Linear equation (differential equations) problem

[aero_zeppelin]

Homework Statement

Solve:

dP/dt + 2tP = P + 4t - 2

The Attempt at a Solution

I've done a couple of these but I'm not sure how to start with this one...

First I have to put it in standard form right?

dP/dt + P(2t - 1) = 4t - 2

Then obtain the integrating factor by integrating (2t - 1)

 

[LCKurtz]

Right so far. Now calculate$$
e^{\int 2t-1 dt}$$for your integrating factor.

 

[aero_zeppelin]

Ok! I get:

e^(t^2 - t)

Then we include it in the equation and integrate both sides?

d [e^(t^2 - t) P ] / dt = e^(t^2 - t) * (4t - 2)

The right hand side integral is giving me problems though...

 

[LCKurtz]

Ok! I get:

e^(t^2 - t)

Then we include it in the equation and integrate both sides?

d [e^(t^2 - t) P ] / dt = e^(t^2 - t) * (4t - 2)

The right hand side integral is giving me problems though...

Try a u substitution: $u = t^2-t$ as your first step.

 

[aero_zeppelin]

Oh god, I can't believe I missed that one hehe I was overcomplicating it...

I end up with:

P = [2 e^(t^2 - t) + C ] / e^(t^2 - t)

Interval of definition would be (-inf, inf) correct?

 

[LCKurtz]

Oh god, I can't believe I missed that one hehe I was overcomplicating it...

I end up with:

P = [2 e^(t^2 - t) + C ] / e^(t^2 - t)

Interval of definition would be (-inf, inf) correct?

Yes, since the denominator is never zero.

 

[aero_zeppelin]

Thanks a lot!