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# NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations - Exercise 9.6

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 9 (Ex 9.6)

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6 (Ex 9.6) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 9 Differential Equations Exercise 9.6 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 12 Subject: Class 12 Maths Chapter Name: Chapter 9 - Differential Equations Exercise: Exercise - 9.6 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

NCERT solutions class 12 maths chapter nine exercise 9.6 differential equations facilitate children to study more about linear differential equations. There are  well-known forms of such equations, i.e., dy/dx + Py = Q and dx/dy + P1x = Q1. Given below are the steps to resolve such differential equations.

• Write the given differential equation in the form of dy/dx + Py = Q wherein P, Q are constants or functions of x only.

• Find the Integrating Factor (I.F), i.e., e ∫p dx

• Write the solution of the given differential equation as y (I.F) = ∫[Q (I.F.)] dx + C.

NCERT solutions class 12 maths chapter nine exercise 9.6 has a pool of nineteen problems requiring students to discover the general and the particular solution of a linear differential equation by using the above method.

The factor to remember while solving the NCERT solutions class 12 maths chapter nine exercise 9.6 differential equations is figuring out the right integrating factor. This is a new idea and students want to recognize a way to successfully use it in the problems.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Chapter 9 – Differential Equations

Exercise 9.6

1. Solve the differential equation $\dfrac{{dy}}{{dx}} + 2y = \sin x$.

Ans: The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = 2$ and $Q = \sin x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {2dx} }}$

$I.F. = {e^{2x}}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y{e^{2x}} = \int {\sin x.{e^{2x}}dx} + C.......\left( 1 \right)$

Let $I = \int {\sin x.{e^{2x}}dx}$

Using integration by parts,

$I = \sin x\int {{e^{2x}}dx} - \int {\left( {\dfrac{{d\left( {\sin x} \right)}}{{dx}}.\int {{e^{2x}}dx} } \right)} dx$

$I = \dfrac{{{e^{2x}}\sin x}}{2} - \int {\cos x.\dfrac{{{e^{2x}}}}{2}dx}$

Again using integration by parts,

$I = \dfrac{{{e^{2x}}\sin x}}{2} - \cos x\int {\dfrac{{{e^{2x}}}}{2}dx} + \int {\left( {\dfrac{{d\left( {\cos x} \right)}}{{dx}}.\int {\dfrac{{{e^{2x}}}}{2}dx} } \right)} dx$

$I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} + \int {\left( { - \sin x} \right).\dfrac{{{e^{2x}}}}{4}dx}$

$I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} - \dfrac{1}{4}\int {\sin x.{e^{2x}}dx}$

As $I = \int {\sin x.{e^{2x}}dx}$,

Therefore,

$I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} - \dfrac{1}{4}I$

$I + \dfrac{1}{4}I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}$

$\dfrac{5}{4}I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}$

$I = \dfrac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right)$

Substituting this value of I in equation 1.

$y{e^{2x}} = \dfrac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + C$

$y = \dfrac{1}{5}\left( {2\sin x - \cos x} \right) + C{e^{ - 2x}}$

This required differential equation .

2. Solve the differential equation $\dfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}$.

Ans: The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = 3$ and $Q = {e^{ - 2x}}$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {3dx} }}$

$I.F. = {e^{3x}}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y{e^{3x}} = \int {\left( {{e^{3x}} \times {e^{ - 2x}}} \right)dx} + C$

$y{e^{3x}} = \int {{e^x}dx} + C$

$y{e^{3x}} = {e^x} + C$

$y = {e^{ - 2x}} + C{e^x}$

This required differential equation.

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

Ans: The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \dfrac{1}{x}$ and $Q = {x^2}$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {\dfrac{1}{x}dx} }}$

$I.F. = {e^{\log x}}$

$I.F. = x$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$yx = \int {\left( {{x^2}.x} \right)dx} + C$

$yx = \int {\left( {{x^3}} \right)dx} + C$

$yx = \dfrac{{{x^4}}}{4} + C$

This required differential equation.

4. Solve the differential equation $\dfrac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x{\text{ }}\left( {0 \leqslant x \leqslant \dfrac{\pi }{2}} \right)$

Ans: The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \sec x$ and $Q = \tan x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {\sec xdx} }}$

$I.F. = {e^{\log \left( {\sec x + \tan x} \right)}}$

$I.F. = \sec x + \tan x$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y\left( {\sec x + \tan x} \right) = \int {\tan x\left( {\sec x + \tan x} \right)dx} + C$

$y\left( {\sec x + \tan x} \right) = \int {\sec x\tan xdx} + \int {{{\tan }^2}xdx} + C$

$y\left( {\sec x + \tan x} \right) = \sec x + \int {\left( {{{\sec }^2}x - 1} \right)dx} + C$

$y\left( {\sec x + \tan x} \right) = \sec x + \tan x - x + C$

This required differential equation.

5. Solve the differential equation ${\cos ^2}x\dfrac{{dy}}{{dx}} + y = \tan x{\text{ }}\left( {0 \leqslant x\dfrac{\pi }{2}} \right)$

Ans: On rearranging the given equation,

$\dfrac{{dy}}{{dx}} + {\sec ^2}x.y = {\sec ^2}x\tan x$

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = {\sec ^2}x$ and $Q = {\sec ^2}x\tan x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {{{\sec }^2}xdx} }}$

$I.F. = {e^{\tan x}}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y{e^{\tan x}} = \int {\tan x{{\sec }^2}x{e^{\tan x}}dx + C}$

Put $\tan x = t$

Differentiating w.r.t. $t$.

${\sec ^2}xdx = dt$

Therefore, above equation become,

$y{e^{\tan x}} = \int {t{e^t}dt + C}$

Using integration by parts,

$y{e^{\tan x}} = t{e^t} - \int {{e^t}dt + C}$

$y{e^{\tan x}} = t{e^t} - {e^t} + C$

Substituting the value of $t$ .

$y{e^{\tan x}} = \tan x{e^{\tan x}} - {e^{\tan x}} + C$

$y = \tan x - 1 + C{e^{ - \tan x}}$

This is required differential equation.

6. Solve the differential equation $x\dfrac{{dy}}{{dx}} + 2y = {x^2}\log x$

Ans: On rearranging given equation can be written as,

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

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \dfrac{2}{x}$ and $Q = x\log x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {\dfrac{2}{x}dx} }}$

$I.F. = {e^{\log {x^2}}}$

$I.F. = {x^2}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y{x^2} = \int {x\log x.{x^2}dx} + C$

$y{x^2} = \int {{x^3}\log xdx} + C$

Using integration by parts,

$y{x^2} = \log x\int {{x^3}dx} - \int {\left( {\dfrac{d}{{dx}}\left( {\log x} \right)\int {{x^3}dx} } \right)dx} + C$

$y{x^2} = \dfrac{{{x^4}\log x}}{4} - \int {\left( {\dfrac{1}{x}.\dfrac{{{x^4}}}{4}} \right)dx + C}$

$y{x^2} = \dfrac{{{x^4}\log x}}{4} - \dfrac{1}{4}.\dfrac{{{x^4}}}{4} + C$

$y{x^2} = \dfrac{{{x^4}\log x}}{4} - \dfrac{{{x^4}}}{{16}} + C$

$y = \dfrac{{{x^2}\log x}}{4} - \dfrac{{{x^2}}}{{16}} + C{x^{ - 2}}$

$y = \dfrac{{{x^2}}}{{16}}\left( {4\log x - 1} \right) + C{x^{ - 2}}$

This is required differential equation.

7. Solve the differential equation $x\log x\dfrac{{dy}}{{dx}} + y = \dfrac{2}{x}\log x$

Ans: On rearranging the given equation can be written as,

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

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \dfrac{1}{{x\log x}}$ and $Q = \dfrac{2}{{{x^2}}}$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$



$I.F. = {e^{\log \left( {\log x} \right)}}$

$I.F. = \log x$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

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

Using integration by parts,

$y\log x = 2\left[ {\log x\int {\dfrac{1}{{{x^2}}}dx} - \int {\left( {\dfrac{d}{{dx}}\left( {\log x} \right)\int {\dfrac{1}{{{x^2}}}dx} } \right)dx} } \right] + C$

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

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

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

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

This is required differential equation.

8. Solve the differential equation $\left( {1 + {x^2}} \right)dy + 2xydx = \cot xdx$

Ans: On rearranging the given equation can be written as,

$\dfrac{{dy}}{{dx}} + \dfrac{{2xy}}{{1 + {x^2}}} = \dfrac{{\cot x}}{{1 + {x^2}}}$

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \dfrac{{2x}}{{1 + {x^2}}}$ and $Q = \dfrac{{\cot x}}{{1 + {x^2}}}$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {\dfrac{{2x}}{{1 + {x^2}}}dx} }}$

$I.F. = {e^{\log \left( {1 + {x^2}} \right)}}$

$I.F. = 1 + {x^2}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y\left( {1 + {x^2}} \right) = \int {\dfrac{{\cot x}}{{1 + {x^2}}} \times \left( {1 + {x^2}} \right)dx} + C$

$y\left( {1 + {x^2}} \right) = \int {\cot xdx} + C$

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

This is required differential equation.

9. Solve the differential equation $x\dfrac{{dy}}{{dx}} + y - x + xy\cot x = 0$

Ans: On rearranging given equation can be written as,

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

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \dfrac{1}{x} + \cot x$ and $Q = 1$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {\left( {\dfrac{1}{x} + \cot x} \right)dx} }}$

$I.F. = {e^{\log x + \log \sin x}}$

$I.F. = x\sin x$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$yx\sin x = \int {\left( {1.x\sin x} \right)dx} + C$

$yx\sin x = x\int {\sin xdx - \int {\left( {\dfrac{d}{{dx}}\left( x \right)\int {\sin xdx} } \right)dx} } + C$

$yx\sin x = x\left( { - \cos x} \right) - \int {\left( { - \cos x} \right)dx} + C$

$yx\sin x = - x\cos x + \sin x + C$

$y = \dfrac{{ - x\cos x}}{{x\sin x}} + \dfrac{{\sin x}}{{x\sin x}} + \dfrac{C}{{x\sin x}}$

$y = - \cot x + \dfrac{1}{x} + \dfrac{C}{{x\sin x}}$

This is required differential equation.

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

Ans: On rearranging the given equation can be written as,

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

The given equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$

Where, $P = - 1$ and $Q = y$

Calculating integration factor,

$I.F. = {e^{\int { - 1dy} }}$

$I.F. = {e^{ - y}}$

Therefore, solution of the given differential equation is given by the relation,

$x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy} + C$

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

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

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

$x{e^{ - y}} = - y{e^{ - y}} - {e^{ - y}} + C$

$x = - y - 1 + C{e^y}$

$x + y + 1 = C{e^y}$

This is required differential equation.

11. Solve the differential equation $ydx + \left( {x - {y^2}} \right)dy = 0$

Ans: On rearranging the given equation can be written as,

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

The given equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$

Where, $P = \dfrac{1}{y}$ and $Q = y$

Calculating integration factor,

$I.F. = {e^{\int {Pdy} }}$

$I.F. = {e^{\int {\dfrac{1}{y}dy} }}$

$I.F. = {e^{\log y}}$

$I.F. = y$

Therefore, solution of the given differential equation is given by the relation,

$x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy} + C$

$xy = \int {\left( {y.y} \right)dy} + C$

$xy = \int {{y^2}dy} + C$

$xy = \dfrac{{{y^3}}}{3} + C$

This is required differential equation.

12. Solve the differential equation $\left( {x + 3{y^2}} \right)\dfrac{{dy}}{{dx}} = y{\text{ }}\left( {y > 0} \right)$

Ans: On rearranging the given equation can be written as,

$\dfrac{{dx}}{{dy}} - \dfrac{x}{y} = 3y$

The given equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$

Where, $P = - \dfrac{1}{y}$ and $Q = 3y$

Calculating integration factor,

$I.F. = {e^{\int {Pdy} }}$

$I.F. = {e^{\int {\left( { - \dfrac{1}{y}} \right)dy} }}$

$I.F. = {e^{\log \left( {\dfrac{1}{y}} \right)}}$

$I.F. = \dfrac{1}{y}$

Therefore, solution of the given differential equation is given by the relation,

$x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy} + C$

$x\left( {\dfrac{1}{y}} \right) = \int {\left( {3y \times \dfrac{1}{y}} \right)dy} + C$

$\dfrac{x}{y} = \int {3dy} + C$

$\dfrac{x}{y} = 3y + C$

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

This is required differential equation.

13. Solve the differential equation $\dfrac{{dy}}{{dx}} + 2y\tan x = \sin x;{\text{ }}y = 0{\text{ when }}x = \dfrac{\pi }{3}$

Ans: The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = 2\tan x$ and $Q = \sin x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {2\tan xdx} }}$

$I.F. = {e^{\log \left( {{{\sec }^2}x} \right)}}$

$I.F. = {\sec ^2}x$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y{\sec ^2}x = \int {\left( {\sin x.{{\sec }^2}x} \right)dx} + C$

$y{\sec ^2}x = \int {\left( {\sec x\tan x} \right)dx} + C$

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

Now, $y = 0{\text{ when }}x = \dfrac{\pi }{3}$

$0 \times {\sec ^2}\left( {\dfrac{\pi }{3}} \right) = \sec \left( {\dfrac{\pi }{3}} \right) + C$

$C = - 2$

Substituting the value of $C = - 2$in equation $\left( 1 \right)$.

$y{\sec ^2}x = \sec x - 2$

$y = \cos x - 2{\sec ^2}x$

This is required differential equation.

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

Ans: On rearranging the given equation can be written as,

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

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \dfrac{{2x}}{{1 + {x^2}}}$ and $Q = \dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int {\dfrac{{2x}}{{1 + {x^2}}}dx} }}$

$I.F. = {e^{\log \left( {1 + {x^2}} \right)}}$

$I.F. = 1 + {x^2}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

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

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

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

Now, $y = 0{\text{ when }}x = 1$

$0 = {\tan ^{ - 1}}1 + C$

$C = - \dfrac{\pi }{4}$

Substituting the value of $C = - \dfrac{\pi }{4}$ in equation$\left( 1 \right)$,

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

This is required differential equation.

15. Solve the differential equation $\dfrac{{dy}}{{dx}} - 3y\cot x = \sin 2x{\text{ }}y = 2{\text{ when }}x = \dfrac{\pi }{2}$

Ans: The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = - 3\cot x$ and $Q = \sin 2x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int { - 3\cot xdx} }}$

$I.F. = {e^{ - 3\log \left( {\sin x} \right)}}$

$I.F. = \dfrac{1}{{{{\sin }^3}x}}$

Therefore, solution of the given differential equation is given by the relation,

$y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx} + C$

$y.\dfrac{1}{{{{\sin }^3}x}} = \int {\left( {\sin 2x.\dfrac{1}{{{{\sin }^3}x}}} \right)dx} + C$

$y.\cos e{c^3}x = 2\int {\left( {\cot x\cos ecx} \right)dx} + C$

$y.\cos e{c^3}x = - 2\cos ecx + C$

$y = - \dfrac{2}{{\cos e{c^2}x}} + \dfrac{C}{{\cos e{c^3}x}}$

$y = - 2{\sin ^2}x + C{\sin ^3}x......\left( 1 \right)$

Now, $y = 2{\text{ when }}x = \dfrac{\pi }{2}$

$2 = - 2{\sin ^2}\left( {\dfrac{\pi }{2}} \right) + C{\sin ^3}\left( {\dfrac{\pi }{2}} \right)$

$2 = - 2 + C$

$C = 4$

Substituting the value of $C = 4$in equation 1,

$y = - 2{\sin ^2}x + 4{\sin ^3}x$

$y = 4{\sin ^3}x - 2{\sin ^2}x$

This is required differential equation.

16. Find the equation of a curve passing through the origin given that the slope of thetangent to the curve at any point $(x,y)$ is equal to the sum of the coordinates ofthe point.

Ans: Let $f\left( {x,y} \right)$ be the curve passing through the origin.

At point $(x,y)$, the slope of curve will be $\dfrac{{dy}}{{dx}}$ .

According to the question,

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

The given equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$

Where, $P = - 1$ and $Q = x$

Calculating integration factor,

$I.F. = {e^{\int { - 1dx} }}$

$I.F. = {e^{ - x}}$

Therefore, solution of the given differential equation is given by the relation,

$x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy} + C$

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

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

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

$y{e^{ - x}} = - x{e^{ - x}} - {e^{ - x}} + C$

$y = - x - 1 + C{e^x}$

$x + y + 1 = C{e^x}......\left( 1 \right)$

As, equation is passing through the origin.

Therefore,

$0 + 0 + 1 = C{e^0}$

$1 = C$

Substituting the value of $1 = C$ in equation 1.

$x + y + 1 = {e^x}$

This is required differential equation of curve passing through origin.

17. Find the equation of a curve passing through the point $\left( {0,2} \right)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope ofthe tangent to the curve at that point by 5.

Ans: Let $f\left( {x,y} \right)$ be the curve passing through the origin.

At point $(x,y)$, the slope of curve will be $\dfrac{{dy}}{{dx}}$ .

According to the question,

$\dfrac{{dy}}{{dx}} + 5 = x + y$

$\dfrac{{dy}}{{dx}} - y = x - 5$

The given equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$

Where, $P = - 1$ and $Q = x - 5$

Calculating integration factor,

$I.F. = {e^{\int { - 1dx} }}$

$I.F. = {e^{ - x}}$

Therefore, solution of the given differential equation is given by the relation,

$x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy} + C$

$y{e^{ - x}} = \int {\left( {x - 5} \right){e^{ - x}}dx} + C$

$y{e^{ - x}} = \left( {x - 5} \right)\int {{e^{ - x}}dx} - \int {\left( {\dfrac{d}{{dx}}\left( {x - 5} \right)\int {{e^{ - x}}dx} } \right)} dx + C$

$y{e^{ - x}} = \left( {5 - x} \right){e^{ - x}} + \int {{e^{ - x}}dx} + C$

$y{e^{ - x}} = \left( {5 - x} \right){e^{ - x}} - {e^{ - x}} + C$

$y{e^{ - x}} = \left( {4 - x} \right){e^{ - x}} + C$

$y = 4 - x + C{e^x}$

$y + x - 4 = C{e^x}......\left( 1 \right)$

As equation is passing through $\left( {0,2} \right)$ .

Therefore,

$0 + 2 - 4 = C{e^0}$

$- 2 = C$

Substituting the value of $- 2 = C$ in equation 1.

$y + x - 4 = - 2{e^x}$

This the required equation of curve passing through $\left( {0,2} \right)$ .

18. The Integrating Factor of the differential equation $x\dfrac{{dy}}{{dx}} - y = 2{x^2}$ is

$\left( A \right){e^{ - x}}$

$\left( B \right){e^{ - y}}$

$\left( C \right)\dfrac{1}{x}$

$\left( D \right)x$

Ans: On rearranging the given equation can be written as,

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

The given equation is in the form of $\dfrac{{dy}}{{dx}} + Py = Q$

Where, $P = - \dfrac{1}{x}$ and $Q = 2x$

Calculating integration factor,

$I.F. = {e^{\int {Pdx} }}$

$I.F. = {e^{\int { - \dfrac{1}{x}dx} }}$

$I.F. = {e^{ - \log x}}$

$I.F. = {e^{\log \dfrac{1}{x}}}$

$I.F. = \dfrac{1}{x}$

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

18. The Integrating Factor of the differential equation $\left( {1 - {y^2}} \right)\dfrac{{dx}}{{dy}} + yx = ay\left( { - 1 < y < 1} \right)$ is

$\left( A \right)\dfrac{1}{{{y^2} - 1}}$

$\left( B \right)\dfrac{1}{{\sqrt {{y^2} - 1} }}$

$\left( C \right)\dfrac{1}{{1 - {y^2}}}$

$\left( D \right)\dfrac{1}{{\sqrt {1 - {y^2}} }}$

Ans: On rearranging the given equation can be written as,

$\dfrac{{dx}}{{dy}} + \dfrac{{xy}}{{1 - {y^2}}} = \dfrac{{ay}}{{1 - {y^2}}}$

The given equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$

Where, $P = \dfrac{y}{{1 - {y^2}}}$ and $Q = \dfrac{{ay}}{{1 - {y^2}}}$

Calculating integration factor,

$I.F. = {e^{\int {Pdy} }}$

$I.F. = {e^{\int {\dfrac{y}{{1 - {y^2}}}dy} }}$

$I.F. = {e^{\log \left( {\dfrac{1}{{\sqrt {1 - {y^2}} }}} \right)}}$

$I.F. = \dfrac{1}{{\sqrt {1 - {y^2}} }}$

Therefore, the correct option is $\left( D \right)$

## NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.6

Opting for the NCERT solutions for Ex 9.6 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.6 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 9 Exercise 9.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 9 Exercise 9.6, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 9 Exercise 9.6 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

 Chapter 9 - Differential Equations Exercises in PDF Format 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 12 Questions & Solutions (5 Short Answers, 5 Long Answers, 2 MCQs) Exercise 9.4 23 Questions & Solutions (10 Short Answers, 12 Long Answers, 1 MCQs) Exercise 9.5 17 Questions & Solutions (15 Short Answers, 2 MCQs) Exercise 9.6 19 Questions & Solutions (15 Short Answers, 2 Long Answers, 2 MCQs)

## FAQs on NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations - Exercise 9.6

1. Is the class 12 maths chapter 9 Differential Equations (Ex 9.6) important

Class 12 maths chapter 9 Differential Equations (Ex 9.6) is important as it depends on the resolution of linear differential equations. A differential equation involving only the function y and its first derivative is known as a linear differential equation of the first order. These equations can be used to physically describe a variety of linear phenomena in physics, population dynamics, economics, and biology. Consequently, it is essential to comprehend the concept presented in this exercise.

2. What is a differential equation?

A differential equation is an equation that connects the derivatives of one or more unknown functions. Applications typically involve functions that represent physical quantities, derivatives that depict the rates at which those quantities change, and a differential equation that establishes a connection between the three.

3. What is the integrating factor?

A function used to solve differential equations is called an integrating factor. It is a function that can be made integrable by multiplying it by an ordinary differential equation. Ordinary differential equations are typically solved using this method. This factor is also applicable to multivariable calculus. An erroneous differential becomes an accurate differential when it is multiplied by an integrating factor (which can be later integrated to give a scalar field).

4. Why should one opt for the NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.6 by Vedantu?

When it comes to exam preparation, the NCERT solutions for Ex. 9.6 from Class 12 Maths from Vedantu is thought to be the best choice for CBSE students. There are numerous exercises in this chapter. On this page, in PDF format, we have the Exercise 9.6 Class 12 Maths NCERT solutions. This solution is available for download at your convenience, or you can access it directly from the Vedantu website or app to study it.

5. In Class 12 maths chapter 9 Differential Equations, how many different kinds of differential equations are there?

There are two types of differential equations and they are ordinary differential equations and partial differential equations. ODEs, or "ordinary differential equations," have a single independent variable. PDEs also referred to as "partial differential equations," have two or more independent variables.