How to solve this 2nd order nonlinear differential equation

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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
 

Answers and Replies

  • #2
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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.
 
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  • #3
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,

--tornado
 
  • #4
Defennder
Homework Helper
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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
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
--tornado
 
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  • #6
Defennder
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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.
 
  • #7
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:

[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
 
  • #8
anyone care to comment on the solution?
 
  • #9
anyone???
 
  • #10
HallsofIvy
Science Advisor
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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
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'= aeC.
 
  • #11
Defennder
Homework Helper
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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'= aeC.
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).
 
  • #12
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).
Defennder,

How do you rewrite u in terms of y only? Can it be done?
 
  • #13
Defennder
Homework Helper
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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
HallsofIvy
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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"
 
  • #15
Defennder
Homework Helper
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Actually he was referring to this post:
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
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
2
0
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..
 
  • #17
3
0
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
Mute
Homework Helper
1,388
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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 [itex]\omega_0^2[/itex]? If it's supposed to be small, you could do perturbation theory to get an approximation solution valid when [itex]ky(t) \ll \omega_0^2[/itex].
 
  • #19
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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  • #20
3
0
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.
 
  • #21
2,967
5
To recap:

[tex]
u \equiv y'
[/tex]

Then, the second derivative became:

[tex]
y'' = \frac{d y'}{d x} = \frac{d u}{d y} \frac{d y}{d x} = u \frac{d u}{d y}
[/tex]

and your equation becomes:

[tex]
u u' + a y u + b y = c
[/tex]

where [itex]u' \equiv du/dy[/itex]. I think some previous posters made a mistake in converting the second derivative.
 
  • #22
3
0
Suggest than use the program wxmaxima or maple!
 
  • #23
3
0
Thanks for advice. However, it turned out that I constructed a wrong differential equation at the beginning. A new one was very simple to solve.
 

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