NCERT Solutions · Class 12 · Chapter 9: Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4

Exercise 9.4: Homogeneous differential equations. If \(\tfrac{dy}{dx} = F\left(\tfrac{y}{x}\right)\) (every term of the same degree), put \(y = vx\), \(\tfrac{dy}{dx} = v + x\tfrac{dv}{dx}\); the equation separates in v and x. For \(\tfrac{dx}{dy} = h\left(\tfrac{x}{y}\right)\) put \(x = vy\). Replace v by \(\tfrac{y}{x}\) at the end.

17 questions
Every answer checked by computer algebra
Free, no sign-in

Try each question first, then open its solution. Our own step-by-step solutions, set out for step marks.

Exercise 9.4 questions and solutions

Exercise 9.4, Question 1

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

Show solution

\(\dfrac{dy}{dx} = \dfrac{1 + v^2}{1 + v}\) with \(y = vx\): \[\begin{aligned}x\dfrac{dv}{dx} &= \dfrac{1 + v^2}{1 + v} - v \\ &= \dfrac{1 - v}{1 + v}\end{aligned}\]
\(\dfrac{1 + v}{1 - v}\,dv = \dfrac{dx}{x}\); \[\dfrac{1 + v}{1 - v} = -1 + \dfrac{2}{1 - v}\]: \(-v - 2\log|1 - v| = \log|x| + C_1\).
With \(v = \tfrac{y}{x}\): \[-\dfrac{y}{x} - 2\log|x - y| + \log|x| = C_1\]

Answer: \((x - y)^2 = Cxe^{-y/x}\)

Practise this: Step 2, Board standard →

Exercise 9.4, Question 2

\(y' = \dfrac{x + y}{x}\)

Show solution

\[\begin{aligned}&v + xv' = 1 + v \\ \Rightarrow\ &dv = \dfrac{dx}{x}\end{aligned}\], so \(v = \log|x| + C\).

Answer: \(y = x\log|x| + Cx\)

Practise this: Step 2, Board standard →

Exercise 9.4, Question 3

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

Show solution

\(\dfrac{dy}{dx} = \dfrac{1 + v}{1 - v}\): \(x\dfrac{dv}{dx} = \dfrac{1 + v^2}{1 - v}\), so \[\dfrac{1 - v}{1 + v^2}\,dv = \dfrac{dx}{x}\]
\[\tan^{-1}v - \tfrac12\log(1 + v^2) = \log|x| + C\], i.e. \[\tan^{-1}\dfrac{y}{x} - \tfrac12\log\dfrac{x^2 + y^2}{x^2} - \log|x| = C\]

Answer: \[\tan^{-1}\dfrac{y}{x} = \tfrac12\log(x^2 + y^2) + C\]

Practise this: Step 2, Board standard →

Exercise 9.4, Question 4

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

Show solution

\(\dfrac{dy}{dx} = \dfrac{v^2 - 1}{2v}\): \(x\dfrac{dv}{dx} = -\dfrac{1 + v^2}{2v}\), so \(\dfrac{2v}{1 + v^2}\,dv = -\dfrac{dx}{x}\).
\[\begin{aligned}&\log(1 + v^2) = -\log|x| + \log C \\ \Rightarrow\ &x\left(1 + \dfrac{y^2}{x^2}\right) = C\end{aligned}\]

Answer: \(x^2 + y^2 = Cx\)

Practise this: Step 2, Board standard →

Exercise 9.4, Question 5

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

Show solution

\[\begin{aligned}&v + xv' = 1 - 2v^2 + v \\ \Rightarrow\ &\dfrac{dv}{1 - 2v^2} = \dfrac{dx}{x}\end{aligned}\]
\[\dfrac{1}{2\sqrt2}\log\left|\dfrac{1 + \sqrt2 v}{1 - \sqrt2 v}\right| = \log|x| + C\]

Answer: \[\dfrac{1}{2\sqrt2}\log\left|\dfrac{x + \sqrt2 y}{x - \sqrt2 y}\right| = \log|x| + C\]

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 6

\(x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx\)

Show solution

For \(x > 0\): \(\dfrac{dy}{dx} = v + \sqrt{1 + v^2}\), so \(x\dfrac{dv}{dx} = \sqrt{1 + v^2}\).
\[\begin{aligned}&\log\left|v + \sqrt{1 + v^2}\right| = \log|x| + \log C \\ \Rightarrow\ &\dfrac{y + \sqrt{x^2 + y^2}}{x} = Cx\end{aligned}\]

Answer: \(y + \sqrt{x^2 + y^2} = Cx^2\)

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 7

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

Show solution

\[\dfrac{dy}{dx} = \dfrac{v(\cos v + v\sin v)}{v\sin v - \cos v}\]; then \[x\dfrac{dv}{dx} = \dfrac{2v\cos v}{v\sin v - \cos v}\]
\[\begin{aligned}&\left(\tan v - \dfrac1v\right)dv = \dfrac{2\,dx}{x} \\ \Rightarrow\ &-\log|\cos v| - \log|v| = 2\log|x| + C_1\end{aligned}\]
So \(x^2v\cos v = C\), i.e. \(xy\cos\dfrac{y}{x} = C\).

Answer: \(xy\cos\dfrac{y}{x} = C\)

Practise this: Step 4, 95+ stretch (HOTS) →

Exercise 9.4, Question 8

\(x\dfrac{dy}{dx} - y + x\sin\dfrac{y}{x} = 0\)

Show solution

\[\begin{aligned}&v + xv' = v - \sin v \\ \Rightarrow\ &\csc v\,dv = -\dfrac{dx}{x}\end{aligned}\]
\[\log|\csc v - \cot v| = -\log|x| + \log C\], and \(\csc v - \cot v = \tan\dfrac{v}{2}\).

Answer: \(x\tan\dfrac{y}{2x} = C\)

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 9

\(y\,dx + x\log\dfrac{y}{x}\,dy - 2x\,dy = 0\)

Show solution

\(\dfrac{dy}{dx} = \dfrac{v}{2 - \log v}\): \[x\dfrac{dv}{dx} = \dfrac{v(\log v - 1)}{2 - \log v}\]
\(u = \log v - 1\): \[\begin{aligned}&\dfrac{1 - u}{u}\,du = \dfrac{dx}{x} \\ \Rightarrow\ &\log|u| - u = \log|x| + C_1\end{aligned}\], so \(\dfrac{u}{x} = C'e^u = C'\dfrac{v}{e}\), i.e. \(\log v - 1 = Cy\).

Answer: \(Cy = \log\left|\dfrac{y}{x}\right| - 1\)

Practise this: Step 4, 95+ stretch (HOTS) →

Exercise 9.4, Question 10

\(\left(1 + e^{x/y}\right)dx + e^{x/y}\left(1 - \dfrac{x}{y}\right)dy = 0\)

Show solution

Put \(x = vy\): \[\dfrac{dx}{dy} = -\dfrac{e^v(1 - v)}{1 + e^v}\], and \(v + y\dfrac{dv}{dy}\) equals that, so \[y\dfrac{dv}{dy} = -\dfrac{v + e^v}{1 + e^v}\]
\[\begin{aligned}&\dfrac{1 + e^v}{v + e^v}\,dv = -\dfrac{dy}{y} \\ \Rightarrow\ &\log|v + e^v| = -\log|y| + \log C\end{aligned}\]

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

Practise this: Step 4, 95+ stretch (HOTS) →

Exercise 9.4, Question 11

\((x + y)\,dy + (x - y)\,dx = 0\); \(y = 1\) when \(x = 1\)

Show solution

\(\dfrac{dy}{dx} = \dfrac{v - 1}{1 + v}\): \[x\dfrac{dv}{dx} = -\dfrac{1 + v^2}{1 + v}\], so \[\dfrac{1 + v}{1 + v^2}\,dv = -\dfrac{dx}{x}\]
\[\tan^{-1}v + \tfrac12\log(1 + v^2) = -\log|x| + C\], i.e. \[\tan^{-1}\dfrac{y}{x} + \tfrac12\log(x^2 + y^2) = C\]; at \((1, 1)\): \(C = \tfrac{\pi}{4} + \tfrac12\log 2\).

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

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 12

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

Show solution

\[\begin{aligned}&v + xv' = -(v + v^2) \\ \Rightarrow\ &\dfrac{dv}{v(v + 2)} = -\dfrac{dx}{x}\end{aligned}\], so \[\tfrac12\log\left|\dfrac{v}{v + 2}\right| = -\log|x| + C\], i.e. \(\dfrac{y}{y + 2x} = \dfrac{K}{x^2}\).
At \((1, 1)\): \(K = \tfrac13\), so \(3x^2y = y + 2x\).

Answer: \(y + 2x = 3x^2y\), i.e. \(y = \dfrac{2x}{3x^2 - 1}\)

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 13

\(\left[x\sin^2\dfrac{y}{x} - y\right]dx + x\,dy = 0\); \(y = \tfrac{\pi}{4}\) when \(x = 1\)

Show solution

\[\begin{aligned}&v + xv' = v - \sin^2 v \\ \Rightarrow\ &\csc^2 v\,dv = -\dfrac{dx}{x}\end{aligned}\], so \(\cot v = \log|x| + C\).
At \((1, \tfrac{\pi}{4})\): \(C = 1\): \[\begin{aligned}\cot\dfrac{y}{x} &= \log|x| + 1 \\ &= \log|ex|\end{aligned}\]

Answer: \(\cot\dfrac{y}{x} = \log|ex|\)

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 14

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

Show solution

\[\begin{aligned}&xv' = -\csc v \\ \Rightarrow\ &\sin v\,dv = -\dfrac{dx}{x}\end{aligned}\], so \(\cos v = \log|x| + C\).
At \((1, 0)\): \(C = 1\).

Answer: \(\cos\dfrac{y}{x} = \log|ex|\)

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 15

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

Show solution

\[\begin{aligned}&v + xv' = v + \dfrac{v^2}{2} \\ \Rightarrow\ &\dfrac{2\,dv}{v^2} = \dfrac{dx}{x}\end{aligned}\], so \(-\dfrac{2}{v} = \log|x| + C\).
At \((1, 2)\): \(C = -1\): \(-\dfrac{2x}{y} = \log|x| - 1\).

Answer: \(y = \dfrac{2x}{1 - \log|x|}\) (\(x \ne 0, e\))

Practise this: Step 3, Full marks on long answers →

Exercise 9.4, Question 16

A homogeneous equation \(\dfrac{dx}{dy} = h\left(\dfrac{x}{y}\right)\) is solved by the substitution: (A) \(y = vx\) (B) \(v = yx\) (C) \(x = vy\) (D) \(x = v\)

Show solution

x is the dependent variable, a function of \(\tfrac{x}{y}\): put \(x = vy\).

Answer: (C)

Practise this: Step 1, Secure the basics →

Exercise 9.4, Question 17

Which is homogeneous? (A) \((4x + 6y + 5)\,dy - (3y + 2x + 4)\,dx = 0\) (B) \(xy\,dx - (x^3 + y^3)\,dy = 0\) (C) \((x^3 + 2y^2)\,dx + 2xy\,dy = 0\) (D) \(y^2\,dx + (x^2 - xy - y^2)\,dy = 0\)

Show solution

In (D) every coefficient has degree 2: \[\dfrac{dy}{dx} = \dfrac{y^2}{xy + y^2 - x^2}\] depends only on \(\tfrac{y}{x}\). (A) has constants, (B) and (C) mix degrees 2 and 3.

Answer: (D)

Practise this: Step 1, Secure the basics →

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.

Differential Equations: notes and questions Create a free account Teachers: set it as homework

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.