NCERT Solutions · Class 12 · Chapter 9: Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Miscellaneous Exercise

The Miscellaneous Exercise on Differential Equations. Mixed practice: identify the type first (separable, homogeneous, linear in y or in x), then use the matching method; for 'verify' questions, differentiate and substitute.

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Miscellaneous Exercise questions and solutions

Miscellaneous Exercise, Question 1

Order and degree (if defined):

(i) \(\dfrac{d^2y}{dx^2} + 5x\left(\dfrac{dy}{dx}\right)^2 - 6y = \log x\)

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Highest derivative \(y''\), first power.

Answer: Order 2; degree 1

(ii) \(\left(\dfrac{dy}{dx}\right)^3 - 4\left(\dfrac{dy}{dx}\right)^2 + 7y = \sin x\)

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Only \(y'\), highest power 3.

Answer: Order 1; degree 3

(iii) \(\dfrac{d^4y}{dx^4} - \sin\left(\dfrac{d^3y}{dx^3}\right) = 0\)

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\(y'''\) inside sine.

Answer: Order 4; degree not defined

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Miscellaneous Exercise, Question 2

Verify that the function is a solution:

(i) \(xy = ae^x + be^{-x} + x^2\) : \(x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} - xy + x^2 - 2 = 0\)

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Differentiate \(xy\) twice: \((xy)'' = xy'' + 2y' = ae^x + be^{-x} + 2\).
\(ae^x + be^{-x} = xy - x^2\), so \(xy'' + 2y' = xy - x^2 + 2\).

Answer: Verified

(ii) \(y = e^x(a\cos x + b\sin x)\) : \(y'' - 2y' + 2y = 0\)

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\(y' = y + e^x(-a\sin x + b\cos x)\), \(y'' = 2y' - 2y\) after expanding.

Answer: Verified

(iii) \(y = x\sin 3x\) : \(y'' + 9y - 6\cos 3x = 0\)

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\(y' = \sin 3x + 3x\cos 3x\), \(y'' = 6\cos 3x - 9x\sin 3x\).

Answer: Verified

(iv) \(x^2 = 2y^2\log y\) : \((x^2 + y^2)\dfrac{dy}{dx} - xy = 0\)

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\(2x = (4y\log y + 2y)y'\), so \(y' = \dfrac{x}{y(2\log y + 1)}\).
\(2y^2\log y = x^2\) gives \(y^2(2\log y + 1) = x^2 + y^2\), so \(y' = \dfrac{xy}{x^2 + y^2}\).

Answer: Verified

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Miscellaneous Exercise, Question 3

Prove \(x^2 - y^2 = c(x^2 + y^2)^2\) is the general solution of \((x^3 - 3xy^2)\,dx = (y^3 - 3x^2y)\,dy\).

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Homogeneous: \[\dfrac{dy}{dx} = \dfrac{1 - 3v^2}{v^3 - 3v}\] with \(y = vx\); \[x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 - 3v}\], so \[\dfrac{v^3 - 3v}{1 - v^4}\,dv = \dfrac{dx}{x}\]
Partial fractions: \[\dfrac{v^3 - 3v}{(1 - v^2)(1 + v^2)} = \dfrac{-v}{1 - v^2} - \dfrac{2v}{1 + v^2}\] (check: \(-v(1 + v^2) - 2v(1 - v^2) = v^3 - 3v\)).
Integrating: \[\tfrac12\log|1 - v^2| - \log(1 + v^2) = \log|x| + \log k\]; doubling, \[\log\dfrac{|1 - v^2|}{(1 + v^2)^2} = \log(k^2x^2)\], so \(\dfrac{1 - v^2}{(1 + v^2)^2} = cx^2\).
Multiply out with \(v = \tfrac{y}{x}\): \[\dfrac{x^2(x^2 - y^2)}{(x^2 + y^2)^2} = cx^2\]

Answer: Proved

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Miscellaneous Exercise, Question 4

\(\dfrac{dy}{dx} + \sqrt{\dfrac{1 - y^2}{1 - x^2}} = 0\)

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\[\dfrac{dy}{\sqrt{1 - y^2}} = -\dfrac{dx}{\sqrt{1 - x^2}}\]

Answer: \(\sin^{-1}y + \sin^{-1}x = C\)

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Miscellaneous Exercise, Question 5

Show the general solution of \(\dfrac{dy}{dx} + \dfrac{y^2 + y + 1}{x^2 + x + 1} = 0\) is \(x + y + 1 = A(1 - x - y - 2xy)\).

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Separate: \[\dfrac{dy}{y^2 + y + 1} + \dfrac{dx}{x^2 + x + 1} = 0\], giving \[\tan^{-1}\dfrac{2y + 1}{\sqrt3} + \tan^{-1}\dfrac{2x + 1}{\sqrt3} = C\] (after the factor \(\tfrac{2}{\sqrt3}\)).
Take tan of both sides: \[\dfrac{\frac{2x + 1}{\sqrt3} + \frac{2y + 1}{\sqrt3}}{1 - \frac{(2x + 1)(2y + 1)}{3}} = \tan C\], i.e. \[\dfrac{2\sqrt3(x + y + 1)}{2(1 - x - y - 2xy)} = \tan C\]
So \(x + y + 1 = A(1 - x - y - 2xy)\) with \(A = \tfrac{\tan C}{\sqrt3}\).

Answer: Shown

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Miscellaneous Exercise, Question 6

The curve through \(\left(0, \tfrac{\pi}{4}\right)\) with \(\sin x\cos y\,dx + \cos x\sin y\,dy = 0\).

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Divide by \(\cos x\cos y\): \[\begin{aligned}&\tan x\,dx + \tan y\,dy = 0 \\ \Rightarrow\ &\log|\sec x| + \log|\sec y| = \log C\end{aligned}\], so \(\cos x\cos y = k\).
At \(\left(0, \tfrac{\pi}{4}\right)\): \(k = \tfrac{1}{\sqrt2}\).

Answer: \(\cos y = \dfrac{\sec x}{\sqrt2}\)

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Miscellaneous Exercise, Question 7

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

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\[\dfrac{dy}{1 + y^2} = -\dfrac{e^x\,dx}{1 + e^{2x}}\]: \(\tan^{-1}y = -\tan^{-1}e^x + C\).
At \((0, 1)\): \(\tfrac{\pi}{4} = -\tfrac{\pi}{4} + C\).

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

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Miscellaneous Exercise, Question 8

Solve \(ye^{x/y}\,dx = \left(xe^{x/y} + y^2\right)dy\) (\(y \ne 0\)).

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\[\dfrac{dx}{dy} = \dfrac{x}{y} + ye^{-x/y}\] Put \(x = vy\): \(v + y\dfrac{dv}{dy} = v + ye^{-v}\), so \(e^v\,dv = dy\).
\(e^v = y + C\).

Answer: \(e^{x/y} = y + C\)

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Miscellaneous Exercise, Question 9

\((x - y)(dx + dy) = dx - dy\); \(y = -1\) when \(x = 0\) (hint: put \(x - y = t\))

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\(t = x - y\): \(dy = dx - dt\), so \(dx + dy = 2\,dx - dt\) and the equation is \(t(2\,dx - dt) = dt\), i.e. \(2\,dx = \left(1 + \dfrac1t\right)dt\).
\[\begin{aligned}&2x = t + \log|t| + C \\ \Rightarrow\ &x + y = \log|x - y| + C\end{aligned}\]; at \((0, -1)\): \(C = -1\).

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

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Miscellaneous Exercise, Question 10

Solve \(\left[\dfrac{e^{-2\sqrt x}}{\sqrt x} - \dfrac{y}{\sqrt x}\right]\dfrac{dx}{dy} = 1\) (\(x \ne 0\)).

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\[\dfrac{dy}{dx} + \dfrac{y}{\sqrt x} = \dfrac{e^{-2\sqrt x}}{\sqrt x}\]: linear, IF \(= e^{\int x^{-1/2}dx} = e^{2\sqrt x}\).
\[\begin{aligned}ye^{2\sqrt x} &= \int\dfrac{1}{\sqrt x}\,dx \\ &= 2\sqrt x + C\end{aligned}\]

Answer: \(ye^{2\sqrt x} = 2\sqrt x + C\)

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Miscellaneous Exercise, Question 11

\(\dfrac{dy}{dx} + y\cot x = 4x\csc x\) (\(x \ne 0\)); \(y = 0\) when \(x = \tfrac{\pi}{2}\)

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IF \(= \sin x\): \(y\sin x = \int 4x\,dx = 2x^2 + C\).
At \(\tfrac{\pi}{2}\): \(0 = \tfrac{\pi^2}{2} + C\).

Answer: \(y\sin x = 2x^2 - \dfrac{\pi^2}{2}\)

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Miscellaneous Exercise, Question 12

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

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\[\begin{aligned}&\dfrac{dy}{2e^{-y} - 1} = \dfrac{e^y\,dy}{2 - e^y} = \dfrac{dx}{x + 1} \\ \Rightarrow\ &-\log|2 - e^y| = \log|x + 1| + C\end{aligned}\]
So \((2 - e^y)(x + 1) = k\); at \((0, 0)\): \(k = 1\), giving \[\begin{aligned}e^y &= 2 - \dfrac{1}{x + 1} \\ &= \dfrac{2x + 1}{x + 1}\end{aligned}\]

Answer: \[y = \log\left|\dfrac{2x + 1}{x + 1}\right|\] (\(x \ne -1\))

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Miscellaneous Exercise, Question 13

The general solution of \(\dfrac{y\,dx - x\,dy}{y} = 0\) is: (A) \(xy = C\) (B) \(x = Cy^2\) (C) \(y = Cx\) (D) \(y = Cx^2\)

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\[\begin{aligned}&y\,dx = x\,dy \\ \Rightarrow\ &\dfrac{dx}{x} = \dfrac{dy}{y}\end{aligned}\], so \(y = Cx\).

Answer: (C)

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Miscellaneous Exercise, Question 14

The general solution of \(\dfrac{dx}{dy} + P_1x = Q_1\) is: (A) \(ye^{\int P_1dy} = \int Q_1e^{\int P_1dy}\,dy + C\) (B) \(ye^{\int P_1dx} = \int Q_1e^{\int P_1dx}\,dx + C\) (C) \(xe^{\int P_1dy} = \int Q_1e^{\int P_1dy}\,dy + C\) (D) \(xe^{\int P_1dx} = \int Q_1e^{\int P_1dx}\,dx + C\)

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Linear in x with y independent: IF \(= e^{\int P_1\,dy}\), multiplying x.

Answer: (C)

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Miscellaneous Exercise, Question 15

The general solution of \(e^x\,dy + (ye^x + 2x)\,dx = 0\) is: (A) \(xe^y + x^2 = C\) (B) \(xe^y + y^2 = C\) (C) \(ye^x + x^2 = C\) (D) \(ye^y + x^2 = C\)

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\(e^x\,dy + ye^x\,dx = d(ye^x)\), so \(d(ye^x) + 2x\,dx = 0\).

Answer: (C)

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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

Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.