Let $y'(x)+y(x) g'(x)=g(x) g'(x)$, $y(0)=0, x \in R$, where $f'(x)$ denotes $\frac{d}{d x}(f(x))$ and $g(x)$ is a given non-constant differentiable function on $R$ with $g(0)=g(2)=0$. Then the value of $y(2)$ is
Answer & explanation
Correct answer: option 1
We have,
$\frac{d y(x)}{d x}+g'(x) y(x)=g(x) g'(x)$
This is a linear differential equation with integraint factor $e^{g(x)}$. So, its solution is given by
$y(x) e^{g(x)} =\int e^{g(x)} g(x) g'(x) d x+C$
$\Rightarrow y(x) e^{g(x)} =e^{g(x)}\{g(x)-1\}+C$ .....(i)
It is given that $y(0)=0$ and $g(0)=0$
∴ $0=e^0(0-1)+C \Rightarrow C=1$
Putting $C=1$ in (i), we get
$y(x) e^{g(x)}=e^{g(x)}\{g(x)-1\}+1$
Putting $x=2$, we get
$y(2) e^0=e^0(0-1)+1 \Rightarrow y(2)=0$ [∵ g(0) = 0]