Solve $$\sin^2 x {d^2y \over dx^2} = 2 y$$
Ans: $$y \left( x \right) ={\frac {c_{{1}} \left( \cos \left( 2\,x \right) +1
\right) }{\sin \left( 2\,x \right) }}+{\frac {c_{{2}} \left( x\cos
\left( 2\,x \right) -\sin \left( 2\,x \right) +x \right) }{\sin
\left( 2\,x \right) }}$$
Or, $ y \tan (x) = c_1(\tan x - x) + c_2 $

Question

Solve $$\sin^2 x {d^2y \over dx^2} = 2 y$$
Ans: $$y \left( x \right) ={\frac {c_{{1}} \left( \cos \left( 2\,x \right) +1
 \right) }{\sin \left( 2\,x \right) }}+{\frac {c_{{2}} \left( x\cos
 \left( 2\,x \right) -\sin \left( 2\,x \right) +x \right) }{\sin
 \left( 2\,x \right) }}$$
Or,  $ y \tan (x) = c_1(\tan x - x) + c_2 $

experimentx · Answer

more on this question on this thread http://math.stackexchange.com/questions/175954/hints-on-solving-sin2-x-d2y-over-dx2-2-y/175961#comment404667_175961

experimentx · Answer

also feel free to update this thread!!

anonymous · Answer

Yes but I want to solve it with my own way.
thanks for your suggestion.

anonymous · Answer

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