#### Need solution for RD Sharma maths class 12 chapte Diffrential Equation exercise 21.4 question 9

Answer:   $y=x e^{x}+e^{x}$ is the solution of given function

Hint:

Differentiate the function

Given:

$y=x e^{x}+e^{x}$  is the function.

Solution:

Differentiate with respect to x

\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x e^{x}+e^{x}\right) \\ &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x e^{x}\right)+\frac{d}{d x}\left(e^{x}\right) \\ &\Rightarrow \frac{d y}{d x}=x e^{x}+e^{x}+e^{x} \\ &\Rightarrow \frac{d y}{d x}=x e^{x}+2 e^{x} \cdots(i) \end{aligned}

Differentiate (i) with respect to  x

\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(x e^{x}+2 e^{x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=x e^{x}+e^{x}+2 e^{2 x} \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=x e^{x}+3 e^{2 x} \cdots(i i) \end{aligned}

\begin{aligned} &\text { Put value (i) and (ii) in given equation }\\ &\begin{aligned} \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y &=0 \\ L H S &=\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y \\ &=\left(x e^{x}+3 e^{2 x}\right)-2\left(x e^{x}+2 e^{x}\right)+\left(x e^{x}+e^{x}\right) \\ &=x e^{x}+3 e^{x}-2 x e^{x}-4 e^{x}+x e^{x}+e^{x} \\ &=0 \\ &=R H S \end{aligned} \end{aligned}

Thus,$y=x e^{x}+e^{x}$  satisfies the equation

Now, when $x = 0$

\begin{aligned} \mathrm{y} &=0 e^{0}+e^{0} \\ &=0+1 \\ &=1 \end{aligned}

Now, when $x = 0$

\begin{aligned} \mathrm{y}^{\prime} &=0 e^{0}+2 e^{0} \\ &=0+2 \\ \mathrm{y}^{\prime} &=2 \end{aligned}

Thus $y(0)=1 \text { and } y^{\prime}(0)=2$ satisfies the equation and initial problem