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UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\[\begin{align*} \\\int_0^\infty e^{3t}\int_0^te^{tu}\sin(u)\text du\text dt&=\mathcal L\left\{\int_0^te^{tu}\sin(u)\text du\right\}_{p\rightarrow3}\\ \\&=\mathcal L\left.\left\{\mathcal L\{e^t*\sin(u)\}\right\}\right_{p\rightarrow3}\\ \\&=\mathcal L\left.\left\{\frac{1}{p1}\times\frac{1}{p^2+1^2}\right\}\right_{p\rightarrow3}\\ \\&= \end{align*}\]

geoffb
 2 years ago
Best ResponseYou've already chosen the best response.0I could literally mash the keyboard for 12 straight minutes, and I would understand it better than I would that.

geoffb
 2 years ago
Best ResponseYou've already chosen the best response.0\[u = \frac{1}{2} i (t\log(e^{t}))\] right?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1i have this , but it dosent seam right to me for some reason \[\begin{align*} \\&=\mathcal L\left\{\frac{1}{31}\times\frac{1}{3^2+1}\right\} \\&=\mathcal L\left\{\frac{1}{2}\times\frac{1}{10}\right\} \\&=\frac 1{20} \end{align*}\]

geoffb
 2 years ago
Best ResponseYou've already chosen the best response.0Oh, I honestly have no clue. I put it in Wolfram Alpha and it told me that was a root. It makes no sense to me, but 1/20 is a nice answer. ;)

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1ah, this makes more sense , i see where i was getting confused now , \[\begin{align*} \int_0^\infty e^{3t}\int_0^te^{tu}\sin(u)\text du\text dt \\&=\mathcal L\left\{\int_0^te^{tu}\sin(u)\text du\right\}_{p\rightarrow3} \\&=\mathcal L\left.\left\{e^t*\sin(u)\right\}\right_{p\rightarrow3} \\&=\left(\left.\mathcal L\left\{e^t\right\}\times\mathcal L\left\{\sin(u)\right\}\right)\right_{p\rightarrow3} \\&=\left(\left.\frac{1}{p1}\times\frac{1}{p^2+1^2}\right)\right_{p\rightarrow3} \\&=\frac{1}{31}\times\frac{1}{3^2+1}\\&=\frac{1}{2}\times\frac{1}{10}\\ \\&=\frac 1{20} \end{align*}\]
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