Boo. No kisses from me then.
But anyways, here's the solution:
---------------------------------
y'+2y = te^(-2t)
at y(1) = 0
---------------------------------
y'+2y = te^(-2t)
becomes:
m(t)y'+m(t)2y = m(t)te^(-2t)
now m(t) is an unknown function, so we have to solve it so that the left side of the equation (the y'+2y) becomes integrateable and have the y' be eliminated so we do this:
m(t) = e^( S2dt) = e^(2t)
so now we have:
(e^(2t))y'+(e^(2t))2y = (e^(2t))te^(-2t)
now before you integrate both sides, you have to eliminate the y'. And to do this, you just do the product rule for differentiation backwards (this is not the same thing as integration). So you end up with:
(d/dt)ye^(2t) = t (the e^(2t) and e^(-2t) on the right side cancel each other out)
Now when you integrate both sides, you get:
ye^(2t) = (t^2)/2 + c
Isolate the y now and you get:
y = ((t^2)/2 + c)/(e^(2t))
------------------------------------
now at the beginning, we were given y(1) = 0. Basically this means that when t = 1, y is 0, so let's plug those into the new equation:
0 = ((1^2)/2 + c)/(e^2t))
solve for c algebraically, and you end up with:
c = -1/2
now replace c with -1/2 and you have your answer:
===================
y = (t^2 - 1)/(2(e^(2t)))
===================
None.