#### Explain solution for  RD Sharma Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 22 maths textbook solution.

Answer : $\text { (b) } x y y_{2}+x y_{1}^{2}-y y_{1}=0$

Hint: Eliminate the constants by differentiating the equation with respect to x

Given: $a x^{2}+b y^{2}=1$

Explanation: differentiate both sides with respect to x

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

Again differentiate with respect to x

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

\begin{aligned} &\Rightarrow b\left(y_{1}^{2}+y y_{2}\right)=-a \quad\left[y_{1}=\frac{d y}{d x}, y_{2}=\frac{d^{2} y}{d x^{2}}\right] \\ &\Rightarrow y_{1}^{2}+y y_{2}=-\frac{a}{b} & \ldots(i i) \end{aligned}

Now from (i)

\begin{aligned} &\Rightarrow a x=-b y \frac{d y}{d x} \\ &\Rightarrow y \frac{d y}{d x}=-\frac{a x}{b} \end{aligned}                                    ....(iii)

Compare (ii) and (iii) we get

\begin{aligned} &\Rightarrow y_{1}^{2}+y y_{2}=\frac{y}{x} y_{1} \\ &\Rightarrow x y_{1}^{2}+x y y_{2}=y y_{1} \\ &\Rightarrow x y y_{2}+x y_{1}^{2}-y y_{1}=0 \end{aligned}