How to solve this 2nd order nonlinear differential equation

In summary, a grad student in meteorology is seeking help with a 2nd order non linear differential equation, which can be solved analytically by using the substitution y' = u and reducing it to a first order system of two ODEs. The second equation is separable and solvable, leading to a straightforward analytic solution. Additional assistance is requested for rewriting u in terms of y.
  • #1
tornado681
7
0
Hello all,

This is the first time I've 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
 
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  • #2
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
slider142 said:
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
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
Defennder said:
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 don't 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
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
Defennder said:
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
tornado681 said:
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
HallsofIvy said:
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
Defennder said:
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
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
tornado681 said:
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
Actually he was referring to this post:
tornado681 said:
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
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
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
Miriam100 said:
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.
 
Last edited:
  • #20
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
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
Suggest than use the program wxmaxima or maple!
 
  • #23
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.
 

1. How do I identify a 2nd order nonlinear differential equation?

A 2nd order nonlinear differential equation is one that contains a second derivative of the dependent variable, as well as nonlinear functions of the dependent variable and its derivatives. It can be identified by looking for terms with the form y'', y'^2, or y*y'.

2. What is the general form of a 2nd order nonlinear differential equation?

The general form of a 2nd order nonlinear differential equation is:y'' + P(x, y)y' + Q(x, y) = 0where P and Q are functions of both x and y.

3. How do I solve a 2nd order nonlinear differential equation?

There is no single method for solving all 2nd order nonlinear differential equations. However, some common techniques include using substitution, separation of variables, and applying specific methods for certain types of equations (e.g. Bernoulli or Riccati equations).

4. Can I use a computer to solve a 2nd order nonlinear differential equation?

Yes, there are many software programs and online calculators available that can solve 2nd order nonlinear differential equations. However, it is important to understand the underlying mathematical concepts and check the results for accuracy.

5. What are the applications of 2nd order nonlinear differential equations in science and engineering?

2nd order nonlinear differential equations are commonly used to model complex physical systems, such as fluid dynamics, heat transfer, and electrical circuits. They are also used in the study of population dynamics, chemical reactions, and many other fields of science and engineering.

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