The differential equation of the family of el | vedclass

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Question

(A) Given the equation of the family of ellipses: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = c$. Differentiating with respect to $x$: $\frac{2x}{a^2} + \fra

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$\frac{y''}{y'} + \frac{y'}{y} - \frac{1}{x} = 0$

Vedclass · Answer

(A) Given the equation of the family of ellipses: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = c$. Differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \Rightarrow \frac{x}{a^2} + \frac{yy'}{b^2} = 0$. Differentiating again with respect to $x$: $\frac{1}{a^2} + \frac{y y'' + (y')^2}{b^2} = 0$. From the first derivative,$\frac{1}{a^2} = -\frac{yy'}{xb^2}$. Substituting this into the second derivative equation: $-\frac{yy'}{xb^2} + \frac{yy'' + (y')^2}{b^2} = 0$. Multiplying by $\frac{b^2}{y'}$ (assuming $y' 
eq 0$): $-\frac{y}{x} + \frac{yy'' + (y')^2}{y'} = 0$. Rearranging the terms: $\frac{yy'' + (y')^2}{y'} = \frac{y}{x} \Rightarrow \frac{yy''}{y'} + y' = \frac{y}{x}$. Dividing by $y$: $\frac{y''}{y'} + \frac{y'}{y} = \frac{1}{x} \Rightarrow \frac{y''}{y'} + \frac{y'}{y} - \frac{1}{x} = 0$.

The differential equation of the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = c$ is given by $\left( y' = \frac{dy}{dx}, y'' = \frac{d^2y}{dx^2} \right)$.

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