## anonymous 5 years ago du/dt=2+2u+t+tu ?????

1. anonymous

The right-hand side may be factored as$2+2u+t+tu=2(1+u)+t(1+u)=(2+t)(1+u)$

2. anonymous

i can get it to an e function and a t function on each sides but how to i make it a term of a single u ????

3. anonymous

Your equation is then separable:$\frac{du}{1+u}=(2+t)dt \rightarrow \log (1+u) = 2t+\frac{t^2}{2}+c$

4. anonymous

of course

5. anonymous

Exponentiate both sides:$1+u=e^{2t+t^2/2+c}=Ce^{2t+t^2/2} \rightarrow u = Ce^{2t+t^2/2}-1$

6. anonymous

So, are you going to be a fan :p ?

7. anonymous

i had (1+u).du on the otherside, stupid mistake of not having one over to start with, made it seem a lot more difficult than it really was

8. anonymous

already fanned up last time ;)

9. anonymous

oh! [thumbs up]

10. anonymous

Cool, so you're sorted then...