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barking frog
09-27-2003, 03:45 PM
Could someone please outline the method (or provide a link) on how to solve such equations when they have an initial condition? Without an IC the problem is quite trivial, but we never covered such problems in class, nor is it in the book (and yes, I tried using google, but all I could find was it listed in course syllabi and not actually explained).

09-27-2003, 04:09 PM
I'm sorry, but I don't quite see the problem here. Maybe I am misunderstanding the question.

If you solve a linear homogenous 2nd degree ODE (y'' + ay' + by = 0) you should get a function y(t) with two coefficients C1 and C2.

I assume that you are given something like y(t=0) = A and y'(t=0) = B as initial conditions.
Then you just put t=0 into your solution and set it equal to A.
Then you take the first derivative of your solution, set t=0 and equal it to B. from this you can calculate C1 and C2.

Thats all.

Cabbage
09-27-2003, 04:21 PM
Have you used matrices to solve systems of linear equations before? That will do it, but it's not absolutely necessary, since this only involves two equations with two unknowns, which is pretty straightforward.

Say you have:

y'' - 5y' + 6y = 0

with initial conditions

y(0) = 1, y'(0) = 3

You know how to get the general solution:

y(t) = Ae2t + Be3t

From here, first plug in your IC information:

y(0) = A + B = 1,

and (since y'(t) = 2Ae2t + 3Be3t):

y'(0) = 2A + 3B = 3.

Solve the two bolded equations for A and B:

A = 1 - B

2(1 - B) + 3B = 3

2 + B = 3

B = 1

A = 1 - 1 = 0

And so our final solution is:

y(t) = e3t

barking frog
09-27-2003, 04:46 PM
Originally posted by Cabbage
Have you used matrices to solve systems of linear equations before? That will do it, but it's not absolutely necessary, since this only involves two equations with two unknowns, which is pretty straightforward.

Say you have:

y'' - 5y' + 6y = 0

with initial conditions

y(0) = 1, y'(0) = 3

You know how to get the general solution:

y(t) = Ae2t + Be3t

From here, first plug in your IC information:

y(0) = A + B = 1,

and (since y'(t) = 2Ae2t + 3Be3t):

y'(0) = 2A + 3B = 3.

Solve the two bolded equations for A and B:

A = 1 - B

2(1 - B) + 3B = 3

2 + B = 3

B = 1

A = 1 - 1 = 0

And so our final solution is:

y(t) = e3t Got it. Thanks.