Solution of Differential equation

[jacks]

If \(\displaystyle y'(x)+y(x)\cdot g'(x) = g(x)\cdot g'(x)\;,y(0)=0\;,x\in \mathbb{R}\;,\) If \(\displaystyle f'(x) = \frac{d}{dx}\left(f(x)\right)\) and \(g(x)\) is a non constant differentiable function on \(\displaystyle \mathbb{R}\) with \(g(0) = g(2) = 0\;\). Then the value of \(y(2)\) is?

[Papapetros Vaggelis]

For each \(\displaystyle{x\in\mathbb{R}}\) holds :

\(\displaystyle{\begin{aligned}y^\prime(x)+y(x)\,g^\prime(x)=g(x)\,g^\prime(x)&\iff e^{g(x)}\,y^\prime(x)+y(x)\,e^{g(x)}\,g^\prime(x)=g(x)\,e^{g(x)}\,g^\prime(x)\\&\iff \left[\dfrac{\mathrm{d}}{\mathrm{d}t}\,e^{g(t)}\,y(t)\right]_{t=x}=\left(e^{g(x)} \right )'\,g(x) \\&\iff \left[\dfrac{\mathrm{d}}{\mathrm{d}t}\,e^{g(t)}\,y(t)\right]_{t=x}=\left[\dfrac{\mathrm{d}}{\mathrm{d}t}\, \left( e^{g(t)}\,g(t)-e^{g(t)} \right) \right]_{t=x}\end{aligned}}\)

so:

\(\displaystyle{\left(\exists\,c\in\mathbb{R}\right)\,\left(\forall\,x\in\mathbb{R}\right): e^{g(x)}\,y(x)=e^{g(x)}\,g(x)-e^{g(x)}+c}\) .

Setting \(\displaystyle{x=0}\) : \(\displaystyle{e^{g(0)}\,y(0)=e^{g(0)}\,g(0)-e^{g(0)}+c\iff c=1}\)

and then we get :

\(\displaystyle{\forall\,x\in\mathbb{R}: y(x)=\left(g(x)-1\right)+e^{-g(x)}}\).

Therefore, \(\displaystyle{y(2)=\left(g(2)-1\right)+e^{-g(2)}=\left(0-1\right)+e^{0}=0}\) .