Need Help Solving a 2nd Order Nonlinear Differential Equation!

[frank1234]

Solve the 2nd order nonlinear differential equation, with initial conditions y(0)=0 and y'(0)=1

y''=2ay^3-(a+1)y with a within [0,1]

I am pretty much lost on how to go about solving this. It would be greatly appreciated if someone could point me in the right direction on this. Thanks!

 

[Ishuda]

Solve the 2nd order nonlinear differential equation, with initial conditions y(0)=0 and y'(0)=1

y''=2ay^3-(a+1)y with a within [0,1]

I am pretty much lost on how to go about solving this. It would be greatly appreciated if someone could point me in the right direction on this. Thanks!

Now if that were the first derivative (y') instead of the second derivative (y'') the solution would be much easier. Any chance of that?

 

[HallsofIvy]

Notice that the independent variable, which I will call "x", does not appear in the equation.

If we let v= dy/dx then we have \(\displaystyle \frac{d^2y}{dx^2}= \frac{dv}{dx}= \frac{dv}{dy}\frac{dy}{dx}= v\frac{dv}{dy}\).

In terms of v and y the equation becomes \(\displaystyle v\frac{dv}{dy}= 2ay^3- (a+1)y\) or \(\displaystyle vdv= (2ay^3- (a+ 1)y)dy\)

Integrating, \(\displaystyle \frac{1}{2}v^2= \frac{1}{2}\left(\frac{dy}{dx}\right)^2= \frac{a}{2}y^4- \frac{a+ 1}{2}y^2+ C\)
or \(\displaystyle \frac{dy}{dx}= \pm\sqrt{ay^4- (a+ 1)y+ C}\)
\(\displaystyle \frac{dy}{\pm\sqrt{ay^4- (a+ 1)y+ C}}= dx\)