Miscellaneous Exercise
Chapter 9: Differential Equations – NCERT Solutions
Q1
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For each of the differential equations given below, indicate its order and degree (if defined):
(i) $\frac{d^2y}{dx^2} + 5x(\frac{dy}{dx})^2 - 6y = \log x$
(ii) $(\frac{dy}{dx})^3 - 4(\frac{dy}{dx})^2 + 7y = \sin x$
(iii) $\frac{d^4y}{dx^4} - \sin(\frac{d^3y}{dx^3}) = 0$
(i) $\frac{d^2y}{dx^2} + 5x(\frac{dy}{dx})^2 - 6y = \log x$
(ii) $(\frac{dy}{dx})^3 - 4(\frac{dy}{dx})^2 + 7y = \sin x$
(iii) $\frac{d^4y}{dx^4} - \sin(\frac{d^3y}{dx^3}) = 0$
(i) Highest order derivative is $\frac{d^2y}{dx^2}$ (Order 2). Power is 1 (Degree 1).
(ii) Highest order derivative is $\frac{dy}{dx}$ (Order 1). Highest power is 3 (Degree 3).
(iii) Highest order derivative is $\frac{d^4y}{dx^4}$ (Order 4). Not a polynomial in derivatives due to $\sin(y''')$. Degree not defined.
$\boxed{\text{(i) 2, 1 (ii) 1, 3 (iii) 4, Not defined}}$
Q2
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Find the general solution: $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$
$\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$.
Integrate both sides: $\sin^{-1} y = -\sin^{-1} x + C$.
$\boxed{\sin^{-1} x + \sin^{-1} y = C}$
Q3
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Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2+y+1}{x^2+x+1} = 0$ is given by $(x+y+1) = A(1-x-y-2xy)$.
$\int \frac{dy}{y^2+y+1} = -\int \frac{dx}{x^2+x+1}$. Complete the square: $(y+1/2)^2 + (\sqrt{3}/2)^2$.
$\frac{2}{\sqrt{3}} \tan^{-1}(\frac{2y+1}{\sqrt{3}}) + \frac{2}{\sqrt{3}} \tan^{-1}(\frac{2x+1}{\sqrt{3}}) = C'$.
$\tan^{-1}(\frac{2y+1}{\sqrt{3}}) + \tan^{-1}(\frac{2x+1}{\sqrt{3}}) = \frac{\sqrt{3}C'}{2} = C$.
Use $\tan^{-1} A + \tan^{-1} B = \tan^{-1}(\frac{A+B}{1-AB})$. Simplify to get the result.
$\boxed{\text{Shown}}$
Q4
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Find the equation of the curve passing through the point $(0, \frac{\pi}{4})$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0$.
Divide by $\cos x \cos y$: $\tan x dx + \tan y dy = 0$.
$\log|\sec x| + \log|\sec y| = \log C \Rightarrow \sec x \sec y = C$.
At $(0, \pi/4)$: $1 \cdot \sqrt{2} = C \Rightarrow C = \sqrt{2}$.
$\boxed{\sec x \sec y = \sqrt{2} \text{ or } \cos x \cos y = \frac{1}{\sqrt{2}}}$
Q5
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Find the particular solution: $(1+e^{2x})dy + (1+y^2)e^x dx = 0$, given that $y=1$ when $x=0$.
$\frac{dy}{1+y^2} + \frac{e^x}{1+(e^x)^2} dx = 0$.
$\tan^{-1} y + \tan^{-1}(e^x) = C$.
At $x=0, y=1$: $\frac{\pi}{4} + \frac{\pi}{4} = C \Rightarrow C = \frac{\pi}{2}$.
$\boxed{\tan^{-1} y + \tan^{-1}(e^x) = \frac{\pi}{2}}$
Q6
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Solve the differential equation: $y e^{x/y} dx = (x e^{x/y} + y^2) dy$ ($y \ne 0$).
$\frac{dx}{dy} = \frac{x}{y} + \frac{y}{e^{x/y}}$. Put $x = vy$.
$v + y\frac{dv}{dy} = v + y e^{-v} \Rightarrow \frac{dv}{dy} = e^{-v}$.
$e^v dv = dy \Rightarrow e^v = y + C$.
$\boxed{e^{x/y} = y + C}$
Q7
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Solve the differential equation: $[\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{dx}{dy} = 1$ ($x \ne 0$).
$\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$. Linear in $y$.
I.F. $= e^{\int x^{-1/2} dx} = e^{2\sqrt{x}}$.
$y e^{2\sqrt{x}} = \int \frac{e^{-2\sqrt{x}}}{\sqrt{x}} e^{2\sqrt{x}} dx = \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} + C$.
$\boxed{y = (2\sqrt{x} + C)e^{-2\sqrt{x}}}$
Q8
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Find particular solution: $\frac{dy}{dx} + y \cot x = 4x \csc x$, given $y=0$ when $x=\frac{\pi}{2}$.
I.F. $= e^{\int \cot x dx} = \sin x$.
$y \sin x = \int 4x \csc x \sin x dx = \int 4x dx = 2x^2 + C$.
At $x=\pi/2, y=0 \Rightarrow 0 = 2(\pi^2/4) + C \Rightarrow C = -\pi^2/2$.
$\boxed{y \sin x = 2x^2 - \frac{\pi^2}{2}}$
Q9
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Find particular solution: $(x+1)\frac{dy}{dx} = 2e^{-y} - 1$, given $y=0$ when $x=0$.
$\frac{dy}{2e^{-y}-1} = \frac{dx}{x+1} \Rightarrow \frac{e^y}{2-e^y} dy = \frac{dx}{x+1}$.
$-\log|2-e^y| = \log|x+1| + C$.
At $x=0, y=0 \Rightarrow -\log 1 = \log 1 + C \Rightarrow C=0$.
$\log|2-e^y| + \log|x+1| = 0 \Rightarrow (2-e^y)(x+1) = 1$.
$\boxed{y = \log|\frac{2x+1}{x+1}|}$
Q10
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The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village increases from 20,000 to 25,000 in 5 years, in what time will the population be 40,000?
$\frac{dP}{dt} = kP \Rightarrow P = C e^{kt}$.
$t=0, P=20000 \Rightarrow C=20000$.
$t=5, P=25000 \Rightarrow 1.25 = e^{5k} \Rightarrow 5k = \log(1.25)$.
For $P=40000$: $2 = e^{kt} \Rightarrow kt = \log 2$.
$t = \frac{\log 2}{k} = \frac{5 \log 2}{\log(1.25)}$.
$\boxed{\frac{5 \log 2}{\log(5/4)} \text{ years}}$
Q11
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The general solution of the differential equation $\frac{y dx - x dy}{y} = 0$ is:
(A) $xy = C$ (B) $x = Cy^2$ (C) $y = Cx$ (D) $y = Cx^2$
(A) $xy = C$ (B) $x = Cy^2$ (C) $y = Cx$ (D) $y = Cx^2$
$y dx - x dy = 0 \Rightarrow \frac{dx}{x} = \frac{dy}{y}$.
$\log x = \log y + \log C \Rightarrow x = Cy$ or $y = Cx$.
$\boxed{\text{(C) } y = Cx}$
Q12
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The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is:
(A) $x e^y + x^2 = C$ (B) $x e^y + y^2 = C$ (C) $y e^x + x^2 = C$ (D) $y e^y + x^2 = C$
(A) $x e^y + x^2 = C$ (B) $x e^y + y^2 = C$ (C) $y e^x + x^2 = C$ (D) $y e^y + x^2 = C$
$e^x dy + y e^x dx = -2x dx \Rightarrow d(y e^x) = -2x dx$.
$y e^x = -x^2 + C$.
$\boxed{\text{(C) } y e^x + x^2 = C}$