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If $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$

Then show that $$\left(1-x^2\right)y'' -xy' -a^2y=0$$

Can anyone please help me with this problem?

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  • $\begingroup$ Are you sure about the coefficient $a^2$ ? It is a pity to eliminate only $b$ when you build a second order equation. $\endgroup$
    – user65203
    Apr 26, 2018 at 16:16

2 Answers 2

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Observe that $b$ is not in the equation you need to derive. So isolate $b^2$ on one side of the equation and then differentiate twice.

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  • $\begingroup$ This might yield a different equation, as the first differentiation already eliminates $b$. So this is a hint, not an answer. $\endgroup$
    – user65203
    Apr 26, 2018 at 16:19
  • $\begingroup$ Yeah I wanted to write a comment, but I've not got enough reputation. $\endgroup$
    – Tan
    Apr 26, 2018 at 16:30
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$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ differentiate $$\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} =0 \implies y'y=-x\frac {b^2}{a^2} $$ Multiply by $y (y \neq 0)$ $$y'y^2=-xy\frac {b^2}{a^2} $$ $$y'(1-\frac {x^2}{a^2})b^2=-xy\frac {b^2}{a^2} $$ $$y'(1-\frac {x^2}{a^2})=-\frac {xy}{a^2} $$ $$y'(a^2-x^2)=- {xy} $$ Differentiate again... $$(a^2-x^2)y'' -xy' +y=0$$ Are you sure you wrote the correct differential equation ?

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