Question

The D.E whose solution is $\frac{x^2}{a^2}+\frac{{y^2}^{}}{b^2}=1$

• A. $xy{y}_2+x{y}_1=y{y}_1$
• B. $xy{y}_2+y_1=y$
• C. $x^2{y}_2+x{y}_1=y$
• D. $x^2{y}_2+2x{y}_1={y_1}^2$

Answer 1

The correct option is A

$xy{y}_2+x{y}_1=y{y}_1$

Step 1. Differentiate $\frac{x^2}{a^2}+\frac{{y^2}^{}}{b^2}=1$ with respect to $x$

$$\begin{gathered} \frac{2x}{a^2}+\frac{2y{y}_1}{b^2}=0 \\ \Rightarrow \frac{2x}{a^2}=-\frac{2y{y}_1}{b^2} \\ \Rightarrow -x\frac{b^2}{a^2}=\frac{y{y}_1}{} \\ \Rightarrow y{y}_1=-x\frac{b^2}{a^2}.......................\left(i\right) \end{gathered}$$

Step 2.Differentiating $\left(i\right)$ with respect to $x$.

$$\begin{gathered} y{y}_2+{y_1}^2=-\frac{b^2}{a^2}[(uv)\text{'}=uv\text{'}+vu\text{'}] \\ \Rightarrow y{y}_2+{y_1}^2=\frac{y{y}_1}{x}fromeq\left(i\right) \\ \Rightarrow x(y{y}_2+{y_1}^2)=y{y}_1 \\ \Rightarrow xy{y}_2+{x{y}_1}^2=y{y}_1 \end{gathered}$$

Hence option (A) is the correct option.