Question

Use the Laplace transform to solve differential equation
dx/dt + 7x = 5cos(2t)
Initial conditions: x(0) = 4, x'(0) = -4
Assume forcing functions are 0 prior to t = 0-

Answer 1

anyone?

Answer 2

using Laplace transform we have:
L(x')+7L(x) = 5L(cos(2t))
sL(x)-x(0) + 7L(x) = 5s/(s^2+4)
(s+7)L(x)- 4 = 5s/(s^2+4)
(s+7)L(x) = (5s - 4s^2 -16)/(s^2+4)

=> L(x) = -(4s^2 - 5s +16)/(s^2+4)(s+7)

now the boring part, using partial fractions we separate 1/(s^2+4)(s+7) that is:
(7-s)/[53(s^2+4)] + 1/53(s+7). So:

L(x)= (1/53)[(-28s^2+4s^3-4s^2+35s-5s^2+5s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]
L(x)= (1/53)[(4s^3 -37s^2 +40s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]

denoting T:= L^(-1)
and x= (4/53) T(s^3/(s^2+4)) - (37/53)T(s^2/(s^2+4)) +(40/53) T(s^2+4)
-(4/53) T(s^2/s+7) +(5/53)T(s/s+7) - (16/53) T(1/s+7)

finally, you need to replace the antitransformed known.
Sorry for my bad English and ask me if you have any questions

Answer 3

thanks man appreciate all that but I got the solution...using pretty much the same method however I think on your right hand side you should have 4s^2 + 5s + 16 because you're adding 4 to both sides

Answer 4

oh you're right! but the idea is the same, just "despeja" (i don't know how to say that in English) the Transform, separate by PF and take antitransform. Ah! and replace the initial condition

Answer 5

yeah...agree thanks!