Question

$(x^2+y^2)dy=xydx$. If $y\left(1\right)$ and $y\left(x_0\right)=e$ then find the value of $x_0$.

Answer 1

$(x^2+y^2)dy=xydx\Rightarrow \int \frac{x}{y}dy+\int \frac{y}{x}dy=\int dx\Rightarrow x\log y+\frac{y^2}{2x}=x+c$

Now, at $x=1;y=e$

$x\log y+\frac{y^2}{2x}=x+c\Rightarrow x+\frac{e^2}{2}=x+c\Rightarrow c=\frac{e^2}{2}$

Now at $x=x_0;y=e$

$x_0\log e+\frac{e^2}{2x_0}=x_0+\frac{e^2}{2}\Rightarrow \frac{e^2}{2x_0}=\frac{e^2}{2}\Rightarrow x_0=1$