- Sep 10, 2005
- 7,458
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Ok, the course is Differential Equations. And my question is..
When you have a system of equations, typically its in the form
vector(x(t)) = Matrix[A] * [x1; x2]
I know how to find the generall solution when the form is like that. But when there's an additional term at the end such as: [3; 1] exp(t). How do you account for that?
I have 2 questions that I'd like to throw in if anyone would care to solve em and explain
The formatting is kinda ugly tho, but here goes.
A. [x1; x2] ' = [-2, 0; 1, -1][x1; x2] + [3; 1]exp(t)
B. [x1; x2] ' = [0, 1; 1, 0][x1; x2] - [2; 3]
For question A, I have everything in the solution except for the last part which is exp(t)*[1; 1]. How do you get the last term?
When you have a system of equations, typically its in the form
vector(x(t)) = Matrix[A] * [x1; x2]
I know how to find the generall solution when the form is like that. But when there's an additional term at the end such as: [3; 1] exp(t). How do you account for that?
I have 2 questions that I'd like to throw in if anyone would care to solve em and explain
The formatting is kinda ugly tho, but here goes.A. [x1; x2] ' = [-2, 0; 1, -1][x1; x2] + [3; 1]exp(t)
B. [x1; x2] ' = [0, 1; 1, 0][x1; x2] - [2; 3]
For question A, I have everything in the solution except for the last part which is exp(t)*[1; 1]. How do you get the last term?
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