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

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

Answers (1)

Answer:

\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.

 

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