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EAMCET Engineering 2005 Solved Paper

Section: Mathematics

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Question : 158 of 160

Marks: +1, -0

If

x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x

x^3 y

$\sin x$
$2 \sin x + c$
$-3 \sin x + c$
$3 \cos x + c$

Solution:

Given that

x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x

On dividing by

y^4

, we get

\frac{x^2}{y^3} - \frac{x^3}{y^4} \frac{dy}{dx} = \cos x

\Rightarrow \frac{x^3}{y^4} \frac{dy}{dx} = \frac{x^2}{y^3} - \cos x

\Rightarrow \frac{1}{y^4} \frac{dy}{dx} - \frac{1}{x y^3} = -\frac{1}{x^3} \cos x

\text{ Let } -\frac{1}{y^3}=t

\Rightarrow \frac{1}{y^4} \frac{dy}{dx} = \frac{1}{3} \cdot \frac{dt}{dx}

\Rightarrow \frac{1}{3} \cdot \frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^3} \cos x

\Rightarrow \frac{dt}{dx} + \frac{3}{x} t = \frac{3}{x^3} \cos x

\;

This is a linear differential equation in

t

, on comparing with

\frac{dt}{dx} + P t = Q

, we get

P = \frac{3}{x}, Q = \frac{3}{x^3} \cos x

\therefore \text{ I.F. } = e^{\int P \, dx} = e^{\int \frac{3}{x} \, dx} = e^{\log x^3} = x^3

\therefore

Complete solution is

t x^3 = 3 \int \frac{x^3}{x^3} \cos x \, dx + c_1

\Rightarrow t x^3 = 3 \sin x + c_1

\Rightarrow -\frac{1}{y^3} x^3 = 3 \sin x + c_1

\Rightarrow y^{-3} x^3 = -3 \sin x + c

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