# Discontinuous Forcing terms

1. Mar 24, 2012

### jhendren

1. The problem statement, all variables and given/known data

Find Laplace transformation
H(t-1)t^2

2. Relevant equations

3. The attempt at a solution

[2(e^-s)f(t^2)]/s^3

I'm not sure how to get t^2 in terms of t-1

I know the answer is (2+2s+s^2)/s^3

2. Mar 24, 2012

### LCKurtz

Just express $t^2$ as a finite Taylor series about $t=1$. Alternatively you can use this formula$$\mathcal L(u(t-a)f(t)) = e^{-as}\mathcal L(f(t+a)$$which would give, in your case$$\mathcal L(u(t-1)t^2)=e^{-s}\mathcal L((t+1)^2) = e^{-s}\mathcal L(t^2+2t+1)$$

3. Mar 25, 2012

### jhendren

can you show me the proof on that formula because it is no where in my book

4. Mar 25, 2012

### LCKurtz

$$\mathcal L(u(t-a)f(t)=\int_a^\infty e^{-st}f(t)dt$$Let $v = t-a$$$=\int_0^\infty e^{-s(v+a)}f(v+a)dv =e^{-as}\int_0^\infty e^{-sv}f(v+a)dv =e^{-as}\mathcal L(f(t+a))$$This formula is handy because, as in your problem, the function f(t) isn't usually given in terms of the argument of the step function.