Question

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

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

Answer 1

Answer: (A)

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

The given equation can be written in the linear form as follows:
$\frac{dy}{dx}+y{\varphi }^{{}^{\prime }}\left(x\right)=\varphi \left(x\right){\varphi }^{{}^{\prime }}\left(x\right)$
The integrating factor of this equation is $e^{\int {\varphi }^{{}^{\prime }}\left(x\right)dx}=e^{\varphi \left(x\right)}$
Integrating, we have
$y{e}^{\varphi \left(x\right)}=\int t{e}^{t}dt+c$
where $t=\varphi \left(x\right)=t{e}^t-e^t+c$
Hence $y=(\varphi \left(x\right)-1)+C{e}^{-\varphi \left(x\right)}$