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Y''(x) + A sin(y(x))  B = 0; A,B : positive, real 
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#1
Mar812, 08:14 AM

P: 6

Hello there,
I have no idea how to solve the differential equation y''(x) + A sin(y(x))  B = 0 , where A and B are positive real numbers. I do also have initial conditions: y(0) = 0 and y'(0) = 0. I would be grateful for any help. 


#2
Mar812, 10:08 AM

P: 756

y''+A sin(y)B=0
y''y'+A sin(y)y'B y'=0 y'²/2A cos(y)B y =C y'²=2A cos(y)+2B y +2C dy/dx =y' = (+or)sqrt(2A cos(y)+2B y+2C) dx=(+or) dy/sqrt(2A cos(y)+2B y+2C) x =(+or) integal (dy/sqrt(2A cos(y)+2B y+2C)) +c cannot be expressed with a finit number of usual functions; 


#3
Mar812, 10:52 AM

P: 6

Thank you very much.



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