Linear equation (differential equations) problem

  • #1

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)
 

Answers and Replies

  • #2
LCKurtz
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Right so far. Now calculate$$
e^{\int 2t-1 dt}$$for your integrating factor.
 
  • #3
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...
 
  • #4
LCKurtz
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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.
 
  • #5
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?
 
  • #6
LCKurtz
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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.
 
  • #7
Thanks a lot!
 

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