If $\left(x^2+y^2\right) \mathrm{d} y=x y \mathrm{~d} x$, with $y\left(x_0\right)=e, y(1)=1$, then $x_0$ has the value

Question

Aicharya · Accepted Answer

$\begin{aligned} & \left.\left(x^2+y^2ight) \mathrm{d} y=x y \mathrm{~d} x \quad 	ext { [Let } y=v xight] \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x y}{x^2+y^2} \ & \Rightarrow v+x \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{v}{1+v} \ & \Rightarrow-\int \frac{1+v^2}{v^3} \mathrm{~d} v=\int \frac{\mathrm{d} x}{x} \ & \Rightarrow \frac{1}{2 v^2}-\log v=\log x+\log c \ & \Rightarrow c y=e^{\frac{x^2}{2 y^2}} \end{aligned}$ [Let $y=v x$ ] Putting $x=1$ and $y=1$ we get $c=e^{-\frac{1}{2}}$ $\Rightarrow y=e^{\frac{x^2-y^2}{2 y^2}}$, now putting $x=x_0$ and $y=e$ we get $x_0=\sqrt{3} e$

If $\left(x^2+y^2\right) \mathrm{d} y=x y \mathrm{~d} x$, with $y\left(x_0\right)=e, y(1)=1$, then $x_0$ has…

Solution