#### Provide solution for RD Sharma class 12 chapter 21 Diffrential Equations excercise 21.4 question 8

$y=e^{x}+e^{2 x}$ is the solution of given function

Hint:

Differentiate the function

Given:

$y=e^{x}+e^{2 x}$ is the given function

Solution:

Differentiating with respect to  x

$\Rightarrow \frac{d y}{d x}=e^{x}+e^{2 x} \cdots(i)$

Differentiating with respect to  x

\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(e^{x}+2 e^{2 x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+2\left(2 e^{2 x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+4 e^{2 x} \cdots(i i) \end{aligned}

Put value (i) and (ii) in different equations

\begin{aligned} \frac{d^{2} y}{d x^{2}} &-3 \frac{d y}{d x}+2 y=0 \\ L H S &=\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y \\ &=\left(e^{x}+4 e^{2 x}\right)-3\left(e^{x}+2 e^{2 x}\right)+2\left(e^{x}+e^{2 x}\right) \\ &=e^{x}+4 e^{2 x}-3 e^{x}-6 e^{2 x}+2 e^{x}+2 e^{2 x} \\ &=0 \\ &=R H S \end{aligned}

Thus,

$y=e^{x}+e^{2 x}$ satisfies the diffrential equation

Now,when $x=0$

\begin{aligned} \mathrm{y} &=e^{0}+e^{2(0)} \\ &=1+e^{0} \\ &=1+1 \\ &=2 \end{aligned}

Now,when $x=0$

\begin{aligned} y^{\prime} &=e^{0}+2 e^{2(0)} \\ &=1+2(1) \\ y^{\prime} &=3 \end{aligned}

Thus, $y(0)=2 \text { and } y^{\prime}(0)=3$ satisfies the initial value problem