#### Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 17 textbook solution.

Answer  : $\text { (a) } \frac{y^{\prime}}{y}+\frac{y^{\prime \prime}}{y^{\prime}}-\frac{1}{x}=0$

Hint: eliminate constant from given equation.

Given:  $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=C$

Explanation : Differentiate both sides with respect to x

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

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

Again differentiate with respect to x

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

\begin{aligned} &\Rightarrow-\frac{1}{a^{2}}=\frac{1}{b^{2}}\left[\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d^{2} y}{d x^{2}}\right)\right] \\ &\Rightarrow-\frac{b^{2}}{a^{2}}=\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d^{2} y}{d x^{2}}\right) \\ &\Rightarrow-\frac{b^{2}}{a^{2}}=\left(y^{\prime}\right)^{2}+y y^{\prime \prime} \end{aligned}   .....(ii)

Now from (i) we get

$\Rightarrow-\frac{b^{2}}{a^{2}}=\frac{y}{x} y^{\prime}$

Put in (ii)

$\Rightarrow \frac{y y^{\prime}}{x}=\left(y^{\prime}\right)^{2}+y y^{\prime \prime}$

Divide both sides divided by $yy'$

\begin{aligned} &\Rightarrow \frac{1}{x}=\frac{y^{\prime}}{y}+\frac{y^{\prime \prime}}{y^{\prime}} \\ &\Rightarrow \frac{y^{\prime}}{y}+\frac{y^{\prime \prime}}{y^{\prime}}-\frac{1}{x}=0 \end{aligned}

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