Question

The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is

पर्याय

• \[\frac{y "}{y'} + \frac{y'}{y} - \frac{1}{x} = 0\]

• \[\frac{y "}{y'} + \frac{y'}{y} + \frac{1}{x} = 0\]

• \[\frac{y "}{y'} - \frac{y'}{y} - \frac{1}{x} = 0\]

• none of these

Answer 1

\[\frac{y "}{y'} + \frac{y'}{y} - \frac{1}{x} = 0\]

 

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