Now taking the Laplace transform we get

Now we take the inverse laplace transfrom to get

Attachment 6864

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- Jun 18th 2008, 06:45 PMTheEmptySet

Now taking the Laplace transform we get

Now we take the inverse laplace transfrom to get

Attachment 6864 - Jun 18th 2008, 06:50 PMChris L T521
Excellent. I got something similar to that. You can post your question now if you have one.

- Jun 18th 2008, 06:58 PMTheEmptySet
- Jun 24th 2008, 08:18 AMChris L T521
- Jun 24th 2008, 10:23 AMTheEmptySet
Here is another way to do this. Complete the square on the first term in the denominator to get

Let

Note that

So now...

- Jun 24th 2008, 02:03 PMChris L T521
- Jun 25th 2008, 06:10 PMChris L T521
Instead of ODEs, why not try a PDE?

Find the general solution to the boundary value problem for the 1-D Wave Equation:

where is the displacement function of a vibrating string with fixed ends, is the initial position function, and is the initial velocity function.

Show all steps. - Jul 15th 2008, 06:54 PMpankaj
- Jul 20th 2008, 09:51 PMChris L T521
The best way to solve this would be to set up two boundary value problems [where each one has one nonhomogeneous condition], and then solve the problems using the technique of

**Separation of Variables**:

**Problem A**:

If we let , then the PDE becomes:

Now separate the variables:

The two functions and agree if they both are equal to the same constant. We now let :

What we now have is an eigenvalue problem. We can now solve two ordinary DEs:

Since the endpoint conditions and require that if is non-trivial. Thus, must satisfy the eigenvalue problem

If , then the eivenvalue problem becomes:

When solved, the general solution is . When we apply the endpoint conditions, we have . This gives us the trivial solution . Thus, is NOT an eigenvalue.

If , we will let . The eigenvalue problem now becomes:

When solved, the general solution is:

where

Applying the endpoint condition , we have so that . When we apply the other endpoint , we have . Thus, since and for . Thus the only solution is the trivial solution . As a result, there are no negative eigenvalues.

Now, when , we let . The eigenvalue problem then becomes:

The general solution has the form:

Applying the endpoint condition , we have so that . But if we apply the other condition , we have . This can occur when when is a postive multiple of :

Therefore

Solving for , we get:

Since , we can say that:

We now can see that we have an infinte sequence of eigenvalues defined by:

The associated eigenfunction is

-------

***

The homogeneous initial condition

Implies that . Thus, the solution that is associated with the eigenvalue must also satisfy the conditions

After some calculations [which I have omitted], we get the general solution:

To apply the condition, we need to find :

Applying the condition , we get:

Therefore, . Thus, can be defined as the non-trivial solution

Since , we can say that

Each of these terms satisfy the wave equation and the homogeneous boundary condition given at the start of this part of the problem. Thus, by the principle of superposition, we define as the inifinte series:

All we need to do now is find such that it satisfies the nonhomogeneous condition

for . However, this will be the Fourier Sine Series of if we choose

Therefore, the**Formal Series Solution**to Problem A is

where

----------------------------------------------

**Problem B**

Let us start from ***:

The homogeneous intial condition

implies that . Thus, the solution associated with the eigenvalue must also satisfy the conditions

After some calculations [which I have omitted again...], the general solution takes the form

Applying the initial condition, we get

Thus, can be defined as the non-trivial solution

Since , we can say that

Each of these terms satisfy the wave equation and the homogeneous boundary condition given at the start of this part of the problem. Thus, by the principle of superposition, we define as the inifinte series:

Find :

All we need to do now is find such that it satisfies the nonhomogeneous condition

for . However, this will be the Fourier Sine Series of if we choose

Therefore, the**Formal Series Solution**to Problem B is

where

-------------

Thus, the**Formal Series Solution**to the**One Dimensional Wave Equation**is

(Whew)

That was a lot of work... (Emo)

--Chris - Jul 21st 2008, 02:40 AMMarine

here's my suggestion - easy and simple :D

x''+0,4x'+9,4x = 3e^(-t/5)cos(3t)

ok, x_h is easy so I'll omit it ;)

for x_p, we have to go from R to C and I'll use the differential operator

D = d/dt

x''+0,4x'+9,4x = 3e^(-t/5)*е^(3it)

(D^2 + 0,4D + 9,4)x = 3e^([3i-1/5]t) and let p(D) be the char. polynomial

p(D)x = 3e^([3i-1/5]t)

so, going back to R, a particular solution is:

x_p = Re{ 3([3i-1/5]t)/p(3i-1/5) } , where p(3i-1/5) is not 0, atherwise it's: x_p = Re{ 3t([3i-1/5]t)/p'(3i-1/5) }

then the well-known procedure - finding the general solution and the 2 constants :) - Jul 21st 2008, 10:08 AMThePerfectHacker
Another equation is here.

Have you ever learned Sturm-Liouville Theory? (It is about eigenvalues and eigenfunctions of a boundary value problem). - Jul 21st 2008, 10:15 AMChris L T521
No I haven't. I plan on learning it on my own when I find the time. If I don't learn it, I'm taking a class in the spring on the last part of Diff Equ [Laplace Trans, Fourier Series and Transforms, Power Series, Sturm-Liouville and other topics]...

If I did know how to apply Sturm-Liouville Theory, by how much would the amount of work needed to solve this equation reduce?

--Chris