how to solve this 2nd order nonlinear differential equation

tornado681

Hello all,

This is the first time Ive stumbled across this site, but it appears to be extremely helpful. I am a meteorology grad student, and in my research, I have run across the following 2nd order non linear differential equation. It is of the form:

y'' + a*y*y' + b*y=0

where a and b are constants

Can this equation be solved analytically? If not, what program does one recommend for solving it numerically? There is also a slightly more complex form of this equation:


y'' + a*y*y' + b*y=c

where a, b and c are constants

If anyone could assist me in solving this or direct me to a source for solving it numerically, it would be most appreciated.

Thanks,

--tornado

slider142

Try the substitution y' = u to reduce it to a first order system of two ODEs. (Source:Tenenbaum/Pollard)
Ie., your first ODE becomes the system y' = u, uu' = -ayu - by, where in the second eq. u is treated as u(y) and u' = du/dy, which we can then plug into the first equation to integrate for y(x). The second equation is separable, so there is a straightforward analytic solution.

tornado681

slider,

Im not following you. Could you go into a let more detail if possible. Thanks,

--tornado

Defennder

Assuming y is function of x, y' = u. u' = d^2y/dx^2 = d/dy (dy/dx) dy/dx by the chain rule. This is equivalent to u du/dy. Hence u' = u du/dy.

After substituting for y' = u, the ODE is u' = -ayu - by. Replacing u' with u du/dy gives:
u du/dy = -y(au+b) This equation is separable and hence solvable. Once you have u(y), you have dy/dx = u(y), which is again separable and solvable.

tornado681

defennder,

I understand how you get: u' = -ayu - by when you set y'=u

I dont understand how u'=u du/dy . I appreciate you trying to work me through this. Any additional explanation would be appreciated.

Specifcally, how is this so:

--tornado

Defennder

That follows from the chain rule.

[tex]\frac{d^2y}{dx^2} = \frac{d}{dx} \left ( \frac{dy}{dx} \right ) = \frac{d}{dy} \left ( \frac{dy}{dx} \right ) \ \frac{dy}{dx}[/tex]

Replace [tex]\frac{dy}{dx}[/tex] with u.

tornado681

When I solve u' = -ayu - by I get:

[tex]\frac{u}{a}-\frac{b}{a^2}ln[|{au+b}|]+C_1 =-\frac{1}{2}y^2 + C_2 [/tex]

So then we need to make the above equation u=u(y) correct? Since we have a ln (natural log), is this possible? Any more help is most appreciated. Thanks,

--tornado

tornado681

anyone care to comment on the solution?

tornado681

anyone???

HallsofIvy

I don't get that. From du/dy= -y(au+ b) we can get du/(au+b)= -ydy so, integrating both sides, (1/a)ln(au+ b)= -(1/2)y²+ C. I don't know what you mean by "make u= u(y) correct". Solve for u? Without your additional "u/a" that's easy:
[tex]au+ b= C'e^{-\frac{y^2}{2}}[/tex]
where C'= ae^C.

Defennder

Actually u' isn't du/dy.

[tex]u' = \frac{du}{dx} = \frac{du}{dy} (\frac{dy}{dx}) = u \frac{du}{dy}[/tex]

That's where the u/a term comes, once you do long division of u/(au+b).

tornado681

Defennder,

How do you rewrite u in terms of y only? Can it be done?

Defennder

Honestly I have no idea if it's possible. It never occurred to me earlier because I didn't actually attempted the DE itself, I just noted it would be solvable if such could be done.

HallsofIvy

Yes, that is a simple application of the chain rule. In fact, that application to differential equations is a standard method called "quadrature"

Defennder

Actually he was referring to this post:I really don't see how to write u in terms of y there. And anyway the post which resurrected this thread and which preceded yours appears to have been deleted.

ceramica

Does anyone can help me to solve this second order non linear ODE:

y'' + (2/x)(y') - (1/2y)(y')(y') = K,

y' = dy/dx

y'' = dy'/dx

y = y(x)

I've already guess y=Ax^2 satisty this equation, but I want to solve it analitically..

Please help!
Thank before..

Miriam100

Hi,
I'm trying to solve y''(t) + w₀² y(t) = k y(t)² sin(w t). Does anybody know, how to get a particular solution?

Thanks.