Question

The general solution of the differential equation $\frac{dy}{dx}+y{g}^{\prime }(x)=g(x)\cdot g^{\prime }(x)$ where $g(x)$ is a given function of $x$, is

• A. $g(x)+\log [1+y+g(x)]=c$
• B. $g(x)+\log [1+y-g(x)]=c$
• C. $g(x)-\log [1+y-g(x)]=c$
• D. None of these

Answer 1

Answer: (B)

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