#### Please solve RD Sharma class 12 chapter Differential Equation exercise 21.3 question 1 maths textbook solution

$y=b e^{x}+c e^{2 x}$is a solution of the given differential equation.

Hint:

Differentiate the given solution of differential equation on both sides with respect to $x$

Given:

$y=b e^{x}+c e^{2 x}$

Solution:

Differentiating on both sides with respect to $x$

$\frac{d y}{d x}=\frac{d}{d x}\left(b e^{x}+c e^{2 x}\right) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$

\begin{aligned} &\frac{d y}{d x}=\left[b \frac{d\left(e^{x}\right)}{d x}+e^{x} \frac{d b}{d x}\right]+\left[c \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d c}{d x}\right] \\\\ &\frac{d y}{d x}=\left[b e^{x}+e^{x}(0)\right]+\left[2 c e^{2 x}+e^{2 x}(0)\right] \\\\ &\frac{d y}{d x}=b e^{x}+2 c e^{2 x} \end{aligned}......(i)

Now to obtain the second order derivative, differentiate equation (i) with respect to $x$

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(b e^{x}+2 c e^{2 x}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=\left[b \frac{d\left(e^{x}\right)}{d x}+e^{x} \frac{d b}{d x}\right]+2\left[c \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d c}{d x}\right] \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\left[b e^{x}+e^{x}(0)\right]+2\left[2 c e^{2 x}+e^{2 x}(0)\right] \\\\ &\frac{d^{2} y}{d x^{2}}=b e^{x}+4 c e^{2 x} \end{aligned}........(ii)

Now put both equation (i) and (ii) in given differential equation as follows

$\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0$

LHS

\begin{aligned} &=\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y \\\\ &=\left(b e^{x}+4 c e^{2 x}\right)-3\left(b e^{x}+2 c e^{2 x}\right)+2\left(b e^{x}+c e^{2 x}\right) \end{aligned}

\begin{aligned} &=b e^{x}+4 c e^{2 x}-3 b e^{x}-6 c e^{2 x}+2 b e^{x}+2 c e^{2 x} \\\\ &=0 \end{aligned}

$=$ RHS

Thus, $y=b e^{x}+c e^{2 x}$ is a solution of the given differential equation.