If $\phi(x)$ is a differentiable function, then the solution of the differential equation
$d y+\left\{y \phi'(x)-\phi(x) \phi'(x)\right\} d x=0$, is
Answer & explanation
Correct answer: option 1
We have,
$d y+\left\{y \phi'(x)-\phi(x) \phi'(x)\right\} d x=0$
$\Rightarrow \frac{d y}{d x}+\phi'(x) . y=\phi(x) \phi'(x)$ .....(i)
This is a linear differential equation with
Integrating factor = $e^{\int \phi'(x) d x}=e^{\phi(x)}$
Multiplying (i) by $\phi(x)$ and integrating, we get
$y e^{\phi(x)} =\int \phi(x) \phi'(x) e^{\phi(x)} d x$
$\Rightarrow y e^{\phi(x)} =\int e^{\phi(x)} \phi(x) \phi'(x) d x$
$\Rightarrow y e^{\phi(x)}=\int \phi(x) e^{\phi(x)} \phi'(x) d x$
$\Rightarrow y e^{\phi(x)}=\phi(x) e^{\phi(x)}-\int \phi'(x) e^{\phi(x)} d x$
$\Rightarrow y e^{\phi(x)}=\phi(x) e^{\phi(x)}-e^{\phi(x)}+C$
$\Rightarrow y=(\phi(x)-1)+C e^{-\phi(x)}$