Let y=y(x) be the solution of the differential equation xdydx-y=x^2 x, x(0,), y!(()/(2))=2…

Question

Let $y=y(x)$ be the solution of the differential equation $x\dfrac{dy}{dx}-y=x^2\cot x$, $x\in(0,\pi)$, $y\!\left(\frac{\pi}{2}\right)=\dfrac{\pi}{2}$. If $6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right)=k\pi$, find $k$.

Accepted Answer

$$\frac{dy}{dx}-\frac{1}{x}y = x\cot x.$$

I.F. $=e^{-\ln x}=1/x$.
$$\frac{y}{x} = \int \cot x\,dx = \ln(\sin x)+C.$$

$y(\pi/2)=\pi/2$: $\dfrac{\pi/2}{\pi/2}=\ln(\sin(\pi/2))+C \Rightarrow 1=0+C \Rightarrow C=1$. So $y=x\ln(\sin x)+x$.

$$y\!\left(\frac{\pi}{6}\right) = \frac{\pi}{6}\ln\frac12+\frac{\pi}{6} = \frac{\pi}{6}(1-\ln 2).$$
$$y\!\left(\frac{\pi}{4}\right) = \frac{\pi}{4}\ln\frac{1}{\sqrt2}+\frac{\pi}{4} = \frac{\pi}{4}\left(1-\frac12\ln2\right).$$

$$6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right) = \pi(1-\ln2) - 2\pi\left(1-\frac12\ln2\right) = -\pi.$$

Solutions · 1