Laplace Transform to solve differential?

Question

Solve the differentials by laplace transform
x''''(t) + x'''(t) = cos (t)
x(0) = x'(0) = x'''(0) = 0 x'' (0)=1

So i haven't covered laplace transforms in about 2 years and what i remember was the teacher saying "it wasn't on the final, so we won't cover it."
Need to know it now, and current prof expects us to already know it.....only went over what to do after we transform.

This is what i did.
S^4 x(s) - s^3 x(0) - s^2 x ' (0) - s x''(0) - x'''(0) +
s^3 x(s) - s^2 x(0) - s x'(0) - x''(0) = s^2 / (s^2 + 1)

Sub in the boundy conditions and solve for x(s). Is this correct so far?

thanks in advance

Not asking you guys to solve, just writing the problem statement exactly how it appeared.

Just if i'm doing it right.

Answer 1

Yes, this is correct so far, though usually we distinguish between the function of s and the function of t by writing x(t) and X(s). So you should have
s^4 X(s) - s^3 x(0) - s^2 x'(0) - s x''(0) - x'''(0) + s^3 X(s) - s^2 x(0) - s x'(0) - x''(0) = s^2 / (s^2 + 1)
i.e.
s^4 X(s) - s + s^3 X(s) - 1 = s^2 / (s^2 + 1)

Good luck with the rest of it!

Answer 2

♠ laplace or not laplace, I’d better lower the range first;
♣ x’’’ +x’’ =sint +C; 0+1=0+C; thus x’’’ +x’’ =sint +1;
♦ x’’ +x’ =-cost +t +C; 1+0=-1+0+C; thus x’’ +x’ =-cost +t+2;
♥ x’ +x=-sint +0.5t^2 +2t +C; 0+0=0+0+0+C; thus x’ +x=-sint +0.5t^2 +2t;
▲ and here is a real problem: we are lucky to do this x’+x=0; and here comes laplace: k+1=0, k=-1; thus x=u*exp(-t), u being integration constant to be varied; now return to ♥; x’ +x=-sint +0.5t^2 +2t;
or u’exp(-t) –u·exp(-t) +u·exp(-t) = -sint +0.5t^2 +2t; or u’ =exp(t)·(-sint +0.5t^2 +2t);
▼integrating by parts: u= (0.5(cost -sint) +0.5t^2 +t –1)·exp(t)+C;
x=0.5(cost -sint) +0.5t^2 +t –1 +C*exp(-t); 0=0.5-1+C; thus
x=0.5(cost -sint) +0.5t^2 +t –1 +0.5exp(-t);