Subhotosh Khan's answer is very good (in fact, the way, I probably would do this) but does not use the "Laplace Transform".
Use Laplace transforms to solve the initial value problem
x"+y'−3y=0 x(0)=2 x'(0)=0
y"+x'−3x=0 y(0)=2 y'(0)=0
where primes indicate derivatives with respect to t.
I am so confused. First I tried (s^2)X(s) + sY(s) - 3Y(s) = 0,
(s^2)Y(s) + sX(s) - 3X(s) = 0 but I really don't know what is going on and how to solve this.
Okay, that's good. Now you have two equations for X and Y. Solve them just as you would two algebraic equations
\(\displaystyle s^2X+(s- 2)Y= 0\) and \(\displaystyle s^2Y+ (s- 3)X= 0\). If you multiply the first equation by \(\displaystyle s^2\) you get \(\displaystyle s^4X+ s^2(s- 2)Y= 0\). If you multiply the second equation by s- 2 you get \(\displaystyle (s-2)(s-3)X+ s^2(s- 2)Y= 0\). Now subtract the second of those from the first to eliminate Y.
And, just for fun, here's yet a third way to solve this system. let u= x' and v= y' so x''= u' and y''= v'. Then the two original equations become x''+ y'- 3y= u'+ v- 3y= 0 and y''+ x'- 3x= v'+ u-3x= 0 so we can write this system of two second order equations as a system of four first order equations, x'= u, y'= v, u'= 3y- v, and v'= 3x- u. Those are linear equations so we can write them as matrix equations:
\(\displaystyle \begin{bmatrix}x \\ y \\ u \\ v\end{bmatrix}'=\)\(\displaystyle \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 3 & 0 & -1 \\ 3 & 0 & 0 & -1\end{bmatrix}\)\(\displaystyle \begin{bmatrix}x \\ y \\ u \\ v\end{bmatrix}\)
Find the eigenvalues and eigenvectors of that "coefficient" matrix".
