Intergrating Factor - dy/dt = -2ty + 4e^-t^2

[killersanta]

Homework Statement

Solve the differential equations [Find the general solution y(t)=KYh(t) + Yp (t)]. Use the method of finding the integrating factor.

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

The Attempt at a Solution

S = intergrat

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

P(t) = 2t
b(t) = 4e^-t^2

U(t) = e^(S 2t dt)
U(t) = e ^t^2

Y(t) = 1 / e^t^2 * S e^t^2 * 4e^-t^2

I guess I just don't know how to intergrat e^t^2 * 4e^-t^2

 

[dextercioby]

e^{t^2} e^{-t^2} = 1

, as one function is the inverse of the other.

 

[killersanta]

Would the 4 go down to? Like this:

<br /> e^{t^2} / 4e^{t^2}<br />

Wouldn't that leave you with 1/4?

 

[dextercioby]

No, with a 4. The 4 is in the numerator, if you write 4e^{-t^2} as a fraction.

 

[killersanta]

int_/ 4*e^{t^2} / e^{t^2} dt

Then, the e^{t^2} cancel out and you are left with int_/ 4 dt which is just 4t. Is this right?

 

[dextercioby]

You should be more certain of yourself. You dropped the integration constant.

 

[killersanta]

oh, yeah... I meant 4t + c...So the whole answer is y(t) = {4t + C}/e^{t^2} (The e^{t^2} is under both)...Yeah, I second guess myself way too much. Most the time, after I ask for help, I realize how easy a problem was and can't believe I ask for help.