Question

General solution of differnetial equation $\frac{dy}{dx}+y{g}^{\prime }\left(x\right)=g\left(x\right)g^{\prime }\left(x\right)$ where $g\left(x\right)$ is a function in $x$, is:

• A. $g\left(x\right)+\log (1+y+g(x\left)\right)=c$
• B. $g\left(x\right)+\log (1+y-g(x\left)\right)=c$
• C. $g\left(x\right)-\log (1+y-g(x\left)\right)=c$
• D. $g\left(x\right)+\log (y+g(x\left)\right)=c$

Answer 1

Answer: (B)

$g\left(x\right)+\log (1+y-g(x\left)\right)=c$

I.F$=e^{\int p\left(x\right)dx}=e^{\int g\left(x\right)dx}=e^{g\left(x\right)}$

$\Rightarrow e^{g\left(x\right)}\frac{dy}{dx}+y{g}^{\prime }\left(x\right)egx=g\left(x\right)g^{\prime }\left(x\right)eg\left(x\right)$

$\Rightarrow \frac{d}{dx}yeg\left(x\right)=g\left(x\right)g^{\prime }\left(x\right)eg\left(x\right)$

$\int g\left(x\right)g^{\prime }\left(x\right)eg\left(x\right)=g\left(x\right)eg\left(x\right)-\int g^{\prime }\left(x\right)eg\left(x\right)dx$

$=g\left(x\right)e^{g\left(x\right)}-e^{g\left(x\right)}=e^{g\left(x\right)}\left(g\right(x)-1)$

$\Rightarrow \int d\left(y{e}^{g\left(x\right)}\right)=\int g\left(x\right)g^{\prime }\left(x\right)e^{g\left(x\right)}dx+e^c$

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

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

$\Rightarrow g\left(x\right)+log(1+y-g(x\left)\right)=c$