Question

The general solution of the differential 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 given function of 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. None of the above

Answer 1

The correct option is B $g\left(x\right)+log[1+y-g(x\left)\right]=c$

We have, $\frac{dy}{dx}=\left[g\right(x)-y].g^{\prime }\left(x\right)Putg\left(x\right)-y=V\Rightarrow g^{\prime }\left(x\right)-\frac{dy}{dx}=\frac{dV}{dx}$
Hence, $g^{\prime }\left(x\right)-\frac{dV}{dx}=V.g^{\prime }\left(x\right)\Rightarrow \frac{dV}{dx}=(1-V)g^{\prime }\left(x\right)\Rightarrow \frac{dV}{1-V}=g^{\prime }\left(x\right)dx\int \frac{dV}{1-V}=\int g^{\prime }\left(x\right)dx\Rightarrow -log(1-V)=g\left(x\right)-c\Rightarrow g\left(x\right)+log(1-V)=c\therefore g\left(x\right)+log[1+y-g(x\left)\right]=c$