Question

The differential equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=C$ is
(a) $\frac{y"}{y\text{'}}+\frac{y\text{'}}{y}-\frac{1}{x}=0$

(b) $\frac{y"}{y\text{'}}+\frac{y\text{'}}{y}+\frac{1}{x}=0$

(c) $\frac{y"}{y\text{'}}-\frac{y\text{'}}{y}-\frac{1}{x}=0$

(d) none of these

Answer 1

(a) $\frac{y"}{y\text{'}}+\frac{y\text{'}}{y}-\frac{1}{x}=0$

We have,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=C.....\left(1\right)$
Differentiating with respect to x, we get
$$\begin{gathered} \frac{2x}{a^2}+\frac{2y}{b^2}y\text{'}=0 \\ \Rightarrow \frac{x}{a^2}+\frac{y}{b^2}y\text{'}=0.....\left(2\right) \\ \mathrm{Again}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}{\left(y\text{'}\right)}^2+\frac{y}{b^2}y\text{'}\text{'}=0.....\left(3\right) \\ \mathrm{Multiplying}\mathrm{throughout}\mathrm{by}x,\mathrm{we}\mathrm{get} \\ \frac{x}{a^2}+\frac{x}{b^2}{\left(y\text{'}\right)}^2+\frac{xy}{b^2}y\text{'}\text{'}=0.....\left(4\right) \\ \mathrm{Subtracting}\left(2\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get} \\ \frac{1}{b^2}\left[x{\left(y\text{'}\right)}^2+xyy\text{'}\text{'}-yy\text{'}\right]=0 \\ \Rightarrow x{\left(y\text{'}\right)}^2+xyy\text{'}\text{'}-yy\text{'}=0 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}xyy\text{'},\mathrm{we}\mathrm{get} \\ \frac{y\text{'}}{y}+\frac{y\text{'}\text{'}}{y\text{'}}-\frac{1}{x}=0 \\ \Rightarrow \frac{y\text{'}\text{'}}{y\text{'}}+\frac{y\text{'}}{y}-\frac{1}{x}=0 \end{gathered}$$