1. I've found the same result as in yours. Is it weird or is that the true result? y= Ax+ B/x + C x ln|x|

2. Originally Posted by freakymath
3) ty''+(1-2t)y'-2y=0 y(0)=1 , y'(0)=2 Laplace
For the Laplace question, this should also be straight forward. Just apply the laplace transformation to each term. You can look up the transformations on a table.
Once you get the solution in the transformed space, use the table to get the inverse laplace transform of your solution.

3. Originally Posted by freakymath
I've found the same result as in yours. Is it weird or is that the true result? y= Ax+ B/x + C x ln|x|
You should actually have a value for C (I just left it as a variable since I didn't do the work).
The C x ln|x| is the particular solution.

Why is it weird? You can plug it back in to check.

4. Maybe i mustn't force my limits! Does that laplace question have a long answer?

5. Originally Posted by freakymath
Maybe i mustn't force my limits! Does that laplace question have a long answer?
If you know how to do it, it is straightforward, but requires a bit of work.

For your test at least know how to setup the question.
ty''+(1-2t)y'-2y=0
Apply laplace denoted by L
L(ty'' + (1-2t)y' - 2y) = L(0)
L(ty'' + y' - 2ty' - 2y) = 0
L(ty'') + L(y') - 2L(ty') - 2L(y) = 0

If you have time, you look up the laplace transforms of these individual terms on a table (probably provided with the exam) that looks like this: http://en.wikipedia.org/wiki/Laplace...s_and_theorems

6. Thx for ur all kind replies sir -> Take care

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