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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 \frac{dy}{dx} - y = x^2 \cot x,; x \in (0,\pi) $.
If $ y\left( \frac{\pi}{2} \right) = \frac{\pi}{2} $, then $ 6y\left( \frac{\pi}{6} \right) - 8y\left( \frac{\pi}{4} \right) $ is equal to:
Solution
$ xdy - ydx = x^2 \cot x,dx $
$ x^2 d\left( \frac{y}{x} \right) = x^2 \cot x,dx $
$ d\left( \frac{y}{x} \right) = \cot x,dx $
$ \int d\left( \frac{y}{x} \right) = \int \cot x,dx $
$ \frac{y}{x} = \log_e(\sin x) + C $
given $ y\left( \frac{\pi}{2} \right) = \frac{\pi}{2} $
$ \Rightarrow C = 1 $
$ y = x(\log_e \sin x + 1) $
$ y\left( \frac{\pi}{6} \right) = \frac{\pi}{6}[-\log_e 2 + 1] $
$ y\left( \frac{\pi}{4} \right) = \frac{\pi}{4}\left[ 1 - \frac{1}{2}\log_e 2 + 1 \right] $
$ 6y\left( \frac{\pi}{6} \right) - 8y\left( \frac{\pi}{4} \right) $
$ = \pi\left[ (-\log_e 2 + 1) + 2\left( \frac{1}{2}\log_e 2 - 1 \right) \right] $
$ = -\pi $
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