Exercise 9.5 Practice
Chapter 9: Differential Equations – NCERT Solutions (Rationalised)
Topic: Linear Differential Equations
Q1
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Find the general solution: $\frac{dy}{dx} + 2y = \sin x$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P=2, Q=\sin x$.
Integrating Factor (I.F.) $= e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.
Solution is $y(\text{I.F.}) = \int Q(\text{I.F.}) dx + C$.
$y e^{2x} = \int e^{2x} \sin x dx$. Let $I = \int e^{2x} \sin x dx$.
Using integration by parts: $I = \frac{e^{2x}}{5}(2\sin x - \cos x)$.
$\boxed{y e^{2x} = \frac{e^{2x}}{5}(2\sin x - \cos x) + C \text{ or } y = \frac{1}{5}(2\sin x - \cos x) + Ce^{-2x}}$
Q2
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Find the general solution: $\frac{dy}{dx} + 3y = e^{-2x}$
$P=3, Q=e^{-2x}$. I.F. $= e^{\int 3 dx} = e^{3x}$.
$y e^{3x} = \int e^{-2x} \cdot e^{3x} dx = \int e^x dx$.
$y e^{3x} = e^x + C$.
$\boxed{y = e^{-2x} + Ce^{-3x}}$
Q3
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Find the general solution: $\frac{dy}{dx} + \frac{y}{x} = x^2$
$P = \frac{1}{x}, Q = x^2$. I.F. $= e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
$y(x) = \int x^2 \cdot x dx = \int x^3 dx$.
$xy = \frac{x^4}{4} + C$.
$\boxed{y = \frac{x^3}{4} + Cx^{-1}}$
Q4
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Find the general solution: $\frac{dy}{dx} + y \sec x = \tan x$ ($0 \le x < \pi/2$)
$P = \sec x, Q = \tan x$. I.F. $= e^{\int \sec x dx} = e^{\log(\sec x + \tan x)} = \sec x + \tan x$.
$y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx = \int (\sec x \tan x + \tan^2 x) dx$.
$\int (\sec x \tan x + \sec^2 x - 1) dx = \sec x + \tan x - x + C$.
$\boxed{y(\sec x + \tan x) = \sec x + \tan x - x + C}$
Q5
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Find the general solution: $\cos^2 x \frac{dy}{dx} + y = \tan x$ ($0 \le x < \pi/2$)
Divide by $\cos^2 x$: $\frac{dy}{dx} + y \sec^2 x = \tan x \sec^2 x$.
$P = \sec^2 x, Q = \tan x \sec^2 x$. I.F. $= e^{\int \sec^2 x dx} = e^{\tan x}$.
$y e^{\tan x} = \int \tan x \sec^2 x e^{\tan x} dx$. Let $\tan x = t$.
$\int t e^t dt = e^t(t-1) + C$.
$\boxed{y e^{\tan x} = e^{\tan x}(\tan x - 1) + C \text{ or } y = \tan x - 1 + C e^{-\tan x}}$
Q6
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Find the general solution: $x \frac{dy}{dx} + 2y = x^2 \log x$
Divide by $x$: $\frac{dy}{dx} + \frac{2}{x}y = x \log x$.
$P = \frac{2}{x}$. I.F. $= e^{2\log x} = x^2$.
$y x^2 = \int x \log x \cdot x^2 dx = \int x^3 \log x dx$.
By parts: $\log x (\frac{x^4}{4}) - \int \frac{1}{x} \frac{x^4}{4} dx = \frac{x^4}{4}\log x - \frac{x^4}{16} + C$.
$\boxed{y = \frac{x^2}{16}(4\log x - 1) + Cx^{-2}}$
Q7
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Find the general solution: $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$
Divide by $x \log x$: $\frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2}{x^2}$.
$P = \frac{1}{x \log x}$. I.F. $= e^{\int \frac{dx}{x \log x}} = e^{\log(\log x)} = \log x$.
$y \log x = \int \frac{2}{x^2} \log x dx$. By parts: $\log x (-\frac{2}{x}) - \int \frac{1}{x}(-\frac{2}{x}) dx$.
$y \log x = -\frac{2}{x}\log x - \frac{2}{x} + C$.
$\boxed{y \log x = -\frac{2}{x}(1+\log x) + C}$
Q8
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Find the general solution: $(1+x^2)dy + 2xy dx = \cot x dx$ ($x \ne 0$)
$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{\cot x}{1+x^2}$.
$P = \frac{2x}{1+x^2}$. I.F. $= e^{\log(1+x^2)} = 1+x^2$.
$y(1+x^2) = \int \frac{\cot x}{1+x^2}(1+x^2) dx = \int \cot x dx$.
$y(1+x^2) = \log|\sin x| + C$.
$\boxed{y = (1+x^2)^{-1} \log|\sin x| + C(1+x^2)^{-1}}$
Q9
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Find the general solution: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ ($x \ne 0$)
$\frac{dy}{dx} + y(\frac{1}{x} + \cot x) = 1$.
$P = \frac{1}{x} + \cot x$. I.F. $= e^{\log x + \log \sin x} = x \sin x$.
$y(x \sin x) = \int x \sin x dx$.
$y x \sin x = -x \cos x + \sin x + C$.
$\boxed{y = \frac{1}{x} - \cot x + \frac{C}{x \sin x}}$
Q10
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Find the general solution: $(x+y)\frac{dy}{dx} = 1$
$\frac{dx}{dy} = x+y \Rightarrow \frac{dx}{dy} - x = y$. Linear in $x$.
$P = -1, Q = y$. I.F. $= e^{-y}$.
$x e^{-y} = \int y e^{-y} dy = -y e^{-y} - e^{-y} + C$.
$\boxed{x = -y - 1 + C e^y \text{ or } x+y+1 = Ce^y}$
Q11
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Find the general solution: $y dx + (x-y^2) dy = 0$
$\frac{dx}{dy} + \frac{x}{y} = y$. Linear in $x$.
$P = \frac{1}{y}$. I.F. $= y$.
$x y = \int y \cdot y dy = \frac{y^3}{3} + C$.
$\boxed{x = \frac{y^2}{3} + \frac{C}{y}}$
Q12
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Find the general solution: $(x+3y^2)\frac{dy}{dx} = y$ ($y > 0$)
$\frac{dx}{dy} = \frac{x+3y^2}{y} = \frac{x}{y} + 3y \Rightarrow \frac{dx}{dy} - \frac{1}{y}x = 3y$.
$P = -\frac{1}{y}$. I.F. $= e^{-\log y} = \frac{1}{y}$.
$x \frac{1}{y} = \int 3y \cdot \frac{1}{y} dy = \int 3 dy = 3y + C$.
$\boxed{x = 3y^2 + Cy}$
Q13
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Find particular solution: $\frac{dy}{dx} + 2y \tan x = \sin x; y=0$ when $x=\frac{\pi}{3}$
$P = 2\tan x$. I.F. $= e^{2\log \sec x} = \sec^2 x$.
$y \sec^2 x = \int \sin x \sec^2 x dx = \int \sec x \tan x dx = \sec x + C$.
$y = \cos x + C \cos^2 x$.
At $x=\pi/3, y=0 \Rightarrow 0 = \frac{1}{2} + C(\frac{1}{4}) \Rightarrow C = -2$.
$\boxed{y = \cos x - 2\cos^2 x}$
Q14
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Find particular solution: $(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; y=0$ when $x=1$
$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2}$.
I.F. $= 1+x^2$.
$y(1+x^2) = \int \frac{1}{1+x^2} dx = \tan^{-1} x + C$.
At $x=1, y=0 \Rightarrow 0 = \frac{\pi}{4} + C \Rightarrow C = -\frac{\pi}{4}$.
$\boxed{y(1+x^2) = \tan^{-1} x - \frac{\pi}{4}}$
Q15
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Find particular solution: $\frac{dy}{dx} - 3y \cot x = \sin 2x; y=2$ when $x=\frac{\pi}{2}$
$P = -3\cot x$. I.F. $= e^{-3\log \sin x} = \frac{1}{\sin^3 x}$.
$y \frac{1}{\sin^3 x} = \int \frac{2\sin x \cos x}{\sin^3 x} dx = 2\int \cot x \csc x dx = -2\csc x + C$.
$y = -2\sin^2 x + C \sin^3 x$.
At $x=\pi/2, y=2 \Rightarrow 2 = -2 + C \Rightarrow C=4$.
$\boxed{y = 4\sin^3 x - 2\sin^2 x}$
Q16
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Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.
$\frac{dy}{dx} = x + y \Rightarrow \frac{dy}{dx} - y = x$.
I.F. $= e^{-x}$.
$y e^{-x} = \int x e^{-x} dx = -x e^{-x} - e^{-x} + C$.
$y = -x - 1 + C e^x$.
Passes through $(0,0) \Rightarrow 0 = -1 + C \Rightarrow C=1$.
$\boxed{y + x + 1 = e^x}$
Q17
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The integrating factor of the differential equation $x \frac{dy}{dx} - y = 2x^2$ is:
(A) $e^{-x}$ (B) $e^{-y}$ (C) $1/x$ (D) $x$
(A) $e^{-x}$ (B) $e^{-y}$ (C) $1/x$ (D) $x$
Rewrite: $\frac{dy}{dx} - \frac{1}{x}y = 2x$.
$P = -\frac{1}{x}$. I.F. $= e^{-\log x} = \frac{1}{x}$.
$\boxed{\text{(C) } 1/x}$
Q18
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The integrating factor of the differential equation $(1-y^2)\frac{dx}{dy} + yx = ay$ ($|y|<1$) is:
(A) $\frac{1}{y^2-1}$ (B) $\frac{1}{\sqrt{y^2-1}}$ (C) $\frac{1}{1-y^2}$ (D) $\frac{1}{\sqrt{1-y^2}}$
(A) $\frac{1}{y^2-1}$ (B) $\frac{1}{\sqrt{y^2-1}}$ (C) $\frac{1}{1-y^2}$ (D) $\frac{1}{\sqrt{1-y^2}}$
Rewrite: $\frac{dx}{dy} + \frac{y}{1-y^2}x = \frac{ay}{1-y^2}$.
$P = \frac{y}{1-y^2}$. I.F. $= e^{\int \frac{y}{1-y^2} dy} = e^{-\frac{1}{2}\log(1-y^2)} = (1-y^2)^{-1/2}$.
$\boxed{\text{(D) } \frac{1}{\sqrt{1-y^2}}}$