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- May 18th 2008, 02:19 PMszpengchaosolve these two system of differential equation
- May 18th 2008, 02:41 PMgalactus
#2:

I get the following solutions for the respective DE's:

$\displaystyle x=C_{1}e^{5t}-\frac{1}{3}e^{2t}-\frac{3}{5}y$

$\displaystyle y=2e^{t}+2xt+C_{2}$

$\displaystyle z=e^{t}+t(x+y)+C_{3}$

Perhaps that is something for you to start with. - May 18th 2008, 02:46 PMszpengchaonot sure
i m just not familiar with how to do the second question...

would you post your steps? - May 18th 2008, 03:17 PMgalactus
I am sorry, but I done that with my calculator. I just ran it through the calculator to show you what to shoot for.

- May 18th 2008, 04:31 PMNonCommAlg
for your first question, since $\displaystyle \lambda$ is not an eigenvalue of A, the matrix $\displaystyle \lambda I -A$ is invertible. let $\displaystyle \bold{u}=(\lambda I - A)^{-1} \bold{v},$

and $\displaystyle \bold{x}(t)=e^{\lambda t}\bold{u}.$ then $\displaystyle \frac{d \bold{x}}{dt}=\lambda e^{\lambda t}\bold{u}=\lambda \bold{x},$ and hence: $\displaystyle \frac{d \bold{x}}{dt} - A \bold{x} = (\lambda I - A)\bold{x}=e^{\lambda t}(\lambda I - A)\bold{u}=e^{\lambda t}\bold{v},$ which is

what you were asked to prove. now, the example in your first question should be very easy for you to do!

for the second question, call the first differential equation (1), the second (2) and the third (3). first differentiate

both sides of (1) with respect to $\displaystyle t$ to get: $\displaystyle x'' = 5x' + 3y' + 2e^{2t}.$ call this (4). now plug $\displaystyle y'$ from (2) into (4) to get

$\displaystyle x'' - 5x' - 6x = 6e^t + 2e^{2t},$ which is a simple linear equation and gives you the general solution for $\displaystyle x.$ now that we

have $\displaystyle x,$ we can easily find $\displaystyle y$ and $\displaystyle z,$ because by (1): $\displaystyle y=\frac{x'-5x-e^{2t}}{3},$ and by (3): $\displaystyle z=\int(x + y + e^t)dt. \ \ \ \square$