ODE Problem

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January 11th 2011, 04:27 PM
Mark24

ODE Problem

I have this ODE:
d^2t/dx^2 = -S/K where S and K are constants. Boundary conditions are: T=T1 at x=0 T=T2 at x=L.

It looks so simple, but i'm having some issues. I used the auxiliary equation first to get the complimentary part: m^2=0, real roots at 0,0 means an equation of form: Tc=A+Bx with A,B as unknown constants.

Next, i'm trying to get the particular solution using the method which I believe is MUC. This is where i'm a bit confused. I assumed a solution in the form of: Tp=-S/K (constant). But then my general solution becomes: T(x)=A+Bx-S/K... and I don't think I can have two constants (both A and S/K) in there since one would be absorbed.

In this case, would I assume Tp=-S/K*x^2 since a constant already exists? But if I do this, I wind up with:
T = A + Bx -S/K*x^2
T(0) = T1 = A
T(L) = T2 = T1 + B*L - S/K*L^2 -> B = (T2/L - T1/L + S/K*L)
T(x)=T1 + (T2/L - T1/L + S/K*L)*x - S/K*x^2 which i'm being told is wrong, and that each S/K term should be S/(2K). I can't see why though.
January 11th 2011, 04:56 PM
dwsmith

You were correct, sorry.

$m^2=0$

$t(x)_c=C_1e^{0*x}+C_2xe^{x*0}\Rightarrow t(x)=C_1+C_2x$

The differential operator D annihilates constants.

$D(D^2)=0\Rightarrow D^3=0\Rightarrow m^3=0\Rightarrow m_1=m_2=m_3=0$

$t(x)=C_1+C_2x+C_3x^2$

$\displaystyle t''(x)=-\frac{s}{k}=2C_3\Rightarrow C_3=-\frac{s}{2k}$

$\displaystyle t(x)=C_1+C_2x-\frac{s}{2k}x^2$
January 11th 2011, 07:21 PM
Mark24

edit: Nevermind, I figured it out.

What I was doing wrong was putting Tp=-S/K*x^2 instead of Tp=Cx^2 (leaving a constant in.) At first I thought I couldn't solve it because I was only given 2 initial conditions, but then I figured out that I also had the original DE that I could use to solve for the constant C.
I plugged Tp into the original diff eq after taking the derivative twice, and that gives me the correct answer.

Thanks
January 11th 2011, 07:22 PM
dwsmith

What is MUC?
January 11th 2011, 07:59 PM
Prove It

This can be solved using direct integration...

$\displaystyle \frac{d^2t}{dx^2} = -\frac{S}{K}$

$\displaystyle \frac{dt}{dx} = \int{-\frac{S}{K}\,dx}$

$\displaystyle \frac{dt}{dx} = -\frac{S\,x}{K} + C_1$

$\displaystyle t = \int{-\frac{S\,x}{K} + C_1\,dx}$

$\displaystyle t = -\frac{S\,x^2}{2K} + C_1\,x + C_2$ .
January 11th 2011, 08:38 PM
Chris L T521

Quote:

Originally Posted by dwsmith

What is MUC?

The Method of Undetermined Coefficients.

I noticed you helped the user using the annihilator approach. In essence, the method of the undetermined coefficients and annihilator method are the "same". The thing that makes them different is that in the annihilator method, you get the proper (general) form of the particular solution without having to guess at it. Using MUC, you would have to guess at the general form based off of your solution to the associated homogeneous equation and any nonhomogeneous terms that may appear in the ODE.