Question

x'=-x+y
y'=-4x+3y
Obtain a general solution by seeking solutions in the form qe^rt. One eigenvalue will be defective. Solve for it by seeking a solution in the form:
x(t) = (q+pt)*e^t

I'm getting lost looking for the defective eigenvalue portion.

Answer 1

By the 1st equation:\[\frac{d(q+pt)*e^t}{dt}=y-x\]
\[qe^t+pe^t+pte^t=y-x\]
\[e^t(q+p(1+t))+x=y\]
By the 2nd equation:
\[\frac{d(e^t(q+p(1+t))+x)}{dt}=3(e^t(q+p(1+t))+x)-4x\]
Differentiate it, you've got 2 unknowns (not including t) and 2 equations, so solve them simultaneously

Answer 2

I'm supposed to solve for this using matrices. Because one eigenvalue is defective, I'm having a lot of trouble. We're supposed to solve for the second solution using the modified form:
x(t)=(q+pt)e^rt

I've attached a pdf which shows a similar problem being solved.The solution method is pretty clear to me until we get to that second page. I understand how they get the constants for that set of equations, but I don't understand they're assigning the q1 and q2 values. I also don't understand how they go from that system of equations to solve for alpha and beta in terms of one another. Where is beta coming from? It's like it just materializes! I don't understand.

Answer 3

With regards to solving DE with matrices I am more clueless than yourself. I eagerly await an answer to this question.