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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $y = y(x)$ be the solution of the differential equation
$x^4dy + (4x^3y + 2\sin x)dx = 0,; x > 0,; y\left(\frac{\pi}{2}\right) = 0.$
Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to:
Solution
$(x^4dy + 4x^3y,dx) = -2\sin x,dx$
$\Rightarrow \int d(x^4y) = \int -2\sin x,dx$
$\Rightarrow x^4y = 2\cos x + c$
$\Rightarrow x^4f(x) = 2\cos x + c$
As $f\left(\frac{\pi}{2}\right) = 0$
So, $c = 0$
$\left(\frac{\pi}{3}\right)^4 f\left(\frac{\pi}{3}\right) = 2\cos \frac{\pi}{3}$
$\Rightarrow \pi^4 f\left(\frac{\pi}{3}\right) = 81$
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