# Let y=y(x) be the solution of the differential

Question:

Let $y=y(x)$ be the solution of the differential

equation $x d y-y d x=\sqrt{\left(x^{2}-y^{2}\right)} d x, x \geq 1$, with

$y(1)=0$. If the area bounded by the line $\mathrm{x}=1, \mathrm{x}=\mathrm{e}^{\pi}, \mathrm{y}=0$ and $\mathrm{y}=\mathrm{y}(\mathrm{x})$ is $\alpha \mathrm{e}^{2 \pi}+\beta$, then the value of $10(\alpha+\beta)$ is equal to______

Solution:

$x d y-y d x=\sqrt{x^{2}-y^{2}} d x$

$\Rightarrow \frac{x d y-y d x}{x^{2}}=\frac{1}{x} \sqrt{1-\frac{y^{2}}{x^{2}}} d x$

$\Rightarrow \int \frac{\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}{\sqrt{1-\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}}}=\int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}$

$\Rightarrow \sin ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\ell \mathrm{n}|\mathrm{x}|+\mathrm{c}$

at $x=1, y=0 \Rightarrow c=0$

$y=x \sin (\ell n x)$

$A=\int_{1}^{e^{\pi}} x \sin (\ell n x) d x$

$x=e^{t}, d x=e^{t} d t \Rightarrow \int_{0}^{\pi} e^{2 t} \sin (t) d t=A$

$\alpha e^{2 \pi}+\beta=\left(\frac{e^{2 t}}{5}(2 \sin t-\cos t)\right)_{0}^{\pi}=\frac{1+e^{2 \pi}}{5}$

$\alpha=\frac{1}{5}, \beta=\frac{1}{5}$ so $10(\alpha+\beta)=4$