Laplace transform question, urgently need help!

[jamesf123]

I have an exam tomorrow on advanced discrete maths and there's an exercise question example in my book with no answer because i missed the lecture we went through it, and i'm certain I need to know how to do it, if any one can do this, please answer or at least have a go as it will help me greatly! thank you

Solve the following simultaneous differential equations using the Laplace transform method,

4dx/dt + dy/dt + y = 2t
d²x/dt² - dy/dt = 0

Include in your answer a brief description of how the method is working at each step.

[jemidiah]

I don't know what this has to do with discrete math. In any case, denote the Laplace transform of x by X and of y by Y. Since the Laplace transform is linear, L(t) = 1/s^2, L(df/dt) = sL(f) - f(0), and L(d^2 f / dt^2) = s^2 L(f) - s f(0) - f'(0), the equations become in the s-domain...

4 (sX - x(0)) + (sY - y(0)) + Y = 2/s^2
(s^2 X - sx(0) - x'(0)) - (sY - y(0)) = 0

From the second equation, we can easily solve for X in terms of Y:
X = (sx(0) + x'(0) + sY - y(0))/s^2

Multiplying the first equation by s and then plugging the above in gives
4 (x'(0) + sY - y(0)) + (s^2 Y - sy(0)) + sY = 2/s

Solving for Y gives
(5s + s^2)Y = 2/s + sy(0) - 4 (x'(0) - y(0))
=> Y = 2/[s(5s + s^2)] + y(0)/[5 + s] - 4(x'(0) - y(0))/[5s + s^2]

Compute the inverse Laplace transform of this term-by-term after applying partial fractions, leveraging linearity and the following properties of the inverse Laplace transform L':
L'(1/(s+a)) = e^(-at)
L'(1/s) = 1
L'(1/s^2) = t

From partial fractions, 1/[5s + s^2] = 1/5 * [1/s - 1/(s+5)] and 1/[s(5s+s^2)] = 1/5 * 1/s^2 + 1/25 * [1/(s+5) - 1/s]. Putting everything together gives

y = L'(Y) = 2 * [1/5 * L'(1/s^2) + 1/25 * L'(1/(s+5) - 1/s)] + y(0) L'(1/(5+s)) - 4(x'(0) - y(0)) * 1/5 * L'(1/s - 1/(s+5))
= 2t/5 + 2/25 * (e^(-5t) - 1) + y(0) e^(-5t) - 4(x'(0) - y(0))/5 * (1 - e^(-5t))

These calculations are very error-prone, though I've done my best to be accurate. I also checked the above result with the result of Mathematica, and they appear to agree. The Mathematica output uses constants that aren't in terms of x(0), x'(0), and y(0), though the appearance of an exponential term, a constant term, and 2t/5 appear in both.

I won't do the X calculations, since they're so similar and these problems are so tedious.


Disclaimer: I have hardly worked with Laplace transforms, and even my small experience was something like 7 years ago. In the above, I believed the properties Wikipedia gave in the tables at the Laplace transform article.