Question

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

• A. $g(x)-\log [1-y-g(x)]=C$
• B. $g(x)-\log [1+y-g(x)]=C$
• C. $g(x)+[1+y-\log g(x)]=C$
• D. $g(x)+\log [1+y-g(x)]=C$

Answer 1

Answer: (D)

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