#### Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 15 textbook solution.

Answer  :   $\text { (b) } g(x)+\log (y-g(x)+1)=C$

Hint: If the Differential equation is linear in y i.e. $\frac{d y}{d x}+P(x) y=Q(x) \text { then } I F=e^{\int P(x) d x}$

Given  :  $\frac{d y}{d x}+y g^{\prime}(x)=g(x) g^{\prime}(x)$

Explanation: The given differential equation is linear in y

\begin{aligned} &\frac{d y}{d x}+P(x) y=Q(x) \text { then IF }=e^{\int P(x) d x} \\ &I F=e^{\int g^{\prime}(x) d x} \\ &=e^{g(x)} \end{aligned}

General solution of the given differential equation,

\begin{aligned} &y e^{g(x)}=\int g(x) e^{g(x)} g^{\prime}(x) d x+C_{1} \\ &\text { Let } g(x)=t \Rightarrow g^{\prime}(x) d x=d t \end{aligned}

\begin{aligned} &y e^{g(x)}=\int t e^{t} d t+C_{1}\\ &y e^{g(x)}=t e^{t}-e^{t}+C_{1}\\ &y e^{g(x)}=g(x) e^{g(x)}-e^{g(x)}+C_{1}\\ &\text { Divide by } e^{g(x)} \end{aligned}

\begin{aligned} &y=g(x)-1+\frac{C_{1}}{e^{g(x)}} \\ &y=g(x)-1+C_{1} e^{-g(x)} \\ &y-g(x)+1=C_{1} e^{-g(x)} \end{aligned}

Taking logarithm on both sides

\begin{aligned} \log (y-g(x)+1) &=\log \left(C_{1} e^{-g(x)}\right) \\ &=\log C_{1}+\log e^{-g(x)} \\ &=\log C_{1}-g(x) \log e \\ &=\log C_{1}-g(x) \end{aligned}

$\text { Hence, } g(x)+\log (y-g(x)+1)=C \quad \text { [ taking } \log C_{1}=C \text { ] }$