General solution of a second order nonlinear DE

In summary, the student is trying to solve an equation of motion using a Maple-based solver, but is not sure how to proceed. When they get to a point where they cannot solve the equation using substitution, they realize they need to use the Lagrange method. When they do this, they find that the u(R) function is first order, linear, and nonhomogenous. However, they are lucky since the non-homogenous term consists of powers of R only and can be solved easily using the Lagrange method.
  • #1
ephedyn
170
1
1. Homework Statement + relevant equations

I have to solve
http://home.vs.moe.edu.sg/linl/eqn1.gif

sigma, pinfinity, rho, N are constants. To make things easier for us, we are allowed to treat T as a constant.

2. The attempt at a solution

Treat T as constant:

http://home.vs.moe.edu.sg/linl/eqn2.gif

At this point, this might help... Using a Maple-based solver, I get:

dsolve('Rho*(y*D2y+(3*Dy^2)/2)=-(pinf+(2*sig)/R-(N*T)/(R^3))')
Warning: Explicit solution could not be found; implicit solution returned.
> In dsolve at 312

ans =

Int(Rho*_a^2*R^2*3^(1/2)/(Rho*_a*R*(C1-2*_a^3*pinf*R^3-4*_a^3*sig*R^2+2*_a^3*N*T))^(1/2),_a = .. y)-t-C2 = 0
Int(-Rho*_a^2*R^2*3^(1/2)/(Rho*_a*R*(C1-2*_a^3*pinf*R^3-4*_a^3*sig*R^2+2*_a^3*N*T))^(1/2),_a = .. y)-t-C2 = 0

But I'm not sure how to read '_a = .. y)-t-C2 = 0' in symbolic.

Anyway, moving along... using substitution:

http://home.vs.moe.edu.sg/linl/eqn3.gif

I don't know how to proceed from here, but using the same solver,

y denotes U:
dsolve('Rho*(R*y*Dy+3*y^2/2)=-(pinf+2*sig/y-N*T/y^3)','R')
Warning: Explicit solution could not be found; implicit solution returned.
> In dsolve at 312

ans =

log(R)-Int(2/(2*N*T-3*Rho*_a^5-2*pinf*_a^3-4*sig*_a^2)*Rho*_a^4,_a = .. y)+C1 = 0

Does anyone know how I can get to either solution or at least simplify the problem further from here? I don't know how to introduce an implicit solution into my working, too.

Thanks.
 
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  • #2
When you have a DE like this, with a quadratic form of the derivative and the second derivative, you should always try the following. This is also a de-facto method of solving equations of motion in mechanics. So write the second derivative as:

[tex]\ddot R = \frac{d\dot R}{dR}\frac{dR}{dt}=\frac{d\dot R}{dR}\dot R = \frac12 \frac{d\dot R^2}{dR}[/tex]

Now put the new variable as [tex]u=\dot R^2[/tex]

So our original DE is:

[tex]R\frac12 \frac{du}{dR}+\frac32 u = f(R,T)[/tex]

Dividing out:

[tex]\frac{du}{dR}+\frac{3}{R}u=\frac{2f(R,T)}{R}[/tex]

Now what is this? this is a Differential equation for the [tex]u(R)[/tex] function. As we see it is first order, and linear, nonhomogenous. But we are lucky since the non-homogenous term consists of powers of R only. So this can be solved easily using the Lagrange method, of variating the constants..
After we get solve this we get the u(R) function. And then we just need to take the square root and integrate to get the R(t) function, which looks like it can be given in this case.
 
  • #3
Ahhh! Thanks! I wasn't too confident so I didn't want to jump in for a reply before I got it, but I solved it at last.
 
  • #4
Oh wait, I was wondering what you meant by "which looks like it can be given in this case". Because I verified that

http://home.vs.moe.edu.sg/linl/eqnsol.gif

But its square root cannot be integrated to a form containing standard functions, so I'm certain I have to make assumptions at this point (fluid mechanics problem). Nonetheless, thanks a lot for your help!
 
Last edited by a moderator:
  • #5
Then I was probably wrong, I didnt actually calculate anything just guessed.. :D Looks like it was a wrong guess :D
 
  • #6
Thaakisfox said:
Then I was probably wrong, I didnt actually calculate anything just guessed.. :D Looks like it was a wrong guess :D

:D No problem, I'll just have to cheat a little.
 

1. What is a general solution of a second order nonlinear differential equation?

A general solution of a second order nonlinear differential equation is a function that satisfies the equation and contains two arbitrary constants. It can be used to find specific solutions by substituting values for the constants.

2. How is a general solution of a second order nonlinear differential equation different from a specific solution?

A specific solution is a function that satisfies the equation and contains specific values for the constants, whereas a general solution contains arbitrary constants. A general solution can be used to find specific solutions by substituting values for the constants.

3. What is the process for finding a general solution of a second order nonlinear differential equation?

The process for finding a general solution involves solving the differential equation algebraically, using methods such as separation of variables, substitution, or the method of undetermined coefficients. This results in a general solution containing two arbitrary constants.

4. Can a general solution of a second order nonlinear differential equation be used to find all possible solutions?

Yes, a general solution can be used to find all possible solutions by substituting different values for the arbitrary constants. This allows for a range of solutions to be generated from a single general solution.

5. How can a general solution of a second order nonlinear differential equation be verified?

A general solution can be verified by substituting it into the original differential equation. If the equation is satisfied, then the general solution is correct. It can also be verified by using initial or boundary conditions to find specific solutions and comparing them to the general solution.

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