v}}{{1 - v}} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v - v + {v^2}}}{{1 - v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 - v}}$

$\dfrac{{1 - v}}{{1 + {v^2}}}dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{{1 - v}}{{1 + {v^2}}}dv} = \int {\dfrac{1}{x}dx}$

$\int {\dfrac{1}{{1 + {v^2}}}} dv - \int {\dfrac{v}{{1 + {v^2}}}} dv = \int {\dfrac{1}{x}dx} ......\left( 1 \right)$

Let $I = \int {\dfrac{v}{{1 + {v^2}}}} dv$

Now, let $1 + {v^2} = t$

Differentiating equation w.r.t. $v$, we get

$2vdv = dt$

$vdv = \dfrac{{dt}}{2}$

Substituting $vdv = \dfrac{{dt}}{2}$ in the above equation, we get

$I = \int {\dfrac{1}{2t}} dt$

Substituting this value in equation $\left( 1 \right)$ .

Therefore,

$\int {\dfrac{1}{{1 + {v^2}}}} dv - \int {\dfrac{1}{2t}} dt = \int {\dfrac{1}{x}dx}$

${\tan ^{ - 1}}v - \dfrac{1}{2}\log t = \log x + C$

Substituting the value of $1 + {v^2} = t$ and $v = \dfrac{y}{x}$ .

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

After rearranging the given equation we get,

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

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

${\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right) = \dfrac{1}{2}\left( {2\log x + \log \left( {1 + \dfrac{{{y^2}}}{{{x^2}}}} \right)} \right)$

${\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right) = \log \left( {\dfrac{{{y^2} + {x^2}}}{{{y^2}}} \times {x^2}} \right) + C$

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

This is the required differential equation.

4. Solve the differential equation $\left( {{x^2} - {y^2}} \right)dx + 2xydy = 0$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = - \dfrac{{{x^2} - {y^2}}}{{2xy}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{{x^2} - {{\left( {vx} \right)}^2}}}{{2x.vx}}$

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{1 - {v^2}}}{{2v}}$

$x\dfrac{{dv}}{{dx}} = - \dfrac{{1 - {v^2}}}{{2v}} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 + {v^2} - 2{v^2}}}{{2v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 - {v^2}}}{{2v}}$

$- \dfrac{{2v}}{{1 + {v^2}}}dv = \dfrac{1}{x}dx$

$\dfrac{{2v}}{{1 + {v^2}}}dv = - \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{{2v}}{{1 + {v^2}}}dv} = - \int {\dfrac{1}{x}dx} ......\left( 1 \right)$

Now, let $1 + {v^2} = t$

Differentiating equation w.r.t. $v$, we get

$2vdv = dt$

Substituting $2vdv = dt$ in equation $\left( 1 \right)$, we get

$\int {\dfrac{1}{t}dt} = - \int {\dfrac{1}{x}dx}$

$\log t = - \log x + C$

Substituting the value of $1 + {v^2} = t$ and $v = \dfrac{y}{x}$.

$\log \left( {1 + \dfrac{{{y^2}}}{{{x^2}}}} \right) = - \log x + C$

$\log \left( {1 + \dfrac{{{y^2}}}{{{x^2}}}} \right) + \log x = C$

$\log \left( {\dfrac{{{x^2} + {y^2}}}{{{x^2}}} \times x} \right) = C$

$\dfrac{{{x^2} + {y^2}}}{x} = {e^C}$

$\dfrac{{{x^2} + {y^2}}}{x} = K$

${x^2} + {y^2} = Kx$

This is the required differential equation.

5. Solve the differential equation ${x^2}\dfrac{{dy}}{{dx}} = {x^2} - 2{y^2} + xy$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}}= \dfrac{{x^2} - 2{y^2} + xy}{x^2}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = 1 - 2{v^2} + v$

$x\dfrac{{dv}}{{dx}} = 1 - 2{v^2}$

$\dfrac{1}{{1 - 2{v^2}}}dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{1}{{1 - 2{v^2}}}dv} = \int {\dfrac{1}{x}dx}$

$\int {\dfrac{1}{{{1^2} - {{\left( {\sqrt 2 v} \right)}^2}}}dv} = \int {\dfrac{1}{x}dx}$

On integrating using standard trigonometric identity we get,

$\dfrac{1}{{\sqrt 2 }}.\dfrac{1}{{1.2}}.\log \left| {\dfrac{{1 + \sqrt 2 v}}{{1 - \sqrt 2 v}}} \right| = \log \left| x \right| + C$

Substituting the value of $v = \dfrac{y}{x}$.

$\dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{1 + \sqrt 2 \dfrac{y}{x}}}{{1 - \sqrt 2 \dfrac{y}{x}}}} \right| = \log \left| x \right| + C$

$\dfrac{1}{{2\sqrt 2 }}\log \left| {\dfrac{{x + \sqrt 2 y}}{{x - \sqrt 2 y}}} \right| = \log \left| x \right| + C$

This is the required differential equation.

6. Solve the differential equation $xdy - ydx = \sqrt {{x^2} + {y^2}} dx$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {{x^2} + {y^2}} + y}}{x}$

$\dfrac{{dy}}{{dx}} = \sqrt {1 + \dfrac{{{y^2}}}{{{x^2}}}} + \dfrac{y}{x}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \sqrt {1 + {v^2}} + v$

$x\dfrac{{dv}}{{dx}} = \sqrt {1 + {v^2}}$

$\dfrac{1}{{\sqrt {1 + {v^2}} }}dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{1}{{\sqrt {1 + {v^2}} }}dv} = \int {\dfrac{1}{x}dx}$

Using $\int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }} = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right|} + C$, we get

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

Substituting the value of $v = \dfrac{y}{x}$.

$\log \left| {\dfrac{y}{x} + \sqrt {1 + {{\left( {\dfrac{y}{x}} \right)}^2}} } \right| = \log xC$

$\dfrac{y}{x} + \sqrt {1 + {{\left( {\dfrac{y}{x}} \right)}^2}} = xC$

$\dfrac{y}{x} + \sqrt {{{\dfrac{{{x^2} + y}}{{{x^2}}}}^2}} = xC$

$\dfrac{y}{x} + \dfrac{{\sqrt {{x^2} + {y^2}} }}{x} = xC$

$y + \sqrt {{x^2} + {y^2}} = C{x^2}$

This is the required differential equation.

7. Solve the differential equation $\left\{ {x\cos \left( {\dfrac{y}{x}} \right) + y\sin \left( {\dfrac{y}{x}} \right)} \right\}ydx = \left\{ {y\sin \left( {\dfrac{y}{x}} \right) - x\cos \left( {\dfrac{y}{x}} \right)} \right\}xdy$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{\left\{ {x\cos \left( {\dfrac{y}{x}} \right) + y\sin \left( {\dfrac{y}{x}} \right)} \right\}y}}{{\left\{ {y\sin \left( {\dfrac{y}{x}} \right) - x\cos \left( {\dfrac{y}{x}} \right)} \right\}x}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{\left\{ {x\cos \left( v \right) + vx\sin \left( v \right)} \right\}vx}}{{\left\{ {vx\sin \left( v \right) - x\cos \left( v \right)} \right\}x}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{\left\{ {\cos \left( v \right) + v\sin \left( v \right)} \right\}v}}{{\left\{ {v\sin \left( v \right) - \cos \left( v \right)} \right\}}} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{v\cos \left( v \right) + {v^2}\sin \left( v \right) - {v^2}\sin \left( v \right) + v\cos \left( v \right)}}{{v\sin \left( v \right) - \cos \left( v \right)}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{2v\cos \left( v \right)}}{{v\sin \left( v \right) - \cos \left( v \right)}}$

$\dfrac{{v\sin \left( v \right) - \cos \left( v \right)}}{{2v\cos \left( v \right)}}dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{{v\sin \left( v \right) - \cos \left( v \right)}}{{2v\cos \left( v \right)}}dv} = \int {\dfrac{1}{x}dx}$

$\int {\dfrac{{v\sin \left( v \right)}}{{2v\cos \left( v \right)}}dv - \int {\dfrac{{\cos \left( v \right)}}{{2v\cos \left( v \right)}}dv} } = \int {\dfrac{1}{x}dx}$

$\dfrac{1}{2}\int {\tan vdv - \dfrac{1}{2}\int {\dfrac{1}{v}dv} } = \int {\dfrac{1}{x}dx}$

$\dfrac{1}{2}\log \sec v - \dfrac{1}{2}\log v = \log x + \log C$

$\dfrac{1}{2}\left( {\log \sec v - \log v} \right) = \log xC$

$\log \dfrac{{\sec v}}{v} = 2\log xC$

$\log \dfrac{{\sec v}}{v} = \log {\left( {xC} \right)^2}$

$\dfrac{{\sec v}}{v} = {\left( {xC} \right)^2}$

Substituting the value of $v = \dfrac{y}{x}$.

$\dfrac{{\sec \left( {\dfrac{y}{x}} \right)}}{{\dfrac{y}{x}}} = {\left( {xC} \right)^2}$

$\dfrac{x}{y}\sec \left( {\dfrac{y}{x}} \right) = {\left( {xC} \right)^2}$

$\dfrac{x}{{y{x^2}}}.\dfrac{1}{{\cos \left( {\dfrac{y}{x}} \right)}} = {C^2}$

$\dfrac{1}{{yx}}.\dfrac{1}{{\cos \left( {\dfrac{y}{x}} \right)}} = {C^2}$

$\dfrac{1}{{{C^2}}} = yx\cos \left( {\dfrac{y}{x}} \right)$

$yx\cos \left( {\dfrac{y}{x}} \right) = K$

This is the required differential equation.

8. Solve the differential equation $x\dfrac{{dy}}{{dx}} - y + x\sin \left( {\dfrac{y}{x}} \right) = 0$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{y - x\sin \left( {\dfrac{y}{x}} \right)}}{x}$

$\dfrac{{dy}}{{dx}} = \dfrac{y}{x} - \sin \left( {\dfrac{y}{x}} \right)$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx}}{x} - \sin \left( {\dfrac{{vx}}{x}} \right)$

$v + x\dfrac{{dv}}{{dx}} = v - \sin v$

$x\dfrac{{dv}}{{dx}} = - \sin v$

$\dfrac{1}{{\sin v}}dv = - \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{1}{{\sin v}}dv} = - \int {\dfrac{1}{x}dx}$

$\int {\cos ecvdv} = - \int {\dfrac{1}{x}dx}$

$\log \left( {\cos ecv - \cot v} \right) = - \log x + \log C$

$\log \left( {\cos ecv - \cot v} \right) = \log \dfrac{C}{x}$

$\cos ecv - \cot v = \dfrac{C}{x}$

Substituting the value of $v = \dfrac{y}{x}$.

$\cos ec\left( {\dfrac{y}{x}} \right) - \cot \left( {\dfrac{y}{x}} \right) = \dfrac{C}{x}$

$\dfrac{1}{{\sin \left( {\dfrac{y}{x}} \right)}} - \dfrac{{\cos \left( {\dfrac{y}{x}} \right)}}{{\sin \left( {\dfrac{y}{x}} \right)}} = \dfrac{C}{x}$

$1 - \cos \left( {\dfrac{y}{x}} \right) = \dfrac{C}{x}\sin \left( {\dfrac{y}{x}} \right)$

$x\left( {1 - \cos \left( {\dfrac{y}{x}} \right)} \right) = C\sin \left( {\dfrac{y}{x}} \right)$

This is the required differential equation.

9. Solve the differential equation $ydx + x\log \left( {\dfrac{y}{x}} \right)dy - 2xdy = 0$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{ - y}}{{x\log \left( {\dfrac{y}{x}} \right) - 2x}}$

$\dfrac{{dy}}{{dx}} = \dfrac{y}{{2x - x\log \left( {\dfrac{y}{x}} \right)}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx}}{{2x - x\log \left( {\dfrac{{vx}}{x}} \right)}}$

$v + x\dfrac{{dv}}{{dx}} = \dfrac{v}{{2 - \log v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{v}{{2 - \log v}} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{v - 2v + v\log v}}{{2 - \log v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{ - v + v\log v}}{{2 - \log v}}$

$\dfrac{{2 - \log v}}{{ - v + v\log v}}dv = \dfrac{1}{x}dx$

$\dfrac{{2 - \log v}}{{v\left( {\log v - 1} \right)}}dv = \dfrac{1}{x}dx$

$\dfrac{{1 - \left( {\log v - 1} \right)}}{{v\left( {\log v - 1} \right)}}dv = \dfrac{1}{x}dx$

$\dfrac{1}{{v\left( {\log v - 1} \right)}}dv - \dfrac{{\left( {\log v - 1} \right)}}{{v\left( {\log v - 1} \right)}}dv = \dfrac{1}{x}dx$

$\dfrac{1}{{v\left( {\log v - 1} \right)}}dv - \dfrac{1}{v}dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{1}{{v\left( {\log v - 1} \right)}}dv} - \int {\dfrac{1}{v}dv} = \int {\dfrac{1}{x}dx}$                                            ………. (1)

Let $I = \int {\dfrac{1}{{v\left( {\log v - 1} \right)}}dv}$

Put $\log v - 1 = t$

Differentiating w.r.t. $v$ .

$\dfrac{1}{v} = \dfrac{{dt}}{{dv}}$

$\dfrac{1}{v}dv = dt$

Put this value in above equation and we get

$I = \int {\dfrac{1}{t}dt}$

Put this in equation $\left( 1 \right)$

$\int {\dfrac{1}{t}dt} - \int {\dfrac{1}{v}dv} = \int {\dfrac{1}{x}dx}$

$\log t - \log v = \log x + \log c$

Substituting the value of $\log v - 1 = t$ and $v = \dfrac{y}{x}$ .

$\log \left( {\log v - 1} \right) - \log \left( {\dfrac{y}{x}} \right) = \log x + \log c$

$\log \left( {\dfrac{{\log \left( {\dfrac{y}{x}} \right) - 1}}{{\left( {\dfrac{y}{x}} \right)}}} \right) = \log xC$

$\dfrac{{\log \left( {\dfrac{y}{x}} \right) - 1}}{{\left( {\dfrac{y}{x}} \right)}} = xC$

$\log \left( {\dfrac{y}{x}} \right) - 1 = \dfrac{y}{x}.xC$

$\log \left( {\dfrac{y}{x}} \right) - 1 = yC$

This is the required differential equation.

10. Solve the differential equation $\left( {1 + {e^{\dfrac{x}{y}}}} \right)dx + {e^{\dfrac{x}{y}}}\left( {1 - \dfrac{x}{y}} \right)dy = 0$

Ans: After rearranging the given equation we get,

$\dfrac{{dx}}{{dy}} = \dfrac{{ - {e^{\dfrac{x}{y}}}\left( {1 - \dfrac{x}{y}} \right)}}{{1 + {e^{\dfrac{x}{y}}}}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$x = vy$

Differentiating equation w.r.t. $y$, we get

$\dfrac{{dx}}{{dy}} = v + y\dfrac{{dv}}{{dy}}$

Substituting $x = vy$ and $\dfrac{{dx}}{{dy}}$ in the above equation, we get

$v + y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^{\dfrac{{vy}}{y}}}\left( {1 - \dfrac{{vy}}{y}} \right)}}{{1 + {e^{\dfrac{{vy}}{y}}}}}$

$v + y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v}\left( {1 - v} \right)}}{{1 + {e^v}}}$

$v + y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} + v{e^v}}}{{1 + {e^v}}}$

$y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} + v{e^v}}}{{1 + {e^v}}} - v$

$y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} + v{e^v} - v - v{e^v}}}{{1 + {e^v}}}$

$y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} - v}}{{1 + {e^v}}}$

$y\dfrac{{dv}}{{dy}} = - \left[ {\dfrac{{{e^v} + v}}{{1 + {e^v}}}} \right]$

$\left[ {\dfrac{{1 + {e^v}}}{{{e^v} + v}}} \right]dv = - \dfrac{1}{y}dy$

Taking integration on both side,

$\int {\left[ {\dfrac{{1 + {e^v}}}{{{e^v} + v}}} \right]dv} = - \int {\dfrac{1}{y}dy}$

On integrating both side,

$\log \left( {v + {e^v}} \right) = - \log y + \log C$

Substituting the value of $v = \dfrac{x}{y}$.

$\log \left( {\dfrac{x}{y} + {e^{\dfrac{x}{y}}}} \right) = - \log y + \log C$

$\log \left( {\dfrac{x}{y} + {e^{\dfrac{x}{y}}}} \right) = \log \left( {\dfrac{C}{y}} \right)$

$\dfrac{x}{y} + {e^{\dfrac{x}{y}}} = \dfrac{C}{y}$

$x + y{e^{\dfrac{x}{y}}} = C$

This is the required differential equation.

11. Solve the differential equation $\left( {x + y} \right)dy + \left( {x - y} \right)dx = 0;{\text{ }}y = 1;x = 1$

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = - \dfrac{{\left( {x - y} \right)}}{{\left( {x + y} \right)}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{\left( {x - vx} \right)}}{{\left( {x + vx} \right)}}$

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{\left( {1 - v} \right)}}{{\left( {1 + v} \right)}}$

$x\dfrac{{dv}}{{dx}} = - \dfrac{{\left( {1 - v} \right)}}{{\left( {1 + v} \right)}} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 + v - v - {v^2}}}{{1 + v}}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 - {v^2}}}{{1 + v}}$

$\dfrac{{1 + v}}{{1 + {v^2}}}dv = - \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{{1 + v}}{{1 + {v^2}}}dv} = - \int {\dfrac{1}{x}dx}$

$\int {\dfrac{1}{{1 + {v^2}}}dv} + \int {\dfrac{v}{{1 + {v^2}}}dv} = - \int {\dfrac{1}{x}dx}$

${\tan ^{ - 1}}v + \dfrac{1}{2}\log \left( {1 + {v^2}} \right) = - \log x + C$

Substituting the value of $v = \dfrac{y}{x}$.

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

$y = 1$

When

$x = 1$

${\tan ^{ - 1}}\left( {\dfrac{1}{1}} \right) + \dfrac{1}{2}\log \left( {1 + {{\left( {\dfrac{1}{1}} \right)}^2}} \right) = - \log 1 + C$

$\dfrac{\pi }{4} + \dfrac{1}{2}\log \left( 2 \right) = C$

Therefore, the final solution becomes,

${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + \dfrac{1}{2}\log \left( {1 + {{\left( {\dfrac{y}{x}} \right)}^2}} \right) = - \log x + \dfrac{\pi }{4} + \dfrac{1}{2}\log \left( 2 \right)$

$2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + \log \left( {\dfrac{{{x^2} + {y^2}}}{{{x^2}}}} \right) = - 2\log x + 2 \times \dfrac{\pi }{4} + 2 \times \dfrac{1}{2}\log \left( 2 \right)$

$2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + \log \left( {\dfrac{{{x^2} + {y^2}}}{{{x^2}}}} \right) = - \log {x^2} + \dfrac{\pi }{2} + \log \left( 2 \right)$

$2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + \log \left( {\dfrac{{{x^2} + {y^2}}}{{{x^2}}}} \right) + \log {x^2} = + \dfrac{\pi }{2} + \log \left( 2 \right)$

$2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + \log \left( {\dfrac{{{x^2} + {y^2}}}{{{x^2}}} \times {x^2}} \right) = + \dfrac{\pi }{2} + \log \left( 2 \right)$

$2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + \log \left( {{x^2} + {y^2}} \right) = \dfrac{\pi }{2} + \log \left( 2 \right)$

This is the required differential equation.

12. Solve the differential equation ${x^2}dy + \left( {xy + {y^2}} \right)dx = 0$; $y = 1$ when $x = 1$.

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = - \dfrac{{xy + {y^2}}}{{{x^2}}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{xvx + {{\left( {xv} \right)}^2}}}{{{x^2}}}$

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{v{x^2} + {x^2}{v^2}}}{{{x^2}}}$

$v + x\dfrac{{dv}}{{dx}} = - v - {v^2}$

$x\dfrac{{dv}}{{dx}} = - v - {v^2} - v$

$x\dfrac{{dv}}{{dx}} = - 2v - {v^2}$

$\dfrac{1}{{2v + {v^2}}}dv = - \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{1}{{2v + {v^2}}}dv} = - \int {\dfrac{1}{x}dx}$

$\int {\dfrac{1}{{v\left( {2 + v} \right)}}dv} = - \int {\dfrac{1}{x}dx}$

Dividing and multiplying above equation by 2.

$\dfrac{1}{2}\int {\dfrac{2}{{v\left( {2 + v} \right)}}dv} = - \int {\dfrac{1}{x}dx}$

$\dfrac{1}{2}\int {\dfrac{{2 + v - v}}{{v\left( {2 + v} \right)}}dv} = - \int {\dfrac{1}{x}dx}$

$\dfrac{1}{2}\int {\dfrac{{2 + v}}{{v\left( {2 + v} \right)}}dv} - \dfrac{1}{2}\int {\dfrac{v}{{v\left( {2 + v} \right)}}dv} = - \int {\dfrac{1}{x}dx}$

$\dfrac{1}{2}\int {\dfrac{1}{v}dv} - \dfrac{1}{2}\int {\dfrac{1}{{\left( {2 + v} \right)}}dv} = - \int {\dfrac{1}{x}dx}$

$\dfrac{1}{2}\log v - \dfrac{1}{2}\log \left( {2 + v} \right) = - \log x + \log C$

$\dfrac{1}{2}\log \left( {\dfrac{v}{{2 + v}}} \right) = \log \dfrac{C}{x}$

$\log \left( {\dfrac{v}{{2 + v}}} \right) = 2\log \dfrac{C}{x}$

$\log \left( {\dfrac{v}{{2 + v}}} \right) = \log {\left( {\dfrac{C}{x}} \right)^2}$

$\dfrac{v}{{2 + v}} = {\left( {\dfrac{C}{x}} \right)^2}$

Substituting the value of $v = \dfrac{y}{x}$.

$\dfrac{{\dfrac{y}{x}}}{{2 + \dfrac{y}{x}}} = {\left( {\dfrac{C}{x}} \right)^2}$

$\dfrac{{\dfrac{y}{x}}}{{\dfrac{{2x + y}}{x}}} = {\left( {\dfrac{C}{x}} \right)^2}$

$\dfrac{y}{{2x + y}} = {\left( {\dfrac{C}{x}} \right)^2}$

$y = 1$

When

$x = 1$

$\dfrac{1}{{2.1 + 1}} = {\left( {\dfrac{C}{1}} \right)^2}$

$\dfrac{1}{3} = {C^2}$

Therefore, the final solution becomes,

$\dfrac{y}{{2x + y}} = {\left( {\dfrac{C}{x}} \right)^2}$

$\dfrac{y}{{2x + y}} = \dfrac{{{C^2}}}{{{x^2}}}$

$\dfrac{{y{x^2}}}{{2x + y}} = {C^2}$

$\dfrac{{y{x^2}}}{{2x + y}} = \dfrac{1}{3}$

$2x + y = 3y{x^2}$

This is the required differential equation.

13. Solve the differential equation $\left[ {x{{\sin }^2}\left( {\dfrac{y}{x}} \right) - y} \right]dx + xdy = 0$; $y = \dfrac{\pi }{4}$ when $x = 1$.

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = - \dfrac{{\left[ {x{{\sin }^2}\left( {\dfrac{y}{x}} \right) - y} \right]}}{x}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = - \dfrac{{\left[ {x{{\sin }^2}\left( {\dfrac{{vx}}{x}} \right) - vx} \right]}}{x}$

$v + x\dfrac{{dv}}{{dx}} = - {\sin ^2}\left( v \right) + v$

$x\dfrac{{dv}}{{dx}} = - {\sin ^2}\left( v \right) + v - v$

$x\dfrac{{dv}}{{dx}} = - {\sin ^2}\left( v \right)$

$\dfrac{1}{{{{\sin }^2}\left( v \right)}}dv = - \dfrac{1}{x}dx$

$\cos e{c^2}vdv = - \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\cos e{c^2}vdv} = - \int {\dfrac{1}{x}dx}$

$- \cot v = - \log x - \log C$

$\cot v = \log x + \log C$

$\cot v = \log xC$

Substituting the value of $v = \dfrac{y}{x}$.

$\cot \left( {\dfrac{y}{x}} \right) = \log xC$

$y = \dfrac{\pi }{4}$

When

$x = 1$

$\cot \left( {\dfrac{{\dfrac{\pi }{4}}}{1}} \right) = \log 1C$

$\cot \left( {\dfrac{\pi }{4}} \right) = \log C$

$1 = \log C$

${e^1} = C$

$e = C$

Therefore, final solution becomes,

$\cot \left( {\dfrac{y}{x}} \right) = \log \left| {xe} \right|$

This is the required differential equation.

14. Solve the differential equation $\dfrac{{dy}}{{dx}} - \dfrac{y}{x} + \cos ec\left( {\dfrac{y}{x}} \right) = 0$; $y = 0$ when $x = 1$.

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{y}{x} - \cos ec\left( {\dfrac{y}{x}} \right)$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx}}{x} - \cos ec\left( {\dfrac{{vx}}{x}} \right)$

$v + x\dfrac{{dv}}{{dx}} = v - \cos ec\left( v \right)$

$x\dfrac{{dv}}{{dx}} = - \cos ec\left( v \right)$

$\dfrac{1}{{\cos ec\left( v \right)}}dv = - \dfrac{1}{x}dx$

$\sin vdv = - \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\sin vdv} = - \int {\dfrac{1}{x}dx}$

$- \cos v = - \log x + C$

Substituting the value of $v = \dfrac{y}{x}$.

$- \cos \left( {\dfrac{y}{x}} \right) = - \log x + C$

$y = 0$

When

$x = 1$

$- \cos \left( {\dfrac{0}{1}} \right) = - \log 1 + C$

$- 1 = C$

Therefore, the final solution becomes,

$- \cos \left( {\dfrac{y}{x}} \right) = - \log x - 1$

$\cos \left( {\dfrac{y}{x}} \right) = \log x + 1$

$\cos \left( {\dfrac{y}{x}} \right) = \log x + \log e$

$\cos \left( {\dfrac{y}{x}} \right) = \log \left| {xe} \right|$

This is the required differential equation.

15. Solve the differential equation $2xy + {y^2} - 2{x^2}\dfrac{{dy}}{{dx}} = 0$; $y = 2$ when $x = 1$.

Ans: After rearranging the given equation we get,

$\dfrac{{dy}}{{dx}} = \dfrac{{2xy + {y^2}}}{{2{x^2}}}$

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

$y = vx$

Differentiating equation w.r.t. $x$, we get

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Substituting $y = vx$ and $\dfrac{{dy}}{{dx}}$ in the above equation, we get

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{2xvx + {{\left( {vx} \right)}^2}}}{{2{x^2}}}$

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{2{x^2}v + {v^2}{x^2}}}{{2{x^2}}}$

$v + x\dfrac{{dv}}{{dx}} = \dfrac{{2v + {v^2}}}{2}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{2v + {v^2}}}{2} - v$

$x\dfrac{{dv}}{{dx}} = \dfrac{{2v + {v^2} - 2v}}{2}$

$x\dfrac{{dv}}{{dx}} = \dfrac{{{v^2}}}{2}$

$\dfrac{2}{{{v^2}}}dv = \dfrac{1}{x}dx$

Taking integration on both side,

$\int {\dfrac{2}{{{v^2}}}dv} = \int {\dfrac{1}{x}dx}$

$- \dfrac{2}{v} = \log x + C$

Substituting the value of $v = \dfrac{y}{x}$.

$- \dfrac{2}{{\dfrac{y}{x}}} = \log x + C$

$- \dfrac{{2x}}{y} = \log x + C$

$y = 2$

When

$x = 1$

$- \dfrac{{2.1}}{2} = \log 1 + C$

$- 1 = C$

Therefore, final solution becomes,

$- \dfrac{{2x}}{y} = \log x - 1$

$\dfrac{{2x}}{y} = 1 - \log x$

$y = \dfrac{{2x}}{{1 - \log x}}:x \ne e$

This is the required differential equation.

16. A homogeneous differential equation of the from $\dfrac{{dx}}{{dy}} = h\left( {\dfrac{x}{y}} \right)$can be solved by

making the substitution.

$\left( A \right)y = vx{\text{ }}\left( B \right)v = yx$

$\left( C \right)x = vy{\text{ }}\left( D \right)x = v$

Ans: As $h\left( {\dfrac{x}{y}} \right)$ is function of $\dfrac{x}{y}$

Therefore, we have to substitute, $x = vy$ .

So, the correct option is $\left( C \right)$ .

17. Which of the following is a homogeneous differential equation?

$\left( A \right)\left( {4x + 6y + 5} \right)dy - \left( {3y + 2x + 4} \right)dx = 0$

$\left( B \right)\left( {xy} \right)dx - \left( {{x^3} + {y^3}} \right)dy = 0$

$\left( C \right)\left( {{x^3} + 2{y^2}} \right)dx + 2xydy = 0$

$\left( D \right){y^2}dx + \left( {{x^2} - xy - {y^2}} \right)dy = 0$

Ans: The correct option is $\left( D \right)$ .

Explanation:

${y^2}dx + \left( {{x^2} - xy - {y^2}} \right)dy = 0$

After rearranging the given equation we get,

$\dfrac{{dx}}{{dy}} = - \dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}$

Let $f\left( {x,y} \right) = - \dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}$

Now, put $x = kx$ and $y = ky$

$f\left( {kx,ky} \right) = - \dfrac{{{{\left( {kx} \right)}^2} - kxky - {{\left( {ky} \right)}^2}}}{{{{\left( {ky} \right)}^2}}}$

$f\left( {kx,ky} \right) = - \dfrac{{{k^2}}}{{{k^2}}}\dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}$

$f\left( {kx,ky} \right) = {k^0}\left( { - \dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}} \right)$

$f\left( {kx,ky} \right) = {k^0}f\left( {x,y} \right)$

Hence, the given differential equation is homogeneous.

Conclusion

In conclusion, Exercise 9.4 of Chapter 9, "Differential Equations," is a critical part of your understanding of differential equations, focusing on advanced methods of solving these equations. This class 12 ex 9.4 deals with higher-order differential equations and various techniques such as the method of undetermined coefficients and variation of parameters. Through consistent practice, you have strengthened your ability to approach complex differential equations methodically, applying appropriate solution strategies to each unique problem.

Class 12 Maths Chapter 9: Exercises Breakdown

 Exercise Number of Questions Exercise 9.1 12 Questions & Solutions (10 Short Answers, 2 MCQs) Exercise 9.2 12 Questions & Solutions (10 Short Answers, 2 MCQs) Exercise 9.3 23 Questions & Solutions (10 Short Answers, 12 Long Answers, 1 MCQs) Exercise 9.5 19 Questions & Solutions (17 Short Answers, 2 MCQs)

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations Exercise 9.4

1. What is the focus of Exercise 9.4 in Chapter 9 - Differential Equations?

Exercise 9.4 Maths Class 12 focuses on solving higher-order differential equations using methods such as the method of undetermined coefficients and variation of parameters.

2. What is the method of undetermined coefficients?

In Ex 9.4 Class 12 Maths NCERT Solutions, The method of undetermined coefficients is a technique used to find particular solutions to linear differential equations with constant coefficients. It involves guessing the form of the particular solution based on the type of non-homogeneous term.

3. How does the variation of parameters method work?

The variation of parameters method is a technique used to solve non-homogeneous linear differential equations. It involves finding a particular solution by varying the constants in the complementary solution.

4. What types of differential equations are covered in Exercise 9.4?

Exercise 9.4 covers higher-order linear differential equations, including those with constant coefficients and non-homogeneous terms.

5. What are some real-world applications of the techniques learned in Exercise 9.4?

The techniques learned in Exercise 9.4 are applied in various fields such as engineering, physics, economics, and biology. They are used to model and solve problems related to motion, heat transfer, population dynamics, and financial markets.