y''(x) + A sin(y(x)) - B = 0; A,B : positive, real

tarquinius

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.

JJacquelin

y''+A sin(y)-B=0
y''y'+A sin(y)y'-B y'=0
y'²/2-A 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;

tarquinius

Thank you very much.