Question

If $x\varphi \left(x\right)=\underset{5}{\overset{x}{\int }}(3t^2-2{\varphi }^{\prime }(t\left)\right)dt,x>-2,$ and $\varphi \left(0\right)=4,$ then $\varphi \left(2\right)$ is

Answer 1

$x\varphi \left(x\right)=\underset{5}{\overset{x}{\int }}(3t^2-2{\varphi }^{\prime }(t\left)\right)dt,x>-2,$
Differentiating both side w.r.t.$x$
$\varphi \left(x\right)+x\cdot {\varphi }^{\prime }\left(x\right)=3x^2-2{\varphi }^{\prime }\left(x\right)$
Let $\varphi \left(x\right)=y$
$\Rightarrow (x+2)\cdot \frac{dy}{dx}+y=3x^2\Rightarrow \frac{dy}{dx}+\frac{y}{x+2}=\frac{3x^2}{x+2}I.F.=e^{\int \frac{1}{x+2}dx}=e^{\ln (x+2)}=(x+2)$
So, the solution of differential equation is
$y\cdot (x+2)=\int \left(\frac{3x^2}{x+2}\right)(x+2)dx+c\Rightarrow y(x+2)=x^3+c$
We know $y\left(0\right)=4$, then
$4(0+2)=0^3+c\Rightarrow c=8\therefore y(x+2)=x^3+8$
Putting $x=2$, we get
$y\left(4\right)=8+8\therefore y=\varphi \left(2\right)=4$