[quote=Hollysti;49174]I have been working on this hw set all weekend and it just isn't clicking!! Any help would be greatly appreciated.

Solve the given initial value problems by Laplace transforms. Show the details of your work.

1.

y" + 9y = {

6 sin t if 0 <= t <= pi

0 if t > pi, y(0) = 0, y'(0) = 0

}

Take the LT of the DE:

(s^2 + 9)Y = 6(e^{-pi s}+1)/(s^2+1)

(note the implicit assumption that y(0)=0, and y'(0)=0 here since

Ly'' = s^2 Y - s y(0) - y'(0) )

so:

Y = 6(e^{-pi s}+1)/[(s^2+1)(s^2+9)

.. = (6/8)(e^{-pi s}+1)/(s^2+1) - (6/8)(e^{-pi s}+1)/(s^2+9)

Both of the terms on the right are LT's of single sin arches, so:

y_1(t) = (6/8) f(t) - (6/8) g(t)

where:

f(t) = sin(t) 0<=t<=pi

f(t) = 0 otherwise.

and:

ERROR, this is not the inverse LT of (e^{-pi s}+1)/(s^2+9)

g(t) = (1/3) sin(3 t) 0<= t <= pi/3

g(t) = 0 otherwise

This y is a particular integral of the DE (which as I assumed that y(0)=0 and y'(0)=0

when I tool the LT is actualy the answere required the following is not strikly necessary)

the general solution is this plus a solution of the homogeneous equation:

y'' + 9y = 0

which is Asin(3t) + Bcos(3t).

Hence the general solution is:

y(t) = y_1(t) + Asin(3t) + Bcos(3t).

Now y(0)=0, forces B=0, and:

y'(0) means that:

cos(0) - (1/3) 3 cos(0) + A 3 cos(0) = 0

so A = 0.

RonL

(check this very carefully there is almost certainly an error in the algebra)