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First Differential Equation

  • Thread starter TG3
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  • #1
TG3
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Homework Statement



Find the general solution for y' +3y = t + e^(-2t) for y.

The Attempt at a Solution



At first I thought that since the equation was already separated, I could simply integrate both sides and get a solution easily:

That results in 1.5 y^2 + y = .5t^2 - .5e^(-2t)
(Unless I made a simple error, which is quite possible.)

However, I quickly realized that this is not the approach to take, since it still contains multiple powers of y.

So, I suspect that I will need to find an integrating factor to multiply by, (commonly called mu, I believe) but I'm not sure how you're supposed to find that.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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For an equation

[tex]\frac{dy}{dt}+P(t)y=Q(t)[/tex]


an integrating factor u is given by u=e∫P(t) dt
 
  • #3
33,181
4,860

Homework Statement



Find the general solution for y' +3y = t + e^(-2t) for y.

The Attempt at a Solution


At first I thought that since the equation was already separated, I could simply integrate both sides and get a solution easily:
It might look separated to the most casual observer, but it's not, and further, it's not separable. You have
dy/dt + 3y = t + e-3t

As it sits, there's no way to get all of the terms involving y and dy on one side, and the other terms involving t and dt on the other side. An integrating factor, as rock.freak667 suggested, is the way to go.
 

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