Solving a Differential Equation with a Constant and Initial Conditions

[alejandrito29]

Hello

I need help with the following differential equation:

(1-\frac{gh}{c^2}) A(u) - \frac{h^2}{3} A''(u) - \frac{3}{2h} A(u)^2 =0

with g,h,c=constant

the answer has a \sech^2 with A(0)=A_0 and A'(0)=0

thanksution[/b]

 

[tiny-tim]

hello alejandrito29!

that's A'' = pA - qA² with p and q constant

start by multiplying both sides by A', and then integrating

 

[alejandrito29]

differential equation1

i have the differential equation

A''=p A - q A^2

i multiplying by A' both sides then

A' A''=p A A' - q A^2A' then

(\frac{1}{2}(A')^2)'=\frac{p}{2} (A^2)' - \frac{q}{3} (A^3)'
then i integer and:

(\frac{1}{2}(A')^2)=\frac{p}{2} (A^2) - \frac{q}{3} (A^3)

but i write in maple this differential equation and i don't obtain the solution. This solution must have of the way A(x)=k_1 sech ^2 (k_2 x)

help please