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.

How can the following differential equation can be solved? $$ \frac{dy}{dt}=3+e^{-t} -\frac{1}{2}y $$ I proceeded by by rearranging the equation as follows $$ \frac{dy}{dt}+\frac{1}{2}y=3+e^{-t} $$My idea was to make the LHS a derivatives of two variables so that it could be integrated. But apparently I could not do that. How should i proceed now?
Your help is much appreciated.Thankyou.

share|improve this question
    
If $y$ is a solution, then multiplying by $e^{\frac t2}$ we have $\dfrac{d(y(t)e^{\frac t2})}{dt} =3e^{\frac t2}+e^{-\frac t2}$. –  Davide Giraudo Jul 23 '11 at 9:57
    
@nihilisticgeek: please see: Earl A. Coddington: An Introduction to Ordinary Differential Equations, p39. It can be very useful. –  leo Jul 25 '11 at 1:07

1 Answer 1

You need to use what's called an integrating factor. Since the coefficient of $y$ is simply the constant $1/2$, the factor is simple: $\mu = e^{\int 1/2 dt} = e^{t/2}$. If you multiply both sides of the differential equation by $\mu$, you can "factor" the left-hand side as an implicit differentiation like so:

$$ \mu y' + 1/2\cdot\mu y = \mu\cdot(3+e^{-t}); $$ $$(\mu y)' = \mu\cdot(3+e^{-t});$$ $$ (e^{t/2} u)' = 3e^{t/2}+e^{-t/2}. $$

This can be seen with the product rule and because of the fact we chose $\mu$ so that $\mu' = 1/2 \cdot\mu$.

From here you can integrate both sides and then isolate the function $y$,

$$ e^{t/2} y = 6e^{t/2} -2e^{-t/2}+C;$$ $$ y = 6-2e^{-t}+Ce^{-t/2}. $$

share|improve this answer
    
Thank you for your answer. But I got my final answer as $y=\frac{3}{2}-\frac{1}{2}e^{-t} +Ce^{-0.5t}$. I think i need to recheck, Thanks again. –  nihilisticgeek Jul 23 '11 at 10:12
1  
If that's what you got as a final answer, then your error was that you differentiated $3e^{t/2}+e^{-t/2}$ instead of integrated. –  anon Jul 23 '11 at 10:26

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.