Question

system of differential equations using laplace transforms:

x' = 3x - 2y x(0)=1
y' = 3y - 2x y(0)=1

I can take the laplace transform of both eqtns but I'm at a loss for what to do next

Answer 1

you should be getting y = e^t, and x = e^t i think.

Answer 2

\[\mathcal{L}\left \{ \frac{dx}{dt} \right \}=3\mathcal{L}\left \{ x \right \}-2\mathcal{L}\left \{ y \right \}\]\[\mathcal{L}\left \{ \frac{dy}{dt} \right \}=3\mathcal{L}\left \{ y \right \}-2\mathcal{L}\left \{ x \right \}\]
\[sX(s)-1=3X(s)-2Y(s)\]\[sY(s)-1=3Y(s)-2X(s)\]
\[sX(s)-3X(s)=1-2Y(s)\]\[sY(s)-3Y(s)=1-2X(s)\]
\[X(s)(s-3)=1-2Y(s)\]\[Y(s)(s-3)=1-2X(s)\]
\[X(s)=\frac{1}{s-3}-2Y(s)\frac{1}{s-3}\]\[Y(s)=\frac{1}{s-3}-2X(s)\frac{1}{s-3}\]
\[X(s)=\frac{1}{s-3}-2\left ( \frac{1}{s-3}-2X(s)\frac{1}{s-3} \right )\frac{1}{s-3}\]\[Y(s)=\frac{1}{s-3}-2\left ( \frac{1}{s-3}-2Y(s)\frac{1}{s-3} \right )\frac{1}{s-3}\]
\[X(s)=\frac{1}{s-3}-2\frac{1}{(s-3)^2}+4X(s)\frac{1}{(s-3)^2}\]\[Y(s)=\frac{1}{s-3}-2\frac{1}{(s-3)^2}+4Y(s)\frac{1}{(s-3)^2}\]
\[X(s)-4X(s)=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]\[Y(s)-4Y(s)=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]
\[-3X(s)=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]\[-3Y(s)=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]
\[X(s)=-\frac{1}{3}\frac{1}{s-3}-\frac{2}{3}\frac{1}{(s-3)^2}\]\[Y(s)=-\frac{1}{3}\frac{1}{s-3}-\frac{2}{3}\frac{1}{(s-3)^2}\]

Answer 3

Sorry for the length: I was actually working it all out here. You need to take the inverse Laplace transform of the last two expressions. I'm going to recheck for mistakes.

Answer 4

for the second pic, since the system of equations is very symmetric, i just added the two equations together to get a third equation, then subtracted them to get a 4th. Then just looking at the 3rd and fourth, i thought i would be easier to get the inverse transform of X+Y and X-Y, rather than figuring out X and Y individually.

Answer 5

I noticed I made one mistake: The last term in both the last expressions should be positive, not negative (division by -3).

Answer 6

when you moved the 4X(s)(1/(s-3)^2) over to the other side of the equation,the (1/(s-3)^2) got lost in the process.

Answer 7

I just noticed I made a huge mistake near the end. Oh, my!

Answer 8

Yes, that one!

Answer 9

I hate my internet...

Forgoing the last 6 lines above:
\[X(s)-4X(s)\frac{1}{(s-3)^2}=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]\[Y(s)-4Y(s)\frac{1}{(s-3)^2}=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]
\[X(s)\left [1-4\frac{1}{(s-3)^2}\right ]=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]\[Y(s)\left [1-4\frac{1}{(s-3)^2}\right ]=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]
\[X(s)\left [\frac{(s-3)^2-4}{(s-3)^2}\right ]=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]\[Y(s)\left [\frac{(s-3)^2-4}{(s-3)^2}\right ]=\frac{1}{s-3}-2\frac{1}{(s-3)^2}\]
\[X(s)=\frac{(s-3)^2}{(s-3)[(s-3)^2-4]}-2\frac{(s-3)^2}{(s-3)^2[(s-3)^2-4]}\]\[Y(s)=\frac{(s-3)^2}{(s-3)[(s-3)^2-4]}-2\frac{(s-3)^2}{(s-3)^2[(s-3)^2-4]}\]
\[X(s)=\frac{s-3}{(s-3)^2-4}-\frac{2}{(s-3)^2-4}\]\[Y(s)=\frac{s-3}{(s-3)^2-4}-\frac{2}{(s-3)^2-4}\]Now take the inverse Laplace transform of those two babies and see what you get. ;]

I'll recheck again for mistakes.

Answer 10

ok thanks to both of you. i understand the work behind the problem and the solution. however what general strategy would you say you use when approaching these types of problems?

Answer 11

I couldn't spot any this time. Good luck!