Let y=y(x) solve x,dy-y,dx=(x^2-y^2),dx, x1, y(1)=0.

Question

Let $y=y(x)$ solve $x\,dy-y\,dx=(x^2-y^2)\,dx$, $x\ge1$, $y(1)=0$. If the area bounded by $x=1$, $x=e^\pi$, $y=0$, $y=y(x)$ is $\alpha e^{2\pi}+\beta$, then $10(\alpha+\beta)$ is \_\_\_\_\_\_.

Answer: 4

Accepted Answer

Dividing by $x^2$ and integrating: $\sin^{-1}(y/x)=\ln|x|+C$; with $x=1,y=0$: $C=0$, so $y=x\sin(\ln x)$.

The area is $\int_1^{e^\pi}x\sin(\ln x)\,dx$; with $x=e^t$: $\int_0^\pi e^{2t}\sin t\,dt = \dfrac{1+e^{2\pi}}{5}$.

So $\alpha=\dfrac15,\beta=\dfrac15$, and $10(\alpha+\beta)=4$.

Solutions · 1