# Math Help - Help required? Solve the Differential equation by the Laplace transform method"

1. ## Help required? Solve the Differential equation by the Laplace transform method"

Hello

I wonder if you can help? I believe I'm doing this correctly, but iv got caught up in the algebra. Ill list all the question and then as far as I have got.

Kind regards
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You are required to solve the following differential equation by the Laplace transform method:

d^2x/dt^2 + 5*dx/dt - 24x = 4t

with

x(0) = 2
x'(0)=9

(i)
Take the Laplace transform of the differential equation and solve the resulting equation for X, the Laplace transform of the solution
Give your answer as a function of only. Omit the "X ="

d^2x/dt^2 + 5*dx/dt - 24x = 4t
Taking the Laplace transform

L(d^2x/dt^2) + 5L(dx/dt) -24L (x) = 4L(t)

(s^2X - sx(0)-x'(0)) + 5( sX - x(0)) - 24X = 4/s^2
Sub in values given within questions for x(0) and x'(0)

(s^2X - 2s - 9) + (5sX - 10) - 24X = 4/s^2
Expanding brackets out and collecting like terms

X(s^2+5s-24) = 4/s^2 + 2s +19
Factorising LHS

X((s-3)(s+8)) = 4/s^2 + 2s +19

I can't see where to go next to, iv looked at brining the (s-3)(s+8) to the RHS but I get muddled with all the numbers and it doesn't seem to look correct.

Thanks again

2. Originally Posted by Davidc2233
Hello

I wonder if you can help? I believe I'm doing this correctly, but iv got caught up in the algebra. Ill list all the question and then as far as I have got.

Kind regards
----------------------------------------------------------------------
You are required to solve the following differential equation by the Laplace transform method:

d^2x/dt^2 + 5*dx/dt - 24x = 4t

with

x(0) = 2
x'(0)=9

(i)
Take the Laplace transform of the differential equation and solve the resulting equation for X, the Laplace transform of the solution
Give your answer as a function of only. Omit the "X ="

d^2x/dt^2 + 5*dx/dt - 24x = 4t
Taking the Laplace transform

L(d^2x/dt^2) + 5L(dx/dt) -24L (x) = 4L(t)

(s^2X - sx(0)-x'(0)) + 5( sX - x(0)) - 24X = 4/s^2
Sub in values given within questions for x(0) and x'(0)

(s^2X - 2s - 9) + (5sX - 10) - 24X = 4/s^2
Expanding brackets out and collecting like terms

X(s^2+5s-24) = 4/s^2 + 2s +19
Factorising LHS

X((s-3)(s+8)) = 4/s^2 + 2s +19

I can't see where to go next to, iv looked at brining the (s-3)(s+8) to the RHS but I get muddled with all the numbers and it doesn't seem to look correct.

Thanks again
So far your algebra looks correct and your idea is sound. You need to isolate $X(s)$

This gives
$\displaystyle X(s)=\frac{4}{s^2(s-3)(s+8)}+\frac{2s}{(s-3)(s+8)}+\frac{19}{(s-3)(s+8)}$

Now you need to use partial fraction decomposition on the RHS side. Then you will be able to find the inverse transform.