Solve each of the following initial value problems-i y-y-e x- y 0-12ii x d y d x-y-log x- y 1-0iii d y d x-2 y-e-2 x sin x- y 0-0iv x d y d x-y-x-1 e-x- y 1-0v 1- y 2 dx - x - e -tan -1 y d x -0- y 0-0vi d y d x-y tan x-2 x-x 2 tan x- y 0-1vii x d y d x-y-x cos x-sin x- y- 2-1viii dydx - y x -4 x cosec x - y- 2-0ix d y d x-2 y tan x-sin x - y-0 when x- 3x d y d x-3 y x-sin 2 x - y-2 when x- 2xi d y d x-y x-2 cos x- y - 2-0xii d y-cos x 2-y cosecxdxxiii tanxdyd -2 xtan x - x 2- y - tan x - 0 given that y-0 when x - 2-

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Byju's Answer

Standard XII

Mathematics

Equation of Normal at a Point (x,y) in Terms of f'(x)

Solve each of...

Question

Solve each of the following initial value problems:
(i) $y' + y = e^{x}, y (0) = \frac{1}{2}$

(ii) $x \frac{d y}{d x} - y = \log x, y (1) = 0$

(iii) $\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0$

(iv) $x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0$

(v) $(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0$

(vi) $\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1$

(vii) $x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1$

(viii) $\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0$

(ix) $\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}$

(x) $\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}$
(xi) $\frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0$
(xii) $d y = \cos x (2 - y \cos e c x) d x$
(xiii) $\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y; \tan x \neq 0$ given that y = 0 when $x = \frac{π}{2}$ .

Open in App

Solution

$(i) We have, y' + y = e^{x} \Rightarrow \frac{d y}{d x} + y = e^{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 1 and Q = e^{x} ∴ I . F . = e^{\int P d x} = e^{\int 1 d x} = e^{x} Multiplying both sides of (1) by I . F . = e^{x}, we get e^{x} (\frac{d y}{d x} + y) = e^{x} e^{x} \Rightarrow e^{x} \frac{d y}{d x} + e^{x} y = e^{2 x} Integrating both sides with respect to x, we get y e^{x} = \int e^{2 x} d x + C \Rightarrow y e^{x} = \frac{e^{2 x}}{2} + C . . . . . (2) Now, y (0) = \frac{1}{2} ∴ \frac{1}{2} e^{0} = \frac{e^{0}}{2} + C \Rightarrow C = 0 Putting the value of C in (2), we get y e^{x} = \frac{e^{2 x}}{2} \Rightarrow e^{x} = \frac{e^{x}}{2} Hence, y = \frac{e^{x}}{2} is the required solution .$

$(ii) We have, x \frac{d y}{d x} - y = \log x \Rightarrow \frac{d y}{d x} - \frac{y}{x} = \frac{\log x}{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} and Q = \frac{\log x}{x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by I . F . = \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} \times \frac{\log x}{x} \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = \frac{\log x}{x^{2}} Integrating both sides with respect to x, we get y \frac{1}{x} = \int \underset{II}{\frac{1}{x^{2}}} \underset{I}{\times \log x} d x + C \Rightarrow \frac{y}{x} = \log x \int \frac{1}{x^{2}} d x - \int [\frac{d}{d x} (\log x) \int \frac{1}{x^{2}} d x] d x + C \Rightarrow \frac{y}{x} = - \frac{\log x}{x} + \int \frac{1}{x^{2}} d x + C \Rightarrow \frac{y}{x} = - \frac{\log x}{x} - \frac{1}{x} + C \Rightarrow y = - \log x - 1 + C x . . . . . (2) Now, y (1) = 0 ∴ 0 = - 0 - 1 + C (1) \Rightarrow C = 1 Putting the value of C in (2), we get y = - \log x - 1 + x \Rightarrow y = x - 1 - \log x Hence, y = x - 1 - \log x is the required solution .$

$(iii) We have, \frac{d y}{d x} + 2 y = e^{- 2 x} \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 and Q = e^{- 2 x} \sin x ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by I . F . = e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} e^{- 2 x} \sin x \Rightarrow e^{2 x} (\frac{d y}{d x} + 2 y) = \sin x Integrating both sides with respect to x, we get y e^{2 x} = \int \sin x d x + C \Rightarrow y e^{2 x} = - \cos x + C . . . . . (2) Now, y (0) = 0 ∴ 0 \times e^{0} = - \cos 0 + C \Rightarrow C = 1 Putting the value of C in (2), we get y e^{2 x} = - \cos x + 1 \Rightarrow y e^{2 x} = 1 - \cos x Hence, y e^{2 x} = 1 - \cos x is the required solution .$

$(iv) We have, x \frac{d y}{d x} - y = (x + 1) e^{- x} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = (\frac{x + 1}{x}) e^{- x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} and Q = \frac{x + 1}{x} e^{- x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by I . F . = \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} (\frac{x + 1}{x}) e^{- x} \Rightarrow \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = (\frac{x + 1}{x^{2}}) e^{- x} Integrating both sides with respect to x, we get \frac{1}{x} y = \int (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x + C Putting \frac{1}{x} e^{- x} = t \Rightarrow (- \frac{1}{x} e^{- x} - \frac{1}{x^{2}} e^{- x}) d x = d t \Rightarrow (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x = - d t ∴ \frac{1}{x} y = \int - d t + C \Rightarrow \frac{y}{x} = - t + C \Rightarrow \frac{y}{x} = - \frac{e^{- x}}{x} + C \Rightarrow y = - e^{- x} + C x . . . . . (2) Now, y (1) = 0 ∴ 0 = - e^{- 1} + C \Rightarrow C = e^{- 1} Putting the value of C in (2), we get y = - e^{- x} + x e^{- 1} \Rightarrow y = x e^{- 1} - e^{- x} Hence, y = x e^{- 1} - e^{- x} is the required solution .$

$(v) We have, (1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d y = 0 \Rightarrow (x - e^{- \tan^{- 1} y}) \frac{d y}{d x} = - (1 + y^{2}) \Rightarrow (1 + y^{2}) \frac{d x}{d y} = - (x - e^{- \tan^{- 1} y}) \Rightarrow \frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{e^{- \tan^{- 1} y}}{1 + y^{2}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = \frac{1}{1 + y^{2}} and Q = \frac{e^{- \tan^{- 1} y}}{1 + y^{2}} ∴ I . F . = e^{\int P d y} = e^{\int \frac{1}{1 + y^{2}} d y} = e^{\tan^{- 1} y} Multiplying both sides of (1) by I . F . = e^{\tan^{- 1} y}, we get e^{\tan^{- 1} y} (\frac{d x}{d y} + \frac{x}{1 + y^{2}}) = e^{\tan^{- 1} y} \frac{e^{- \tan^{- 1} y}}{1 + y^{2}} \Rightarrow e^{\tan^{- 1} y} (\frac{d x}{d y} + \frac{x}{1 + y^{2}}) = \frac{1}{1 + y^{2}} Integrating both sides with respect to y, we get e^{\tan^{- 1} y} x = \int \frac{1}{1 + y^{2}} d y + C \Rightarrow x e^{\tan^{- 1} y} = \tan^{- 1} y + C . . . . . (2) Now, y (0) = 0 ∴ 0 \times e^{0} = 0 + C \Rightarrow C = 0 Putting the value of C in (2), we get x e^{\tan^{- 1} y} = \tan^{- 1} y + 0 \Rightarrow x e^{\tan^{- 1} y} = \tan^{- 1} y Hence, x e^{\tan^{- 1} y} = \tan^{- 1} y is the required solution .$

$(vi) We have, \frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \tan x and Q = x^{2} \cot x + 2 x ∴ I . F . = e^{\int P d x} = e^{\int \tan x d x} = e^{\log |\sec x|} = \sec x Multiplying both sides of (1) by I . F . = \sec x, we get \sec x (\frac{d y}{d x} + y \tan x) = \sec x (x^{2} \tan x + 2 x) \Rightarrow \sec x (\frac{d y}{d x} + y \tan x) = x^{2} \tan x \sec x + 2 x \sec x Integrating both sides with respect to x, we get y \sec x = \int x^{2} \tan x \sec x d x + 2 \int \underset{II}{x} \underset{I}{\sec x} d x + C \Rightarrow y \sec x = \int x^{2} \tan x \sec x d x + 2 s e c x \int x d x - 2 \int [\frac{d}{d x} (s e c x) \int x d x] d x + C \Rightarrow y \sec x = \int x^{2} \tan x \sec x d x + x^{2} \sec x - \int x^{2} \tan x \sec x d x + C \Rightarrow y \sec x = x^{2} \sec x + C \Rightarrow y = x^{2} + C \cos x . . . . . (2) Now, y (0) = 1 ∴ 1 = 0 + C \cos 0 \Rightarrow C = 1 Putting the value of C in (2), we get y = x^{2} + \cos x Hence, y = x^{2} + \cos x is the required solution .$

$(vii) We have, x \frac{d y}{d x} + y = x \cos x + \sin x \Rightarrow \frac{d y}{d x} + \frac{1}{x} y = \cos x + \frac{\sin x}{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \tan x and Q = x^{2} \cot x + 2 x ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log x} = x Multiplying both sides of (1) by I . F . = x, we get x (\frac{d y}{d x} + \frac{1}{x} y) = x (\cos x + \frac{\sin x}{x}) \Rightarrow x (\frac{d y}{d x} + \frac{1}{x} y) = x \cos x + \sin x Integrating both sides with respect to x, we get x y = \int x \cos x d x + \int \sin x d x + C \Rightarrow x y = [x \sin x - \int 1 (\sin x) d x] - \cos x + C \Rightarrow x y = x \sin x + \cos x - \cos x + C \Rightarrow x y = x \sin x + C . . . . . (2) Now, y (\frac{π}{2}) = 1 ∴ 1 \times \frac{π}{2} = \frac{π}{2} \sin \frac{π}{2} + C \Rightarrow C = 0 Putting the value of C in (2), we get x y = x \sin x \Rightarrow y = \sin x Hence, y = \sin x is the required solution .$

$(viii) We have, \frac{d y}{d x} + y \cot x = 4 x cosec x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x and Q = 4 x cosec x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{\log |\sin x|} = \sin x Multiplying both sides of (1) by I . F . = \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = \sin x (4 x cosec x) \Rightarrow \sin x (\frac{d y}{d x} + y \cot x) = 4 x Integrating both sides with respect to x, we get y \sin x = 4 \int x d x + C \Rightarrow y \sin x = 2 x^{2} + C . . . . . (2) Now, y (\frac{π}{2}) = 0 ∴ 0 \times \sin (\frac{π}{2}) = 2 {(\frac{π}{2})}^{2} + C \Rightarrow C = - \frac{π^{2}}{2} Putting the value of C in (2), we get y \sin x = 2 x^{2} - \frac{π^{2}}{2} Hence, y \sin x = 2 x^{2} - \frac{π^{2}}{2} is the required solution .$

$(ix) We have, \frac{d y}{d x} + 2 y \tan x = \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 \tan x and Q = \sin x ∴ I . F . = e^{\int P d x} = e^{2 \int \tan x d x} = e^{2 \log |\sec x|} = \sec^{2} x Multiplying both sides of (1) by I . F . = \sec^{2} x, we get \sec^{2} x (\frac{d y}{d x} + 2 y \tan x) = \sec^{2} x \times \sin x \Rightarrow \sec^{2} x (\frac{d y}{d x} + 2 y \tan x) = \tan x \sec x Integrating both sides with respect to x, we get y \sec^{2} x = \int \tan x \sec x d x + C \Rightarrow y \sec^{2} x = \sec x + C . . . . . (2) Now, y (\frac{π}{3}) = 0 ∴ 0 {(\sec \frac{π}{3})}^{2} = \sec \frac{π}{3} + C \Rightarrow C = - 2 Putting the value of C in (2), we get y \sec^{2} x = \sec x - 2 \Rightarrow y = \cos x - 2 \cos^{2} x Hence, y = \cos x - 2 \cos^{2} x is the required solution .$

$(x) We have, \frac{d y}{d x} - 3 y \cot x = \sin 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 3 \cot x and Q = \sin 2 x ∴ I . F . = e^{\int P d x} = e^{- 3 \int \cot x d x} = e^{- 3 \log |\sin x|} = {cosec}^{3} x Multiplying both sides of (1) by I . F . = {cosec}^{3} x, we get {cosec}^{3} x (\frac{d y}{d x} - 3 y \cot x) = \sin 2 x ({cosec}^{3} x) \Rightarrow {cosec}^{3} x (\frac{d y}{d x} - 3 y \cot x) = 2 \cot x cosec x Integrating both sides with respect to x, we get y {cosec}^{3} x = 2 \int \cot x cosec x d x + C \Rightarrow y {cosec}^{3} x = - 2 cosec x + C \Rightarrow y = - 2 \sin^{2} x + C \sin^{3} x . . . . . (2) Now, y (\frac{π}{2}) = 2 ∴ 2 = - 2 \sin^{2} \frac{π}{2} + C \sin^{3} \frac{π}{2} \Rightarrow C = 4 Putting the value of C in (2), we get y = - 2 \sin^{2} x + 4 \sin^{3} x \Rightarrow y = 4 \sin^{3} x - 2 \sin^{2} x Hence, y = 4 \sin^{3} x - 2 \sin^{2} x is the required solution .$

$(xi) \frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0 \frac{d y}{d x} + y \cot x = 2 \cos x . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x and Q = 2 \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{logsin x} = \sin x Multiplying both sides of (1) by I . F . = \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = 2 \sin x \cos x \Rightarrow \sin x \frac{d y}{d x} + y \cos x = \sin 2 x Integrating both sides with respect to x, we get y \sin x = \int \sin 2 x d x + C \Rightarrow y \sin x = - \frac{\cos 2 x}{2} + C . . . . . (2) Now, y (\frac{π}{2}) = 0 ∴ 0 \times \sin (\frac{π}{2}) = - \frac{cosπ}{2} + C \Rightarrow C = - \frac{1}{2} Putting the value of C in (2), we get y \sin x = - \frac{\cos 2 x}{2} - \frac{1}{2} \Rightarrow 2 y \sin x = - (1 + \cos 2 x) \Rightarrow 2 y \sin x = - 2 \cos^{2} x \Rightarrow y = - \cot x \cos x Hence, y = - \cot x \cos x is the required solution .$

$(x i i) d y = \cos x (2 - y \cos e c x) d x \Rightarrow \frac{d y}{d x} = 2 \cos x - y c o t x \Rightarrow \frac{d y}{d x} + y \cot x = 2 \cos x . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x and Q = 2 \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{logsin x} = \sin x Multiplying both sides of (1) by I . F . = \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = 2 \sin x \cos x \Rightarrow \sin x \frac{d y}{d x} + y \cos x = \sin 2 x Integrating both sides with respect to x, we get y \sin x = \int \sin 2 x d x + C \Rightarrow y \sin x = - \frac{\cos 2 x}{2} + C Hence, y \sin x = - \frac{\cos 2 x}{2} + C is the required solution .$

(xiii)
$\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y \Rightarrow \frac{d y}{d x} + \frac{1}{\tan x} y = \frac{2 x \tan x + x^{2}}{\tan x} \Rightarrow \frac{d y}{d x} + (\cot x) y = 2 x + x^{2} \cot x$
This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$ .

Integrating factor, I.F. = $e^{\int P d x} = e^{\int \cot x d x} = e^{\log \sin x} = \sin x$

The solution of the given differential equation is given by

$y \times (I . F .) = \int Q \times (I . F .) d x + C \Rightarrow y \times \sin x = \int (2 x + x^{2} \cot x) \sin x d x + C \Rightarrow y \sin x = \int 2 x \sin x d x + \int x^{2} \cos x d x + C \Rightarrow y \sin x = \int 2 x \sin x d x + [x^{2} \int \cos x d x - \int (\frac{d}{d x} x^{2} \times \int \cos x d x) d x] + C$
$\Rightarrow y \sin x = \int 2 x \sin x d x + x^{2} \sin x - \int 2 x \sin x d x + C \Rightarrow y \sin x = x^{2} \sin x + C \Rightarrow y = x^{2} + cosec x \times C . . . . . (1)$
It is given that, y = 0 when $x = \frac{π}{2}$ .

$∴ 0 = {(\frac{π}{2})}^{2} + cosec \frac{π}{2} \times C \Rightarrow C = - \frac{π^{2}}{4}$
Putting $C = - \frac{π^{2}}{4}$ in (1), we get

$y = x^{2} - \frac{π^{2}}{4} cosec x$

Hence, $y = x^{2} - \frac{π^{2}}{4} cosec x$ is the required solution.

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