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