Let the solution curve of the differential equation xdy-ydx=x^2+y^2dx,x0, y(1)=0, be y=y(…

Question

Let the solution curve of the differential equation $xdy-ydx=\sqrt{{x}^{2}+{y}^{2}}dx,x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to

Accepted Answer

$\begin{gathered} xdy-ydx=\sqrt{{x}^{2}+{y}^{2}}dx \ \frac{xdy-ydx}{{x}^{2}}=\frac{\sqrt{{x}^{2}+{y}^{2}}}{{x}^{2}}dx \end{gathered}$

$\begin{gathered} \int \frac{d(y/x)}{\sqrt{1+(y/x{)}^{2}}}=\int \frac{1}{x}dx \ \log (\frac{y}{x}+\sqrt{1+{(\frac{y}{x})}^{2}})=\log x+\log C \ y+\sqrt{{x}^{2}+{y}^{2}}={x}^{2} \ y-\sqrt{{x}^{2}+{y}^{2}}=-1 \ 2y={x}^{2}-1 \ y(3)=4 \end{gathered}$

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