NCERT Solutions · Class 12 · Chapter 9: Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

Exercise 9.5: Linear differential equations. For \(\tfrac{dy}{dx} + Py = Q\) (P, Q functions of x): integrating factor \(\text{IF} = e^{\int P\,dx}\), and \(y \cdot \text{IF} = \int Q \cdot \text{IF}\,dx + C\). If the equation is linear in x instead, use \(\tfrac{dx}{dy} + P_1x = Q_1\) with \(\text{IF} = e^{\int P_1\,dy}\).

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Exercise 9.5 questions and solutions

Exercise 9.5, Question 1

\(\dfrac{dy}{dx} + 2y = \sin x\)

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IF \(= e^{2x}\): \[\begin{aligned}ye^{2x} &= \int e^{2x}\sin x\,dx \\ &= \dfrac{e^{2x}}{5}(2\sin x - \cos x) + C\end{aligned}\]

Answer: \[y = \tfrac15(2\sin x - \cos x) + Ce^{-2x}\]

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Exercise 9.5, Question 2

\(\dfrac{dy}{dx} + 3y = e^{-2x}\)

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IF \(= e^{3x}\): \(ye^{3x} = \int e^x\,dx = e^x + C\).

Answer: \(y = e^{-2x} + Ce^{-3x}\)

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Exercise 9.5, Question 3

\(\dfrac{dy}{dx} + \dfrac{y}{x} = x^2\)

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IF \(= e^{\log x} = x\): \(xy = \int x^3\,dx = \dfrac{x^4}{4} + C\).

Answer: \(y = \dfrac{x^3}{4} + \dfrac{C}{x}\)

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Exercise 9.5, Question 4

\(\dfrac{dy}{dx} + (\sec x)y = \tan x\) (\(0 \le x < \tfrac{\pi}{2}\))

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IF \[= e^{\log(\sec x + \tan x)} = \sec x + \tan x\]
\[\begin{aligned}y(\sec x + \tan x) &= \int(\sec x\tan x + \tan^2 x)\,dx \\ &= \sec x + \tan x - x + C\end{aligned}\]

Answer: \[y(\sec x + \tan x) = \sec x + \tan x - x + C\]

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Exercise 9.5, Question 5

\(\cos^2 x\dfrac{dy}{dx} + y = \tan x\) (\(0 \le x < \tfrac{\pi}{2}\))

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\[\dfrac{dy}{dx} + \sec^2 x\,y = \tan x\sec^2 x\]; IF \(= e^{\tan x}\).
\[\begin{aligned}ye^{\tan x} &= \int\tan x\sec^2 x\,e^{\tan x}\,dx \\ &= (\tan x - 1)e^{\tan x} + C\end{aligned}\] (put \(t = \tan x\), then parts).

Answer: \(y = \tan x - 1 + Ce^{-\tan x}\)

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Exercise 9.5, Question 6

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

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\(\dfrac{dy}{dx} + \dfrac2x y = x\log x\); IF \(= x^2\).
\[\begin{aligned}x^2y &= \int x^3\log x\,dx \\ &= \dfrac{x^4}{4}\log x - \dfrac{x^4}{16} + C\end{aligned}\]

Answer: \[y = \dfrac{x^2}{16}(4\log x - 1) + \dfrac{C}{x^2}\]

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Exercise 9.5, Question 7

\(x\log x\dfrac{dy}{dx} + y = \dfrac2x\log x\)

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\[\dfrac{dy}{dx} + \dfrac{y}{x\log x} = \dfrac{2}{x^2}\]; IF \(= e^{\log(\log x)} = \log x\).
\[\begin{aligned}y\log x &= \int\dfrac{2\log x}{x^2}\,dx \\ &= -\dfrac{2\log x}{x} - \dfrac2x + C\end{aligned}\] (parts).

Answer: \(y\log x = -\dfrac2x(1 + \log x) + C\)

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Exercise 9.5, Question 8

\((1 + x^2)\,dy + 2xy\,dx = \cot x\,dx\) (\(x \ne 0\))

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\[\dfrac{dy}{dx} + \dfrac{2x}{1 + x^2}y = \dfrac{\cot x}{1 + x^2}\]; IF \(= 1 + x^2\).
\[\begin{aligned}(1 + x^2)y &= \int\cot x\,dx \\ &= \log|\sin x| + C\end{aligned}\]

Answer: \(y = \dfrac{\log|\sin x| + C}{1 + x^2}\)

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Exercise 9.5, Question 9

\(x\dfrac{dy}{dx} + y - x + xy\cot x = 0\) (\(x \ne 0\))

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\[\dfrac{dy}{dx} + \left(\dfrac1x + \cot x\right)y = 1\]; IF \(= e^{\log x + \log\sin x} = x\sin x\).
\[\begin{aligned}xy\sin x &= \int x\sin x\,dx \\ &= -x\cos x + \sin x + C\end{aligned}\]

Answer: \[y = \dfrac1x - \cot x + \dfrac{C}{x\sin x}\]

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Exercise 9.5, Question 10

\((x + y)\dfrac{dy}{dx} = 1\)

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\(\dfrac{dx}{dy} = x + y\), i.e. \(\dfrac{dx}{dy} - x = y\): linear in x, IF \(= e^{-y}\).
\[\begin{aligned}xe^{-y} &= \int ye^{-y}\,dy \\ &= -(y + 1)e^{-y} + C\end{aligned}\]

Answer: \(x + y + 1 = Ce^y\)

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Exercise 9.5, Question 11

\(y\,dx + (x - y^2)\,dy = 0\)

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\(\dfrac{dx}{dy} + \dfrac{x}{y} = y\); IF \(= y\): \(xy = \int y^2\,dy = \dfrac{y^3}{3} + C\).

Answer: \(x = \dfrac{y^2}{3} + \dfrac{C}{y}\)

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Exercise 9.5, Question 12

\((x + 3y^2)\dfrac{dy}{dx} = y\) (\(y > 0\))

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\(\dfrac{dx}{dy} - \dfrac{x}{y} = 3y\); IF \(= \dfrac1y\): \(\dfrac{x}{y} = \int 3\,dy = 3y + C\).

Answer: \(x = 3y^2 + Cy\)

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Exercise 9.5, Question 13

\(\dfrac{dy}{dx} + 2y\tan x = \sin x\); \(y = 0\) when \(x = \tfrac{\pi}{3}\)

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IF \(= e^{2\log\sec x} = \sec^2 x\): \[\begin{aligned}y\sec^2 x &= \int\sin x\sec^2 x\,dx \\ &= \int\sec x\tan x\,dx \\ &= \sec x + C\end{aligned}\]
\(y = \cos x + C\cos^2 x\); at \(\tfrac{\pi}{3}\): \(0 = \tfrac12 + \tfrac{C}{4}\), \(C = -2\).

Answer: \(y = \cos x - 2\cos^2 x\)

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Exercise 9.5, Question 14

\((1 + x^2)\dfrac{dy}{dx} + 2xy = \dfrac{1}{1 + x^2}\); \(y = 0\) when \(x = 1\)

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The left side is \(\dfrac{d}{dx}[(1 + x^2)y]\): \((1 + x^2)y = \tan^{-1}x + C\).
At \(x = 1\): \(0 = \tfrac{\pi}{4} + C\).

Answer: \((1 + x^2)y = \tan^{-1}x - \dfrac{\pi}{4}\)

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Exercise 9.5, Question 15

\(\dfrac{dy}{dx} - 3y\cot x = \sin 2x\); \(y = 2\) when \(x = \tfrac{\pi}{2}\)

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IF \(= e^{-3\log\sin x} = \dfrac{1}{\sin^3 x}\): \[\begin{aligned}\dfrac{y}{\sin^3 x} &= \int\dfrac{2\cos x}{\sin^2 x}\,dx \\ &= -\dfrac{2}{\sin x} + C\end{aligned}\]
\(y = -2\sin^2 x + C\sin^3 x\); at \(\tfrac{\pi}{2}\): \(2 = -2 + C\).

Answer: \(y = 4\sin^3 x - 2\sin^2 x\)

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Exercise 9.5, Question 16

A curve through the origin has slope equal to the sum of the coordinates at every point. Find it.

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\(\dfrac{dy}{dx} = x + y\), i.e. \(\dfrac{dy}{dx} - y = x\); IF \(= e^{-x}\).
\[\begin{aligned}ye^{-x} &= \int xe^{-x}\,dx \\ &= -(x + 1)e^{-x} + C\end{aligned}\]; at \((0, 0)\): \(C = 1\).

Answer: \(y = e^x - x - 1\) (i.e. \(x + y + 1 = e^x\))

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Exercise 9.5, Question 17

A curve through \((0, 2)\): at every point the sum of the coordinates exceeds the slope of the tangent by 5. Find it.

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\(x + y - 5 = \dfrac{dy}{dx}\), i.e. \(\dfrac{dy}{dx} - y = x - 5\); IF \(= e^{-x}\).
\[\begin{aligned}ye^{-x} &= \int(x - 5)e^{-x}\,dx \\ &= -(x - 4)e^{-x} + C\end{aligned}\], so \(y = 4 - x + Ce^x\); at \((0, 2)\): \(C = -2\).

Answer: \(y = 4 - x - 2e^x\)

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Exercise 9.5, Question 18

The integrating factor of \(x\dfrac{dy}{dx} - y = 2x^2\) is: (A) \(e^{-x}\) (B) \(e^{-y}\) (C) \(\dfrac1x\) (D) x

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\(\dfrac{dy}{dx} - \dfrac1x y = 2x\): IF \(= e^{-\log x} = \dfrac1x\).

Answer: (C)

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Exercise 9.5, Question 19

The integrating factor of \((1 - y^2)\dfrac{dx}{dy} + yx = ay\) (\(-1 < y < 1\)) is: (A) \(\dfrac{1}{y^2 - 1}\) (B) \(\dfrac{1}{\sqrt{y^2 - 1}}\) (C) \(\dfrac{1}{1 - y^2}\) (D) \(\dfrac{1}{\sqrt{1 - y^2}}\)

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\[\dfrac{dx}{dy} + \dfrac{y}{1 - y^2}x = \dfrac{ay}{1 - y^2}\]: IF \[= e^{\int\frac{y}{1 - y^2}dy} = e^{-\frac12\log(1 - y^2)}\]

Answer: (D)

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Done the NCERT exercises? The board paper asks more

Differential Equations has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

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Differential Equations in our sample papers: Sample paper 1 (question 28) · Sample paper 2 (questions 15, 28, 37) · Sample paper 3 (questions 12, 29) · Sample paper 4 (questions 15, 28) · Sample paper 5 (questions 12, 34).

Also useful: free MCQs and case studies for Differential Equations · formulas for this chapter (Class 12 formula sheet, free PDF) · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

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