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inverse laplace question
- Thread starter smoothda
- Start date
D
Deleted member 4993
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Hello!
Can anyone help me with this problem.
What is inverse laplace y" + y = t ???
when..
y(0) = 1
y`(0) = -2
would be nice if i can see entire calculation...please help me.
/mike
if the Laplace Transform of y(x) = f(s)
then
what is the Laplace Transform of y'(x)?
what is the Laplace Transform of y"(x)?
HallsofIvy
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- Jan 27, 2012
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I'm not sure what you mean by "inverse Laplace" here. If you mean "use the Laplace transform method to solve this problem", then you start by finding the Laplace transform of both sides. And typically, we find Laplace tranforms and their inverses by looking them up in table. A good one is at http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf.
If we call the Laplace transform of y, "Y(s)" then the Laplace transform of y'', according to that table, is \(\displaystyle s^2Y(s)- sy(0)- y'(0)\) and the Laplace transform of \(\displaystyle t\) is \(\displaystyle \frac{1}{s^2}\) so that your differential equation transforms to the algebraic equation \(\displaystyle s^2Y(s)- s+ 2+ Y(s) = \frac{1}{s^2}\). Then \(\displaystyle (s^2+ 1)Y(s)= \frac{1}{s^2}+ s- 2= \frac{s- 2s^2+ 1}{s^2}\) so that \(\displaystyle Y(s)= \frac{s- 2s^2+ 1}{s^2(s^2+ 1)}\).
Reduce that using "partial fractions", then look up the inverse transform in the table of transforms.
If we call the Laplace transform of y, "Y(s)" then the Laplace transform of y'', according to that table, is \(\displaystyle s^2Y(s)- sy(0)- y'(0)\) and the Laplace transform of \(\displaystyle t\) is \(\displaystyle \frac{1}{s^2}\) so that your differential equation transforms to the algebraic equation \(\displaystyle s^2Y(s)- s+ 2+ Y(s) = \frac{1}{s^2}\). Then \(\displaystyle (s^2+ 1)Y(s)= \frac{1}{s^2}+ s- 2= \frac{s- 2s^2+ 1}{s^2}\) so that \(\displaystyle Y(s)= \frac{s- 2s^2+ 1}{s^2(s^2+ 1)}\).
Reduce that using "partial fractions", then look up the inverse transform in the table of transforms.
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