#### Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 19

\begin{aligned} &y=2\left(x^{2}-1\right)+c e^{-x^{2}}\\ \end{aligned}  is a solution of differential equation

Hint:

Differentiate both sides and add $2x^{2}$ and $-2x^{2}$

Given:

\begin{aligned} &y=2\left(x^{2}-1\right)+c e^{-x^{2}}\\ \end{aligned}

Solution:

Differentiating on both sides with respect to $x$

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

Now taking common and then adding $2x^{2}$ and $-2x^{2}$

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

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

\begin{aligned} &\frac{d y}{d x}=-2 x y+4 x^{3} \\\\ &\frac{d y}{d x}+2 x y=4 x^{3} \end{aligned}

Thus, \begin{aligned} &y=2\left(x^{2}-1\right)+c e^{-x^{2}}\\ \end{aligned}is a solution to the given differential equation.