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NCERT Solutions · Class 12 · Chapter 9: Differential Equations
NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3
Exercise 9.3: Variables separable. If \(\tfrac{dy}{dx} = g(x)h(y)\), write \(\tfrac{dy}{h(y)} = g(x)\,dx\) and integrate both sides (one constant is enough). For a particular solution, put the given point in to find C. Growth problems: \(\tfrac{dP}{dt} = kP\) gives \(P = P_0e^{kt}\).
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Exercise 9.3 questions and solutions
Exercise 9.3, Question 1
\(\dfrac{dy}{dx} = \dfrac{1 - \cos x}{1 + \cos x}\)
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\[\begin{aligned}\dfrac{1 - \cos x}{1 + \cos x} &= \tan^2\dfrac{x}{2} \\ &= \sec^2\dfrac{x}{2} - 1\end{aligned}\] \[y = \int\left(\sec^2\dfrac{x}{2} - 1\right)dx\] Answer: \(y = 2\tan\dfrac{x}{2} - x + C\)
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Exercise 9.3, Question 2
\(\dfrac{dy}{dx} = \sqrt{4 - y^2}\) (\(-2 < y < 2\))
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\[\begin{aligned}&\dfrac{dy}{\sqrt{4 - y^2}} = dx \\ \Rightarrow\ &\sin^{-1}\dfrac{y}{2} = x + C\end{aligned}\] Answer: \(y = 2\sin(x + C)\)
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Exercise 9.3, Question 3
\(\dfrac{dy}{dx} + y = 1\) (\(y \ne 1\))
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\[\begin{aligned}&\dfrac{dy}{1 - y} = dx \\ \Rightarrow\ &-\log|1 - y| = x + C_1\end{aligned}\], so \(1 - y = \pm e^{-C_1}e^{-x}\). Answer: \(y = 1 + Ae^{-x}\)
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Exercise 9.3, Question 4
\(\sec^2 x\tan y\,dx + \sec^2 y\tan x\,dy = 0\)
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Divide by \(\tan x\tan y\): \[\dfrac{\sec^2 x}{\tan x}\,dx + \dfrac{\sec^2 y}{\tan y}\,dy = 0\] \(\log|\tan x| + \log|\tan y| = \log|C|\). Answer: \(\tan x\tan y = C\)
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Exercise 9.3, Question 5
\((e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0\)
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\[dy = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}\,dx\]: the top is the derivative of the bottom. Answer: \(y = \log(e^x + e^{-x}) + C\)
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Exercise 9.3, Question 6
\(\dfrac{dy}{dx} = (1 + x^2)(1 + y^2)\)
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\(\dfrac{dy}{1 + y^2} = (1 + x^2)\,dx\). Answer: \(\tan^{-1}y = x + \dfrac{x^3}{3} + C\)
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Exercise 9.3, Question 7
\(y\log y\,dx - x\,dy = 0\)
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\[\begin{aligned}&\dfrac{dy}{y\log y} = \dfrac{dx}{x} \\ \Rightarrow\ &\log|\log y| = \log|x| + \log|C|\end{aligned}\], so \(\log y = Cx\). Answer: \(y = e^{Cx}\)
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Exercise 9.3, Question 8
\(x^5\dfrac{dy}{dx} = -y^5\)
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\[\begin{aligned}&y^{-5}\,dy = -x^{-5}\,dx \\ \Rightarrow\ &-\dfrac{y^{-4}}{4} = \dfrac{x^{-4}}{4} + C_1\end{aligned}\] Answer: \(x^{-4} + y^{-4} = C\)
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Exercise 9.3, Question 9
\(\dfrac{dy}{dx} = \sin^{-1}x\)
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\(y = \int\sin^{-1}x\,dx\): parts with \(v = 1\): \[x\sin^{-1}x - \int\dfrac{x}{\sqrt{1 - x^2}}\,dx\] Answer: \(y = x\sin^{-1}x + \sqrt{1 - x^2} + C\)
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Exercise 9.3, Question 10
\(e^x\tan y\,dx + (1 - e^x)\sec^2 y\,dy = 0\)
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\[\begin{aligned}&\dfrac{\sec^2 y}{\tan y}\,dy = \dfrac{e^x}{e^x - 1}\,dx \\ \Rightarrow\ &\log|\tan y| = \log|e^x - 1| + \log|C|\end{aligned}\] Answer: \(\tan y = C(1 - e^x)\)
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Exercise 9.3, Question 11
\((x^3 + x^2 + x + 1)\dfrac{dy}{dx} = 2x^2 + x\); \(y = 1\) when \(x = 0\)
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\(x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1)\); \[\dfrac{2x^2 + x}{(x + 1)(x^2 + 1)} = \dfrac{1/2}{x + 1} + \dfrac{\frac32x - \frac12}{x^2 + 1}\] \[y = \tfrac12\log|x + 1| + \tfrac34\log(x^2 + 1) - \tfrac12\tan^{-1}x + C\]; \(y(0) = 1\) gives \(C = 1\). Answer: \[y = \tfrac14\log\left[(x + 1)^2(x^2 + 1)^3\right] - \tfrac12\tan^{-1}x + 1\]
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Exercise 9.3, Question 12
\(x(x^2 - 1)\dfrac{dy}{dx} = 1\); \(y = 0\) when \(x = 2\)
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\[\dfrac{1}{x(x - 1)(x + 1)} = -\dfrac1x + \dfrac{1/2}{x - 1} + \dfrac{1/2}{x + 1}\], so \[y = \tfrac12\log\left|\dfrac{x^2 - 1}{x^2}\right| + C\] \(y(2) = \tfrac12\log\tfrac34 + C = 0\). Answer: \[y = \tfrac12\log\left(\dfrac{x^2 - 1}{x^2}\right) - \tfrac12\log\tfrac34\]
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Exercise 9.3, Question 13
\(\cos\left(\dfrac{dy}{dx}\right) = a\) (\(a \in \mathbb R\)); \(y = 1\) when \(x = 0\)
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\(\dfrac{dy}{dx} = \cos^{-1}a\), a constant: \(y = x\cos^{-1}a + C\), and \(C = 1\). So \(\dfrac{y - 1}{x} = \cos^{-1}a\). Answer: \(\cos\left(\dfrac{y - 1}{x}\right) = a\) (i.e. \(y = x\cos^{-1}a + 1\))
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Exercise 9.3, Question 14
\(\dfrac{dy}{dx} = y\tan x\); \(y = 1\) when \(x = 0\)
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\[\begin{aligned}&\dfrac{dy}{y} = \tan x\,dx \\ \Rightarrow\ &\log|y| = \log|\sec x| + \log|C|\end{aligned}\], so \(y = C\sec x\); \(C = 1\). Answer: \(y = \sec x\)
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Exercise 9.3, Question 15
Curve through \((0, 0)\) with \(y' = e^x\sin x\).
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\[\begin{aligned}y &= \int e^x\sin x\,dx \\ &= \tfrac12e^x(\sin x - \cos x) + C\end{aligned}\] At \((0, 0)\): \(0 = -\tfrac12 + C\). Answer: \(2y - 1 = e^x(\sin x - \cos x)\)
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Exercise 9.3, Question 16
\(xy\dfrac{dy}{dx} = (x + 2)(y + 2)\): the solution curve through \((1, -1)\).
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\[\dfrac{y}{y + 2}\,dy = \dfrac{x + 2}{x}\,dx\], i.e. \[\left(1 - \dfrac{2}{y + 2}\right)dy = \left(1 + \dfrac2x\right)dx\] \(y - 2\log|y + 2| = x + 2\log|x| + C\); at \((1, -1)\): \(-1 = 1 + C\), \(C = -2\). Answer: \[y - x + 2 = \log\left[x^2(y + 2)^2\right]\]
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Exercise 9.3, Question 17
Curve through \((0, -2)\) on which (slope of tangent) × (y-coordinate) = x-coordinate.
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\[\begin{aligned}&y\dfrac{dy}{dx} = x \\ \Rightarrow\ &y\,dy = x\,dx \\ \Rightarrow\ &y^2 - x^2 = C\end{aligned}\] At \((0, -2)\): \(C = 4\). Answer: \(y^2 - x^2 = 4\)
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Exercise 9.3, Question 18
The slope of the tangent at \((x, y)\) is twice the slope of the line to \((-4, -3)\); the curve passes through \((-2, 1)\).
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\[\begin{aligned}&\dfrac{dy}{dx} = \dfrac{2(y + 3)}{x + 4} \\ \Rightarrow\ &\dfrac{dy}{y + 3} = \dfrac{2\,dx}{x + 4}\end{aligned}\], so \(y + 3 = C(x + 4)^2\). At \((-2, 1)\): \(4 = 4C\). Answer: \((x + 4)^2 = y + 3\)
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Exercise 9.3, Question 19
A spherical balloon's volume changes at a constant rate. Its radius is 3 units at first and 6 units after 3 s. Find the radius after t seconds.
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\[\begin{aligned}&\dfrac{dV}{dt} = k \\ \Rightarrow\ &\tfrac43\pi r^3 = kt + C\end{aligned}\] \(t = 0, r = 3\): \(C = 36\pi\). \(t = 3, r = 6\): \(288\pi = 3k + 36\pi\), \(k = 84\pi\). \[\begin{aligned}&\tfrac43\pi r^3 = 84\pi t + 36\pi \\ \Rightarrow\ &r^3 = 63t + 27\end{aligned}\] Answer: \(r = (63t + 27)^{1/3}\)
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Exercise 9.3, Question 20
Principal grows continuously at r% a year. Rs 100 doubles in 10 years; find r (use \(\log_e 2 = 0.6931\)).
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\[\begin{aligned}&\dfrac{dP}{dt} = \dfrac{r}{100}P \\ \Rightarrow\ &P = 100e^{rt/100}\end{aligned}\] \[\begin{aligned}&200 = 100e^{10r/100} \\ \Rightarrow\ &\dfrac{r}{10} = \log 2 = 0.6931\end{aligned}\] Answer: \(r = 6.931\) (% per year)
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Exercise 9.3, Question 21
Principal grows continuously at 5% a year. What is Rs 1000 worth after 10 years? (use \(e^{0.5} = 1.648\))
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\(P = 1000e^{0.05t}\); at \(t = 10\): \(1000e^{0.5} = 1000 \times 1.648\). Answer: Rs 1648
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Exercise 9.3, Question 22
A bacteria count of 1,00,000 grows by 10% in 2 hours, at a rate proportional to the count. When will it reach 2,00,000?
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\(N = N_0e^{kt}\). After 2 h: \(1.1 = e^{2k}\), so \(k = \tfrac12\log\tfrac{11}{10}\). \[\begin{aligned}&2 = e^{kt} \\ \Rightarrow\ &t = \dfrac{\log 2}{k}\end{aligned}\] Answer: \(t = \dfrac{2\log 2}{\log\frac{11}{10}}\) hours
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Exercise 9.3, Question 23
The general solution of \(\dfrac{dy}{dx} = e^{x + y}\) is: (A) \(e^x + e^{-y} = C\) (B) \(e^x + e^y = C\) (C) \(e^{-x} + e^y = C\) (D) \(e^{-x} + e^{-y} = C\)
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\[\begin{aligned}&e^{-y}\,dy = e^x\,dx \\ \Rightarrow\ &-e^{-y} = e^x + C_1\end{aligned}\] Answer: (A)
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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 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
← Exercise 9.2 All Chapter 9 exercises Exercise 9.4 → 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 .