General Solution of dy/dt = -2ty + 4e^(-t^2) w/ Initial Value Problem

[KevinL]

For the first question, I am only supposed to find the general solution of the differential equation.

1) dy/dt = -2ty + 4e^(-t^2)
dy/dt + 2ty = 4e^(-t^2)

Integrating factor = e^(Integral of 2t) = e^(t^2)

Multiply both sides by IF:

e^(t^2) * (dy/dt +2ty) = 4e^(t^2-t)

e^(t^2)*dy/dt +2te^(t^2)y = 4e^(t^2-t)

e^(t^2)y' + (e^(t^2))'y = 4e^(t^2-t)

(e^(t^2)y)' = 4e^(t^2-t)

Take integral of both sides:

e^(t^2)y = (integral of) 4e^(t^2-t)

The right hand side is impossible to integrate (unless I'm missing something?). So we just divide both sides by e^(t^2) and leave that as our answer. There was an example in the book where a problem was left like this because the integral was impossible to do, so I'm assuming this is a 'legal' thing to do.

2) Solve the initial value problem of dy/dt = -2ty + 4e^(-t^2)

This is the exact same problem, just with an initial value problem. I don't understand how to do this with an integral on the right hand side. I have a feeling the answer would involve an integral with bounds but I am kinda grasping at straws on that one.

 

[rock.freak667]

when you multiply both sides by e^t2, the right side is


$$4*e^{-t^2} * e^{t^2}=4e^{-t^2+t^2}$$