Question

du/dt=2+2u+t+tu ?????

Answer 1

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

Answer 2

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
????

Answer 3

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

Answer 4

of course

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

already fanned up last time ;)

Answer 9

oh! [thumbs up]

Answer 10

Cool, so you're sorted then...