#### mrambitious

##### New member

- Joined
- Jul 2, 2019

- Messages
- 1

- Thread starter mrambitious
- Start date

- Joined
- Jul 2, 2019

- Messages
- 1

- Joined
- Jan 27, 2012

- Messages
- 5,042

In this case, given x'= 2x- y and y'= 5x- 2y, differentiating, y''= 5x'- 2y'= 5(2x- y)- 2y= 10x- 7y. From y'= 5x- 2y, 5x= y'+ 2y so 10x= 2y'+t 4y. y''= 2y'+ 4y- 7y= 2y'- 3y. y''- 2y'- 3y= 0 has characteristic equation \(\displaystyle r^2- 2r- 3= (r- 3)(r+ 1)= 0\). r= 3 and -1 so the general solution for y is \(\displaystyle y(t)= C_1e^{3t}+ C_2e^{-t}\).

x(t) can then be solved from \(\displaystyle 5x= y'+ 2y\): \(\displaystyle y(t)= C_1e^{3t}+ C_2e^{-t}\) so \(\displaystyle y'(t)= 3C_1r^{3t}- C_2e^{-t}\). \(\displaystyle 5x= 3C_1e^{3t}- C_2e^{-t}+ 2C_1e^{3t}+ 2C_2e^{-t}= 5C_1e^{3t}+ C_2e^{-t}\) so \(\displaystyle x(t)= C_1e^{3t}+ \frac{C_2}{5}e^{-t}\).

Now, \(\displaystyle x(0)= C_1+ \frac{C_2}{5}= 1\) and \(\displaystyle y(0)= C_1+ C_2= 1\). Subtracting the first equation from the second, \(\displaystyle \frac{4C_2}{5}= 0\) so \(\displaystyle C_2= 0\). Then \(\displaystyle C_1= 1\).

So the solution to this system of equations is

\(\displaystyle x(t)= e^{3t}\) and \(\displaystyle y(t)= e^{3t}\).

- Joined
- Jan 27, 2012

- Messages
- 5,042

You arrive at the two equation in X(s) and Y(s):

X(s)(s- 2)= Y(s)+ 1 and Y(s)(s- 2)= 5X(s)+ 1.

You then try to write X(s) in terms of Y(s) and Y(s) in terms of X(s). That doesn't lead any where! The whole point of the Laplace transform is that we have reduced the system of differential equations to a system of algebraic equations which we can solve using algebra.

From X(s)(s- 2)= Y(s)+ 1, we get Y(s)= X(s)(s- 2)- 1. Putting that into the other equation (X(s)(s- 2)- 1)(s- 2)= X(s)(s- 2)^2- (s- 2)= 5X(s)+ 1. X(s)((s- 2)^2- 5)= (s- 2)+ 1= s- 1. \(\displaystyle X(s)= \frac{s- 1}{(s- 2)^2- 5}= \frac{s- 1}{s^2- 4x- 1}\).

And then \(\displaystyle Y(s)= X(s)(s- 2)- 1= \frac{(s- 1)(s- 2)}{s^2- 4x- 1}- 1\).

Can you work out the inverse Laplace transform, to get x(t) and y(t), from that?

That's a rather unnecessary slap in the face to engineers and really I expected better from you.Gosh, I dislike the "Laplace transform". I have never seen a problem that couldn't be done more easily using direct methods rather than "Laplace transform"! I suspect the "Laplace transform" just makes engineers feel better since it only involves "plugging and cranking" rather than thinking!

- Joined
- Jan 27, 2012

- Messages
- 5,042

Sorry, I shall try to behave from now on.

(Some of my best friends are engineers!)

(Some of my best friends are engineers!)