Question

Let $y^{\prime }\left(x\right)+y\left(x\right)g^{\prime }\left(x\right)=g\left(x\right)g^{\prime }\left(x\right),y\left(0\right)=0,x\in R,$ where $f^{\prime }\left(x\right)$ denotes $\frac{df\left(x\right)}{dx},$ and $g\left(x\right)$ is a given non-constant differentiable function on $R$ with $g\left(0\right)=g\left(2\right)=0.$ Then the value of $y\left(2\right)$ is

Answer 1

Given: $y^{\prime }\left(x\right)+y\left(x\right)g^{\prime }\left(x\right)=g\left(x\right)g^{\prime }\left(x\right)$
$\Rightarrow e^{g\left(x\right)}{y}^{\prime }\left(x\right)+e^{g\left(x\right)}{g}^{\prime }\left(x\right)y\left(x\right)=e^{g\left(x\right)}g\left(x\right)g^{\prime }\left(x\right)$
$\Rightarrow \frac{d}{dx}\left(y\right(x\left)e^{g\left(x\right)}\right)=e^{g\left(x\right)}g\left(x\right)g^{\prime }\left(x\right)$
$\therefore y\left(x\right)e^{g\left(x\right)}=\int e^{g\left(x\right)}g\left(x\right)g^{\prime }\left(x\right)dx$
$\Rightarrow y\left(x\right)e^{g\left(x\right)}=\int e^{t}tdt,$ (where $g\left(x\right)=t$)
$\Rightarrow y\left(x\right)e^{g\left(x\right)}=(t-1)e^t+c$
$\therefore y\left(x\right)e^{g\left(x\right)}=\left(g\right(x)-1)e^{g\left(x\right)}+c$
$\because g\left(0\right)=y\left(0\right)=0$
Put $x=0\Rightarrow 0=(0-1)\cdot 1+c\Rightarrow c=1$
Put $x=2\Rightarrow y\left(2\right)\cdot 1=(0-1)\cdot \left(1\right)+1$
$\Rightarrow y\left(2\right)=0.$