Question

If $\varphi \left(x\right)$ is a differential function, then the solution of the differential equation $dy+\left\{y{\varphi }^{\prime }\right(x)-\varphi (x\left){\varphi }^{\prime }\right(x\left)\right\}dx=0$, is

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

Answer 1

The correct option is A $y=\left\{\varphi \right(x)-1\}+C{e}^{-\varphi \left(x\right)}$

Given differential equation is
$\Rightarrow \frac{dy}{dx}+{\varphi }^{\prime }\left(x\right)y=\varphi \left(x\right){\varphi }^{\prime }\left(x\right)$
Which is a linear differential equation with
$P={\varphi }^{\prime }\left(x\right),Q=\varphi \left(x\right).{\varphi }^{\prime }\left(x\right)$
$IF=e^{\int {\varphi }^{\prime }\left(x\right)dx}=e^{{\varphi }^{\prime }\left(x\right)}$
Therefore, solution is $y.e^{{\varphi }^{\prime }\left(x\right)}=\int \varphi \left(x\right).{\varphi }^{\prime }\left(x\right)e^{{\varphi }^{\prime }\left(x\right)}dx+C$
$\Rightarrow y.e^{{\varphi }^{\prime }\left(x\right)}=\int \varphi \left(x\right)e^{{\varphi }^{\prime }\left(x\right)}{\varphi }^{\prime }\left(x\right)dx+C$

$\Rightarrow y.e^{{\varphi }^{\prime }\left(x\right)}=\varphi \left(x\right)e^{{\varphi }^{\prime }\left(x\right)}-e^{{\varphi }^{\prime }\left(x\right)}+C$

$\Rightarrow y=\left[\varphi \right(x)-1]+C{e}^{-{\varphi }^{\prime }\left(x\right)}$