#### Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 33 Maths Textbook Solution.

Answer:  $y=\left(\frac{x^{2}}{2}+C\right) e^{x}$

Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$  where $P,Q$  are constants.

Give:  \begin{aligned} &\frac{d y}{d x}-y=x e^{x} \\ & \end{aligned}

Solution:  $\frac{d y}{d x}+(-1) y=x e^{x}$

\begin{aligned} &=\frac{d y}{d x}+P y=Q \\ &P=-1, Q=x e^{x} \end{aligned}

$If$  of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}

$=e^{\int-1 d x} \\$

$=e^{-\int d x} \\$

$=e^{-x} \\$

$y I f=\int Q I f d x+C \\$

$=y\left(e^{-x}\right)=\int x e^{x} e^{-x} d x+C$

\begin{aligned} &=y\left(e^{-x}\right)=\int x e^{x-x} d x+C \\ & \end{aligned}

$=y\left(e^{-x}\right)=\int x e^{0} d x+C \\$

$=y\left(e^{-x}\right)=\int x d x+C \\$

$=y\left(e^{-x}\right)=\frac{x^{1+1}}{1+1}+C \\$

$=y\left(e^{-x}\right)=\frac{x^{2}}{2}+C$

\begin{aligned} &=y\left(e^{-x}\right) e^{x}=e^{x}\left(\frac{x^{2}}{2}+C\right) \\ & \end{aligned}

$=y\left(e^{-x+x}\right)=e^{x}\left(\frac{x^{2}}{2}+C\right) \\$

$=y\left(e^{0}\right)=e^{x}\left(\frac{x^{2}}{2}+C\right) \\$

$=y=e^{x}\left(\frac{x^{2}}{2}+C\right)$