|Mar8-12, 08:14 AM||#1|
y''(x) + A sin(y(x)) - B = 0; A,B : positive, real
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.
|Mar8-12, 10:08 AM||#2|
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;
|Mar8-12, 10:52 AM||#3|
Thank you very much.
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