# How to solve this 2nd order nonlinear differential equation

1. May 15, 2008

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,

2. May 15, 2008

### 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.

Last edited: May 15, 2008
3. May 15, 2008

slider,

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

4. May 15, 2008

### 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.

5. May 15, 2008

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:

Last edited: May 15, 2008
6. May 15, 2008

### Defennder

That follows from the chain rule.

$$\frac{d^2y}{dx^2} = \frac{d}{dx} \left ( \frac{dy}{dx} \right ) = \frac{d}{dy} \left ( \frac{dy}{dx} \right ) \ \frac{dy}{dx}$$

Replace $$\frac{dy}{dx}$$ with u.

7. May 27, 2008

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

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

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,

8. May 29, 2008

anyone care to comment on the solution?

9. Jun 2, 2008

anyone???

10. Jun 2, 2008

### HallsofIvy

Staff Emeritus
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:
$$au+ b= C'e^{-\frac{y^2}{2}}$$
where C'= aeC.

11. Jun 2, 2008

### Defennder

Actually u' isn't du/dy.

$$u' = \frac{du}{dx} = \frac{du}{dy} (\frac{dy}{dx}) = u \frac{du}{dy}$$

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

12. Jun 4, 2008

Defennder,

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

13. Jun 4, 2008

### 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.

14. Jul 28, 2008

### HallsofIvy

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

15. Jul 29, 2008

### 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.

16. Apr 22, 2010

### 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..

Thank before..

17. Dec 12, 2010

### 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.

18. Dec 12, 2010

### Mute

There's probably no general analytic expression. At least, wolfram alpha doesn't give one, even when I give it initial conditions.

How large is k supposed to be compared to $\omega_0^2$? If it's supposed to be small, you could do perturbation theory to get an approximation solution valid when $ky(t) \ll \omega_0^2$.

19. Dec 14, 2010

### michaelc187

try (f(x+0) - f(x))/0
divide out the zero
and say tadah, solved numerically. whatever the hell that means.

Last edited: Dec 14, 2010
20. Dec 18, 2010

### Miriam100

Thanks for your help. The instructions say 'solve the problem analitically', so I guess I counstructed a wrong equation. It's without y^2, so there's no problem anymore. Tnx.