Question

If $y=y\left(x\right)$ is the solution of $\underset{0}{\overset{x}{\int }}tydt=x^2+y^2$ and $y\left(0\right)=1,$ where point $(a,3)$ satisfies the curve $y=y\left(x\right),$ then the value of $a^2=$

Answer 1

Given:
$\underset{0}{\overset{x}{\int }}tydt=x^2+y^2$
differentiating both sides w.r.t. $x,$
$\Rightarrow xy=2x+2y\frac{dy}{dx}\Rightarrow \frac{2y}{y-2}dy=xdx$
integrating both sides,
$\Rightarrow \int (2+\frac{4}{y-2})dy=\int xdx\Rightarrow 2y+4\ln |y-2|=\frac{x^2}{2}+C\Rightarrow x^2+c=4(y+2\ln |y-2|)$
As, $y\left(0\right)=1;c=4$
Now, point $(a,3)$ lies on this curve
$\Rightarrow a^2+4=4(3+0)\Rightarrow a^2=8$