# Thread: Initial value problems

1. ## Initial value problems

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
}

2.
y" - 2y' + 2y = 8e^(-t)cos t ; y(0) = 16; y'(0) = -16

2. [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)

3. I'm sorry, but would you please show me how you got that very first step?? I can follow what you did after that... Thank you.

4. Originally Posted by Hollysti
I'm sorry, but would you please show me how you got that very first step?? I can follow what you did after that... Thank you.
That would be:

Take the LT of the DE:

(s^2 + 9)Y = 6(e^{-pi s}+1)/(s^2+1)
The left hand side is the LT of y''+9y (with the implicit assumption that
y(0)=0 and y'(0)=0, which is why this will be a particular integral of the
DE rather than a general one, the initial conditions could be left in here
but it will complicate the subsequent algebra, and its the initial conditions
that we need any way).

The right hand side is the LT of the right hand side of the DE, you can
find this by looking up the LT in a table or by doing the required integration
yourself.

Note there is a mistake in my algebra later which I will correct by the time
you read this.

Looking at the solution again there is something about it I don't like, I
will be looking at it again later today

RonL

5. I left this problem with the partial result that the Laplace Transform Y(s) of
the required solution y(t) satisfying the required initial conditions was:

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

Now if we put:

F(s) = 6(e^{-pi s}+1)/(s^2+1),

and:

G(s) = 1/(s^2+9)

we gave that F(s) is the Laplace Transform of:

f(t) = 6.sin(t) 0<=t<=pi
f(t) = 0 otherwise.

and G(s) is the Laplace Transform of:

g(t) = (1/3) sin(3t).

Now we may invoke the convolution theorem to conclude that:

y(t) = integral_{tau=0 to t} f(tau) g(t - tau) dtau.

..... = integral_{tau=0 to min(t, pi)} 2 sin(tau)sin(3(t-tau)) dtau

so:

y(t) = integral_{tau=0 to t} 2 sin(tau)sin(3(t-tau)) dtau for 0<t<pi

..... = integral_{tau=0 to pi} 2 sin(tau)sin(3(t-tau)) dtau for t>=pi

and I will leave these integrals as an exercise for the reader (at least untill I have time
to do them myself)

RonL

6. I left us with the solution in the form:

y(t) = integral_{tau=0 to t} 2 sin(tau)sin(3(t-tau)) dtau for 0<t<pi
..... = integral_{tau=0 to pi} 2 sin(tau)sin(3(t-tau)) dtau for t>=pi

Now using QuickMath to do these integrals (QM does not like multi-character
variable names so we have to use something like u in place of tau there) we get:

integral_{tau=0 to t} 2 sin(tau)sin(3(t-tau)) dtau = sin^3(t)

and:

integral_{tau=0 to pi} 2 sin(tau)sin(3(t-tau)) dtau = 0

so:

y(t) = sin^3(t) for 0<t<pi
..... = 0 for t>=pi,

Which is very satisfying as that agrees to a high level of precission with
the result of my numerical integration of the DE.

RonL