If $y=e^{2\cos^{-1}x}$ also $$(1-x^2)y_2-xy_1-\lambda y=0$$ then the value of $\lambda$ is. I see that the question is incomplete but answer is given as $2$. Am I missing on anything.
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What are $y_1$ and $y_2$? – almagest Jun 25 '16 at 16:03
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Are $y_2$ and $y_1$ supposed to be $y''$ and $y'$? – N. S. Jun 25 '16 at 16:04
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I wasnt sure about it. – Archis Welankar Jun 25 '16 at 16:24
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As answer is $2$. – Archis Welankar Jun 25 '16 at 16:24
2 Answers
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On differentiating;
$y'=$$y\frac{-2}{√(1-x^2)}$ which on squaring gives, $(1-x^2)y^2=4y^2$. Differentiate again and after cancellation of $2y'$ on both sides, you get $(1-x^2)y''-xy'-4y=0$
Nitin Uniyal
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I assume that $y_1$ means $y'$ and $y_2$ means $y''$.
So differentiating we get $y'=-2y\frac{1}{\sqrt{1-x^2}}$. Differentiating again we get $y''=-2xy\frac{1}{(1-x^2)^{3/2}}+4y\frac{1}{1-x^2}$. Hence $$(1-x^2)y''-xy'=-2xy\frac{1}{\sqrt{1-x^2}}+4y+2xy\frac{1}{\sqrt{1-x^2}}=4y$$
So the equation holds with $\lambda=4$.
Note that this does not agree with the answer given in the Question.
almagest
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