Question

If $\phi\left(x\right)$ is a differentiable function, then the solution of the differential equation
$dy+\left\{y\phi'\left(x\right) \phi\left(x\right)\phi'\left(x\right)\right\}dx=0$, is

• A. $y=\left\{\phi\left(x\right)-1\right\}+Ce^{-\phi\left(x\right)}$
• B. $y\phi\left(x\right)=\left\{\phi\left(x\right)\right\}^{2}+C$
• C. $ye^{\phi\left(x\right)}=\phi\left(x\right)e^{\phi\left(x\right)}+C$
• D. $y-\phi\left(x\right)=\phi\left(x\right)e^{-\phi\left(x\right)}$

Answer 1

Answer: (A)

We have,
$dy+\left\{y\phi'\left(x\right)-\phi\left(x\right)\phi'\left(x\right)\right\}dx=0$
$\Rightarrow \frac{dy}{dx}+\phi'\left(x\right)\cdot y=\phi\left(x\right)\phi'\left(x\right)$
This is a linear differential equation. Here,
$I.F.=e^{\int \phi'\left(x\right)dx}=e^{\phi\left(x\right)}$
$\therefore $ Solution is, $ye^{\phi\left(x\right)}=\int $ $\underbrace{\phi(x)}_I$ $\,\underbrace{e^{\phi\left(x\right)}\phi'\left(x\right)}_{II}dx$
$\Rightarrow ye^{\phi\left(x\right)}=\phi\left(x\right)e^{\phi\left(x\right)}-\int \phi'\left(x\right)e^{\phi\left(x\right)}dx$
$\Rightarrow y=\left(\phi\left(x\right)-1\right)+Ce^{-\phi\left(x\right)}$.