#### Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 34 Maths Textbook Solution.

Answer:  $y=\frac{x e^{4 x}}{6}-\frac{1}{36} e^{4 x}+C e^{-2 x}$

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}+2 y=x e^{4 x} \\ & \end{aligned}

Solution:  $\frac{d y}{d x}+(2) y=x e^{4 x}$

\begin{aligned} &=\frac{d y}{d x}+P y=Q \\ &P=2, Q=x e^{4 x} \end{aligned}

$If$  of differential equation is

\begin{aligned} &\text { If }=e^{\int P d x} \\ & \end{aligned}

$=e^{\int 2 d x} \\$

$=e^{2 \int d x} \\$

$=e^{2 x} \\$

$y I f=\int \text { QIf } d x+C \\$

$=y\left(e^{2 x}\right)=\int x e^{4 x} e^{2 x} d x+C \\$

$=y e^{2 x}=\int x e^{6 x} d x+C$

\begin{aligned} &\quad=y e^{2 x}=\int x e^{6 x} d x+C\left[\int f(x) g(x) d c=f(x) \int[g(x) d x] \int f(x)\left[\int g(x) d x\right] d x\right] \\ & \end{aligned}

$=y e^{2 x}=x\left(\int e^{6 x} d x\right)-\int \frac{d}{d x}(x)\left(\int e^{6 x} d x\right) d x+C \\$

$=y e^{2 x}=x\left(\frac{e^{6 x}}{6}\right)-\int 1\left(\frac{e^{6 x}}{6}\right) d x+C \\$

$=y e^{2 x}=x\left(\frac{e^{6 x}}{6}\right)-\frac{1}{6}\left(\frac{e^{6 x}}{6}\right)+C \\$

$=y e^{2 x}=e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C$

\begin{aligned} &=y e^{2 x} e^{-2 x}=e^{-2 x}\left[e^{6 x}\left(\frac{x}{6}\right)-\frac{1}{36}\left(e^{6 x}\right)+C\right] \\ & \end{aligned}

$=y e^{2 x-2 x}=e^{-2 x} e^{6 x}\left(\frac{x}{6}\right)-e^{-2 x} \frac{1}{36}\left(e^{6 x}\right)+C e^{-2 x} \\$

$=y e^{0}=e^{6 x-2 x}\left(\frac{x}{6}\right)-e^{6 x-2 x} \frac{1}{36}+C e^{-2 x} \\$

$=y=e^{4 x}\left(\frac{x}{6}\right)-e^{4 x} \frac{1}{36}+C e^{-2 x}$