Fun integration for a differential equation

Bad-Wolf
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Homework Statement


This is just a piece of a differential equation where I'm just trying to solve this single gross integration and as there are no initial conditions laplace transforms are out so here it is:

\int t^{-2}e^{-2t}


The Attempt at a Solution


I tried doing a integration by parts then another to loop back and equate terms but this resulted in everything canceling kinda like I would expect. I tried mathematica but it does one integration by parts and then leaves it with the second integral there and then proceeds to laugh at me.
 
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I don't think you can integrate that in terms of elementary functions. I'm going to guess you probably made a mistake earlier on.
 
Dick is right
Mathematical gives
-(1/(E^(2*t)*t)) - 2*ExpIntegralEi[-2*t]
 
Alrighty, if you think I made a mistake earlier on then here is the full problem.

Notice that the equation y'' - 4y'+4y' = t^{-2} is equivalent to w'-2w=t^{-2} and y'-2y=w Solve first for w and then for y.

for attacking ole w there I just used an integrating factor which you get as e^-2t which is apparent from just looking at it integrate -2 then the exponential to kill of the natural log on the integrating factor itself yada yada

I'm not sure what my prof is trying to proove with this equation. We're done with nth order differential equations so long as their constant coef and fit undetermined coef format and a few others so what's the dealio with this 2 equations that equals this other equation thing? He just tacked this single problem on the end with no context to anything else :p

Oh ya, that is exactly what mathematica gave me. I wouldn't past the prof to just want us to leave these sick integration lying around or something and leave them in the final solution. I mean if the problem had values we could numerically integrate or even use laplace transforms.
 
Looks like you mean y''-4y'+4=1/t^2. But, no, nothing wrong with your work. It looks like the main point was just to show an example of solving a 2nd order equation by solving two 1st order ones instead. That the homogeneous part requires the Ei function to integrate it may be a mistake.
 
lol I can append a negative sign or I can go through the trouble of making it a fraction with a power; so no, that is what you mean. But you so basically all I found out was that it is as gross as I thought it was.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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