Can someone help me with this question?

Using as a trial solution solve the differential equation ...

given that x = 1 and at t = 0.

Use your solution to prove that for t>=0, x<=1

Thanks in advance

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- Dec 8th 2008, 02:18 AMintomathDifferential Equation
Can someone help me with this question?

Using as a trial solution solve the differential equation ...

given that x = 1 and at t = 0.

Use your solution to prove that for t>=0, x<=1

Thanks in advance - Dec 8th 2008, 02:32 AMmr fantastic
First solve to get the homogenous solution. Can you do this?

You're given that the particular solution has the form (you should think about why that is) so you need to get the value of A. Start by calculating and . Can you do this?

Note: Once you have the value of A and hence the particular solution you can construct the complete solution by adding together the homogenous and particular solutions. The boundary conditions x = 1 and at t = 0 are used to get the value of the arbitrary constants. - Dec 8th 2008, 02:53 AMintomath
This is how i worked it out so far:

Using trial solution

therefore

comparing coefficients:

....Particular Integral

Considering

Let be a trial solution

therefore

u = -1(twice) therefore real equal roots

hence - Complimentary Function

x = Particular Integral + Complimentary Function

therefore ....[A]

putting x = 1, t = 0 in [A]

**This is where i'm stuck, i don't know how to carry on from here.** - Dec 8th 2008, 03:10 AMmr fantastic

.

Carefully substitute the above into the DE and simplify. You will be left with .

So .

Now use the given boundary conditions x = 1 and http://www.mathhelpforum.com/math-he...ffa3a8fe-1.gif at t = 0 to solve for a and B. - Dec 8th 2008, 03:15 AMintomath
Thanks a lot you saved my day