#### Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 3

$y=a e^{2 x}+b e^{-x}$  is a solution  of $\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0$

Hint:

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

Given:

$y=a e^{2 x}+b e^{-x}$  is a solution.

Solution:

Differentiating on both sides with respect to $x$

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

$\frac{d y}{d x}=\left[a \frac{d\left(e^{2 x}\right)}{d x}+e^{2 x} \frac{d a}{d x}\right]+\left[b \frac{d\left(e^{-x}\right)}{d x}+e^{-x} \frac{d b}{d x}\right]$

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

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

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

\begin{aligned} &\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0 \\\\ &\left(4 a e^{2 x}+b e^{-x}\right)-\left(2 a e^{2 x}-b e^{-x}\right)-2\left(a e^{2 x}+b e^{-x}\right)=0 \\\\ &4 a e^{2 x}+b e^{-x}-2 a e^{2 x}+b e^{-x}-2 a e^{2 x}-2 b e^{-x}=0 \end{aligned}

Thus,     $LHS =RHS$