hey

 

this is not so much a problem that i can't do, its problem that seems to have 2 answers, yet i thought laplace transforms are unique....

 

if i have [math] F(s) = \frac{2s-1}{(s-1)^2s^2} [/math]

 

then if i take the inverse transform it straight from here (either by using maple, or relising that [math] \frac{2s-1}{(s-1)^2s^2} = \frac{s^2 - (s-1)^2}{(s-1)^2s^2} [/math] i get

 

[math] f(t) = -t + te^t [/math]

 

however if i split up F(s) to [math] F(s) = \frac{2s-1}{(s-1)^2s^2} = \frac{2}{(s-1)^2s} - \frac{1}{(s-1)^2s^2} [/math]

 

and then take the inverse transform (either by using maple or convolution integrals), i get

 

[math] f(t) = 2te^t-4e^t+2t+t^2 +4 [/math]

 

can anybody give an insight as to why this is happening? (as even maple seems to get 2 quite different answers....)

 

-Sarah

Are you sure you're evaluating the last inverse laplace transform correctly?

 

I get that the inverse laplace transform gives:

[math]f(x) = 2 (xe^x * 1) - (xe^x * x)[/math] where * represents convolution.

 

In other words:

[math]f(x) = 2\int_0^xte^tdt - \int_0^xte^t(x-t)dt[/math]

Which gives:

[math]f(x) = 2\int_0^xte^tdt - x\int_0^xte^tdt + \int_0^xt^2e^tdt[/math]

[math]f(x) = (2-x)\int_0^xte^tdt + \int_0^xt^2e^tdt[/math]

Apply differentiation by parts to the second term:

[math]f(x) = (2-x)\int_0^xte^tdt + \int_0^xt^2d(e^t)[/math]

[math]f(x) = (2-x)\int_0^xte^tdt + (t^2e^t|_0^x - \int_0^xe^td(t^2))[/math]

[math]f(x) = (2-x)\int_0^xte^tdt + (t^2e^t|_0^x - 2\int_0^xte^tdt)[/math]

[math]f(x) = -x\int_0^xte^tdt + x^2e^x[/math]

Apply differentiation by parts again:

[math]f(x) = -x\int_0^xtd(e^t) + x^2e^x[/math]

[math]f(x) = -x(te^t|_0^x - \int_0^xe^tdt) + x^2e^x[/math]

[math]f(x) = -x(xe^x- e^t|_0^x) + x^2e^x[/math]

[math]f(x) = -x(xe^x- (e^x - 1) + x^2e^x[/math]

[math]f(x) = -x^2e^x + xe^x - x + x^2e^x[/math]

[math]f(x) = xe^x - x[/math]

So, I see no contradiction.

for this bit, when i used convolution integrals i used t^2 instead of t, i.e.

 

[math]

f(x) = 2\int_0^xte^tdt - \int_0^xt^2e^{x-t}dt

[/math]

 

 

but i did put it into maple, and it gives the answer in my original post....

 

so i can see what you did is correct, i just want to know why if you used t^2 instead of t(x-t) you get an almost completely different answer....??

for this bit' date=' when i used convolution integrals i used t^2 instead of t, i.e.

 

[math']

f(x) = 2\int_0^xte^tdt - \int_0^xt^2e^{x-t}dt

[/math]

 

 

but i did put it into maple, and it gives the answer in my original post....

 

so i can see what you did is correct, i just want to know why if you used t^2 instead of t(x-t) you get an almost completely different answer....??

 

That's an incorrect usage of the convolution. Here are the correct forms of the convolution:

 

[math](xe^x * x) = \int_0^xte^t(x-t)dt = \int_0^xt(x-t)e^{x-t}dt[/math]

 

What you wrote is a completely different convolution:

[math]\int_0^xt^2e^{x-t}dt = (e^x * x^2)[/math]

ok , fair enough

 

but what about the fact that maple gives 2 different answers?