- #1
KevinL
- 37
- 0
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.
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.