Question
Let $y = y(x)$ be the solution of the differential equation $x\frac{dy}{dx} - y = x^{2}cotx,x \in (0,\pi)$.
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 :
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Solution
$xdy - ydx = x^{2}cotxdx$
$${x^{2}d\left( \frac{y}{x} \right) = x^{2}cotxdx }{d\left( \frac{y}{x} \right) = cotxdx }{\int d\left( \frac{y}{x} \right) = \int cotxdx }{\frac{y}{x} = \log_{e}sinx + C }$$given $y\left( \frac{\pi}{2} \right) = \frac{\pi}{2}$
$${\Rightarrow c = 1 }{y = x\left( \log_{e}sinx + 1 \right) }{y\left( \frac{\pi}{6} \right) = \frac{\pi}{6}\left\lbrack - \log_{e}2 + 1 \right\rbrack }{y\left( \frac{\pi}{4} \right) = \frac{\pi}{4}\left\lbrack - \frac{1}{2}\log_{e}2 + 1 \right\rbrack }{6y\left( \frac{\pi}{6} \right) - 8y\left( \frac{\pi}{4} \right) }{= \pi\left\lbrack \left( - \log_{e}2 + 1 \right) + 2\left( \frac{1}{2}\log_{e}2 - 1 \right) \right\rbrack }{= \pi\lbrack 1 - 2\rbrack = - \pi}$$
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