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How to solve this differentiation equation? $$\sin^2 x {d^2y \over dx^2} = 2 y$$ I don't know how to begin. Can it be any simpler than this?

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Cleaning up Maple's solution, I get

$$ y \left( x \right) ={\frac {c_{{1}} \left( \cos \left( 2\,x \right) +1 \right) }{\sin \left( 2\,x \right) }}+{\frac {c_{{2}} \left( x\cos \left( 2\,x \right) -\sin \left( 2\,x \right) +x \right) }{\sin \left( 2\,x \right) }} $$

I can't imagine that anyone would assign that differential equation as homework to be solved by hand. Are you sure the assignment requires solving the differential equation in closed form?

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    I think it is better to exhibit the solution as $$y\left( x \right) = {c_1}\cot x + {c_2}\left( {x\cot x - 1} \right)$$2012-07-27
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    @PeterTamaroff how did you get that? solution according to book is $$ y \tan (x) = c_1(\tan x - x) + c_2 $$ same as yours2012-07-27
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    I just rewrote Robert's solution using $\cot(2x)+\csc(2x)=\cot(x)$2012-07-27
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    Oh ... any idea on how to arrive at that solution?2012-07-27
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    I'm thinking! =)2012-07-27
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    One way to arrive at this is to rewrite as $y'' = 2 csc^2(x) y$, which suggests that differentiating once should yield a extra factor of $csc(x)$, and derivative of $cot(x)$ is exactly $csc(x) cot(x)$, so substituting $y = cot(x)f(x)$ would be definitely something to try.2012-07-27
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    Another way to come by what @PeterTamaroff got is by using double angle formulas. $$cos(2x) + 1 = 2cos^2(x)$$ $$sin(2x) = 2sin(x)cos(x)$$ Substituting these in will yield the correct answer.2012-07-27
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    @experimentX: Oh, so the book has a solution? Once you know one nontrivial solution (say $y = \cot(x)$), you can get the others by variation of parameters.2012-07-27
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    lol ... how to get the $ y=\cot x $ is one solution? ... ah, not the complete solution. just answer sheet!2012-07-27
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    @RobertIsrael how do i do this by variation of parameters? to do that ... don't i need another solution?2012-07-28
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    Sorry, I meant reduction of order. If $y(x) = u(x) \cot x$, the differential equation becomes $\sin(x) \cos(x) u'' - 2 u' = 0$, which is a first order equation in $u'$.2012-07-29