Reducing final answer of laplace transform

xtipd
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Homework Statement


The problem is not getting the answer to the laplace transform but instead reducing my answer so i dnt lose marks.

If i work out the laplace transform of:
L(t^3 * sinh(4t)) to be
3!/(2(s- 4)^4)- 3!/(s(s+ 4)^4) then how do i add these to get a single fraction? Its doing my head in

The Attempt at a Solution


I know something has to be multiplied but i have no idea what it is...

Thanks in advance!
 
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xtipd said:

Homework Statement


The problem is not getting the answer to the laplace transform but instead reducing my answer so i dnt lose marks.

If i work out the laplace transform of:
L(t^3 * sinh(4t)) to be
3!/(2(s- 4)^4)- 3!/(s(s+ 4)^4) then how do i add these to get a single fraction? Its doing my head in

The Attempt at a Solution


I know something has to be multiplied but i have no idea what it is...

Thanks in advance!

Did you mean \mathcal{L}[t^3 \sinh(4t)]=\frac{3!}{2(s- 4)^4}- \frac{3!}{2(s+ 4)^4}?

If so, the first thing to do would be get rid of the factorials; 3!=3*2 and then cancel the 2 in the denominators.Next, multiply the first fraction (numerator and denominator) by (s+4)^4 and the second fraction (numerator and denominator) by (s-4)^4

Then expand out the numerator and simplify.

You can also simplify the common denominator by noting that (s+4)^4(s-4)^4=[(s+4)(s-4)]^4=(s^2-16)^4
 
Yeh that's what i meant.

Awsome, cheers for the help!

thought it was something like that
 
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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