Diff Eq problem, Laplace Transform

DF19
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Homework Statement


Find the Laplace Transform of the given function
H(t-1)t^2

I'm not sure how to add (t-1) to the t^2 term to solve the problem

Any help would be greatly appreciated.
 
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That notation makes no sense, are you trying to translate the function on the s-axis? If you are what is the original function.
 
If the original function is e^t*t^2 the translation = F(s-1) and since f(t)=t^2 its transform is 2/s^3, therefore you substitute s-1 in for s and get 2/(s-1)^3. Unless you are doing an inverse translation on the t-axis.
 
My book states that it is a H(t) is a heavyside function and I'm suppose to use the proposition:

L{H(t-c)f(t-c)}(s)=e^(-cs)F(s)
 
Oh a unit step therefore you must translate both H(t) and F(t). LH(t-1)=e^s and f(t)=t^2, L(t^2)=2/s^3, so the transform is 2e^s/s^3
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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