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

by tarquinius
Tags: positive, real, sinyx
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tarquinius
#1
Mar8-12, 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.
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JJacquelin
#2
Mar8-12, 10:08 AM
P: 761
y''+A sin(y)-B=0
y''y'+A sin(y)y'-B y'=0
y'/2-A cos(y)-B y =C
y'=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
#3
Mar8-12, 10:52 AM
P: 6
Thank you very much.


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