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Please solve RD Sharma class 12 chapter Differential Equation exercise 21.3 question 17 maths textbook solution

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y=e^{m \cos ^{-1} x}  is a solution of differential equation


Differentiate with respect to x and put values in given equation.


y=e^{m \cos ^{-1} x} 


Differentiating on both sides with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(e^{m \cos ^{-1} x}\right) \\\\ &\frac{d y}{d x}=m e^{m \cos ^{-1} x}\left(\frac{-1}{\sqrt{1+x^{2}}}\right) \\\\ &\frac{d y}{d x}=\left(\frac{-m e^{m \cos ^{-1} x}}{\sqrt{1+x^{2}}}\right) \end{aligned}                        ............(i)

Differentiating equation (i) with respect to x,

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

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

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

Hence proved, the given function is the solution to given differential equation


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