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

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

Answers (1)

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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