Solving a Second Order Differential Equation with Laplace Transform

[ehanes7612]

Homework Statement

this one stumped me..

d^2y/dt^2 +ωy=ksin((√ω)t)

y(∏/4)=0, y'(∏/4)=0

The Attempt at a Solution

→ (s^2 + ω)U(s)= LT {ksin((√ω)(T+∏/4)} is as far as i can get (i know what to do with the left hand side once i get the LT of the right hand side but i don't know what to do with the sin value)

 

[HallsofIvy]

What is "T"? Did you mean "t"?

I presume you know the definition:
L(f(t))= \int_0^\infty e^{-st}f(s)dx

Here that woud be
\int_0^\infty e^{-st}(k sin(\sqrt{\omega}(t+ \pi/4))dt

The substitution v= \sqrt{\omega}(t+ \pi/4) reduces that to a fairy straight forward "integration by parts" but you should also have formulas for the Laplace transform of sin(x) as well as for f(x+ a) in terms of the Laplace transform of f(x).

 

[ehanes7612]

well, i would normally use tau but i couldn't find symbol, so i just used T to distinguish from t

doing this by integration by parts ends with a complicated (a) (from sin at) that is impossible to break down into to a/s^2+a^2 (LT formula) ...if it were just sin√wt then i get it..but i am missing some step