The solution of $x d y-y d x=\sqrt{x^2+y^2} d x$ when $y(\sqrt{3})$ $=1$ is
The solution of $x d y-y d x=\sqrt{x^2+y^2} d x$ when $y(\sqrt{3})$ $=1$ is
$y^2+\sqrt{x^2+y^2}=x^2$
$5 y-\sqrt{x^2+y^2}=x^2$
$y+\sqrt{x^2+y^2}=x$
$5 y^2-\sqrt{x^2+y^2}=x$
Solution
Since, $x d y-y d x=\sqrt{x^2+y^2} d x$
$\begin{aligned}
& \Rightarrow x d y=\left(y+\sqrt{x^2+y^2}\right) d x \\
& \Rightarrow \frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}
\end{aligned}$
Let $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
So, $v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^2+v^2 x^2}}{x}$
$\begin{aligned}
& \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^2} \Rightarrow \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{1}{x} d x \\
& \Rightarrow \log \left(v+\sqrt{v^2+1}\right)=\log x c \\
& \Rightarrow y+\sqrt{x^2+y^2}=xc
\end{aligned}$
Since, $y(\sqrt{3})=1 \Rightarrow 1+\sqrt{4}=3 c \Rightarrow c=1$
So, $y+\sqrt{x^2+y^2}=x$