Question

The general solution of the differential equation, $y'+y \phi'(x)-\phi(x) \cdot \phi'(x)=0,$ where $\phi(x)$ is a known function, is
where $c$ is an arbitrary constant

• A. $y=c e^{-\phi(x)}+\phi(x)-1$
• B. $y=c e^{+\phi(x)}+\phi(x)-1$
• C. $y=c e^{-\phi(x)}-\phi(x)+1$
• D. $y=c e^{-\phi(x)}+\phi(x)+1$

Answer 1

Answer: (A)

$\frac{d y}{d x}+y \phi'(x)=\phi(x) \phi'(x)$
$I.F. =e^{\int \phi'(x) d x}=e^{\phi(x)}$
Hence, the solution is
$y e^{\phi(x)}=\int e^{\phi(x)} \phi(x) \phi'(x) d x$
$=\int e^{t} t \,d t,$ where $\phi(x)=t$
$=t e^{t}-e^{t}+c$
$=\phi(x) e^{\phi(x)}-e^{\phi(x)}+c$
$\therefore y=c e^{-\phi(x)}+\phi(x)-1$