Question

The general solution of the differential equation, $y^{\prime }+y{\varphi }^{\prime }\left(x\right)-\varphi \left(x\right).{\varphi }^{\prime }\left(x\right)=0$ where $\varphi \left(x\right)$ is a known function is

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

Answer 1

Answer: (A)

$y=c{e}^{-\varphi \left(x\right)}+\varphi \left(x\right)-1$

The given equation can be written in the linear form as
$\frac{dy}{dx}+y{\varphi }^{\prime }\left(x\right)=\varphi \left(x\right){\varphi }^{\prime }\left(x\right)$
The integrating factor of this equation is
$\mu =e^{\int {\varphi }^{\prime }\left(x\right)dx}=e^{\varphi \left(x\right)}$
$e^{\varphi \left(x\right)}\frac{dy}{dx}+e^{\varphi \left(x\right)}{\varphi }^{\prime }\left(x\right)y=\varphi \left(x\right){\varphi }^{\prime }\left(x\right)e^{\varphi \left(x\right)}$
Substituting $e^{\varphi \left(x\right)}{\varphi }^{\prime }\left(x\right)=\frac{d}{dx}\left(e^{\varphi \left(x\right)}\right)$
$e^{\varphi \left(x\right)}\frac{dy}{dx}+\frac{d}{dx}\left(e^{\varphi \left(x\right)}\right)y=\varphi \left(x\right){\varphi }^{\prime }\left(x\right)e^{\varphi \left(x\right)}$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left(e^{\varphi \left(x\right)}y\right)=\varphi \left(x\right){\varphi }^{\prime }\left(x\right)e^{\varphi \left(x\right)}\Rightarrow \int \frac{d}{dx}\left(e^{\varphi \left(x\right)}y\right)dx=\int \varphi \left(x\right){\varphi }^{\prime }\left(x\right)e^{\varphi \left(x\right)}dx\Rightarrow y=\varphi \left(x\right)-1+c{e}^{-\varphi \left(x\right)}$