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  • Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 7

Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 7

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Answer:

y=\frac{a}{x}+b  is a solution of differential equation

Hint:

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

Given:

y=\frac{a}{x}+b  is a solution of the equation


Solution:

Differentiating on both sides with respect to x

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

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{a}{x}\right)+\frac{d}{d x}(b) \\\\ &\frac{d y}{d x}=a \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{d}{d x}(b) \end{aligned}

\frac{d y}{d x}=a \frac{d}{d x}\left(x^{-1}\right)+\frac{d}{d x}(b) \quad\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]

\begin{aligned} &\frac{d y}{d x}=a(-1) x^{-1-1}+0 \\\\ &\frac{d y}{d x}=-a x^{-2}+0 \end{aligned}

\frac{d y}{d x}=\frac{-a}{x^{2}}                                    ..............(i)

Differentiate equation (i) with respect to x

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{-a}{x^{2}}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=-a \frac{d}{d x}\left(\frac{1}{x^{2}}\right) \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-a \frac{d}{d x}\left(x^{-2}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=-a\left((-2) x^{-2-1}\right) \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-a\left((-2) x^{-3}\right) \\\\ &\frac{d^{2} y}{d x^{2}}=2 a\left(x^{-3}\right) \end{aligned}

\frac{d^{2} y}{d x^{2}}=\frac{2 a}{x^{3}}                            ............(ii)

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

\begin{aligned} &\frac{d^{2} y}{d x^{2}}+\frac{2}{x}\left(\frac{d y}{d x}\right)=0 \\\\ &L H S=\frac{d^{2} y}{d x^{2}}+\frac{2}{x}\left(\frac{d y}{d x}\right) \end{aligned}

\begin{aligned} &=\frac{2 a}{x^{3}}+\frac{2}{x}\left(\frac{-a}{x^{2}}\right) \\\\ &=\frac{2 a}{x^{3}}-\frac{2 a}{x^{3}} \\\\ &=0 \\\\ &=R H S \end{aligned}

Thus, y=\frac{a}{x}+b is a solution of differential equation.

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