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
where $c$ is an arbitrary constant.

• 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

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

Given differential equation is $y^{{}^{\prime }}+y{\varphi }^{{}^{\prime }}\left(x\right)=\varphi \left(x\right){\varphi }^{{}^{\prime }}\left(x\right)$,

This is a linear differential equation,

The integrating factor is $e^{\int {\varphi }^{{}^{\prime }}\left(x\right)}=e^{\varphi \left(x\right)}$,

The solution to the differential equation is $y{e}^{\varphi \left(x\right)}=\int e^{\varphi \left(x\right)}\varphi \left(x\right){\varphi }^{{}^{\prime }}\left(x\right)dx$,

Solving the integral by method of substitution i.e., taking $e^{\varphi \left(x\right)}=t$,gives,

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