How do I use laplace transforms to determine the solution of a first order differential equation?

Question

I have an exercise with a question that says:

Use the method of Laplace transforms to determine the solution of the first order ordinary differential equation:

4dx/dt + x = 1 + 4cos(2t); x(0)=1

Answer 1

L(x(t)) = X(s)
L(x'(t)) = sX - x(0) = sX - 1
L(cos(2t)) = s / (s^2 + 4)

So altogether:
4*sX - 4 + X = 1/s + 4s/ (s^2 + 4)
X*(4s + 1) = 4 + (5s^2 + 4) / [s*(s^2 + 4)]
X(s) = 4/(4s+1) + (5s^2 + 4) / [s*(s^2 + 4)*(4s+1)]

Now express the RHS in partial fractions and take the inverse transform. I'm using maple to help me through these tedious computations which I really don't want to do.

X(s) = 1/s + (64 + 4s)/[65*(s^2 + 4)] - 16/[65*(4s+1)]
Taking inverse transforms:
x(t) = 1 - 4/65 *exp(-t/4) + (4/65)*cos(2t) + 32/65 *sin(2t)

Answer 2

We're doing Laplace transforms right now in my ODE class too!

This is the answer I got: f(t)= 1+5cos(2t)+8sin(2t). If you have answers in the back of the book, then you can find what I did wrong in my work.

You're looking for the unknown equation f(t). First do a Laplace transform of every term in the equation given. L{f(t)} = F(s), L{f'(t)} = sL{f(t)}-x(0) (look up the other basic forms)

So you get: 4[sF(s)-x(0)]+F(s)= 1/s + 4* s/(s^2+4). The initial conditions state that x(0)=1, so sub that in and solve for F(s). F(s)= (4s^3+6x^2+16s+4)/[ s(s^2 +4).

Then use partial fractions to solve for F(s): set that stuff in the last line = A/s + (Bs+c)/(s^2+4). Multiply everything by s and s^2+4, so 4s^3+6x^2+16s+4 = (A+B)s^2 + Cs + 4A. Equate like coefficients to solve for A, B, C: A=1, B=5, C=16.

Plugging in A, B, C: F(s)= 1/s + (5s+16)/(s^2+4). Separate the last term: F(s) = 1/s + 5[s/(s^2+4)] + 8[2/(s^2+4)]

F(s) = L{f(t)}, so look up the pieces above in a table of Laplace transforms to find f(t): 1/s=L{1},
s/(s^2+k^2)=L{cos (kt)}, k=2
k/(s^2+k^2)= L{sin(kt)}, k=2

so the final answer is: f(t) = 1 + 5cos(2t) + 8 sin(2t)

Answer 3

You take L.T of both sides:

4 ( X(s) - 1 ) + X(s) = 1/s + 4 s/(s^2 + 4 )

4 X(s) - 4 + X(s) = 1/s + 4 s/(s^2 + 4 )

5 X(s) = 4 + 1/s + 4 s/(s^2 + 4 )

the final answer will be in the form of:

x(t) = A + B cos 2t + C sin 2t

Where A,B and C are constants to be found

Answer 4

4[s*x-1)+x=1/s+4 *s/(s^2+4)
(4s+1)x=4+1/s +4s/(s^2+4)
x(s) = 1/(s+1/4) +1/4*1/s(s+1/4)+s/((s+1/4)(s^2+4)
The anti transform of the first summand gives you e^-1/4*t
You must use partial fractions to find the inverse transform
Ex 1/s(s+1/4) = A/s +B/(s+1/4) so
the inverse transform is A +B*e^-1/4 t
s/(s+1/4)(s^2+4) = P/(s+1/4) +(Qs+R)/(s^2+4)