How to solve this 2nd order nonlinear differential equation

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,

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.

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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.

slider,

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

Defennder
Homework Helper
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.

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.

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:

d^2y/dx^2 = d/dy (dy/dx) dy/dx

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Defennder
Homework Helper
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.

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.

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,

anyone care to comment on the solution?

anyone???

HallsofIvy
Homework Helper
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,

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.

Defennder
Homework Helper
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.
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).

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).

Defennder,

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

Defennder
Homework Helper
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
Homework Helper
Defennder,

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

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

Defennder
Homework Helper
Actually he was referring to this post:
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,

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.

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..

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.

Mute
Homework Helper
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.

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 $ky(t) \ll \omega_0^2$.

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

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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.

To recap:

$$u \equiv y'$$

Then, the second derivative became:

$$y'' = \frac{d y'}{d x} = \frac{d u}{d y} \frac{d y}{d x} = u \frac{d u}{d y}$$

$$u u' + a y u + b y = c$$
where $u' \equiv du/dy$. I think some previous posters made a mistake in converting the second derivative.