Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am new to differential equations, to solve $y'+2y=3e^t$ I used the method of variation of constant and I get $y=e^t+Ce^{-2t}$ but when I use another method I get a different answer (I don't know the name of this method)

$(y'+2y=3e^t)*I(t) \Rightarrow Iy'+2yI=3Ie^t$

$(Iy)'=I'y+Iy'=Iy'+2yI \Rightarrow I'y=2yI \Rightarrow I'=2I$

$I=Ce^2t \Rightarrow \int(Iy)'=\int 3Ie^t \Rightarrow ye^{2t}=e^{3t} \Rightarrow y=e^t$

$y=e^t+Ce^{2t} ≠ y=e^t+Ce^{-2t}$

Please Correct me

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The method you are trying to use here is called the integrationg factor method: http://en.wikipedia.org/wiki/Integrating_factor

In this case, the integrating factor can be $$ I(t)=\exp\left( \int_0^t2ds\right)=e^{2t}. $$ You got this right.

Then multiply the ode by $I(t)$: $$e^{2t}y'+2e^{2t}y=3e^{3t}\quad\Leftrightarrow\quad (e^{2t}y)'=3e^{3t}. $$ Now integrate and don't forget the integration constant: $$ e^{2t}y(t)=e^{3t}+C \quad\mbox{hence}\quad y(t)=e^t+Ce^{-2t}. $$

Alternative: the method of undetermined coefficients.

Given the rhs, we know we can look for a particular solution of the form $$ y_p(t)=Ce^t. $$

Plugging this into the ode, we find that for $C=1$, we do get a solution: $$ y_p(t)=e^t. $$

Then add the general solution of the homogeneous equation $y_h(t)=Ae^{-2t}$ to get $$ y(t)=Ae^{-2t}+e^t $$ the general solution of the ode.

share|improve this answer
    
@julianfernandez Then you should have made your comment below the OP's post... not below my answer. –  1015 Feb 19 '13 at 5:14
    
@julien After I substituted into the original equation for positive one I mean, the equation didn't satisfy , so it means positive one doesn't work? Am I correct –  Hooman Feb 19 '13 at 5:15
    
@Hooman No matter what method you use, the general solution of your ode is $y(t)=e^t+Ce^{-2t}$. Anything different doesn't work. –  1015 Feb 19 '13 at 5:17
    
Thanks perfect solution –  Hooman Feb 19 '13 at 5:20
1  
@Hooman You're welcome. I'm glad if I could help. –  1015 Feb 19 '13 at 5:21

You essentially had it. It's correct that $I$ (the integrating factor, a slightly different but essentially equivalent method) is equal to $Ce^{2t}$. The arbitrary constant would just get cancelled (since it can be moved out of the derivative and exists on both sides), so we can ignore it. We then have

$$(ye^{2t})'=3e^{3t}$$

$$ye^{2t}=e^{3t}+C$$

Don't forget the constant of integration! This is the arbitrary constant you care about that will show up in the final solution. Just isolate for $y$:

$$y=e^{t}+Ce^{-2t}$$

Forgetting constants of integration is a common pitfall when solving ODEs. If something bizarre like this shows up again, in my experience "did I forget the constant?" is one of the first things you should be asking yourself.

share|improve this answer
    
Thanks Man,perfect solution –  Hooman Feb 19 '13 at 5:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.