Solving for du/dt: tdt= \frac{2+2u-du}{1+u}

[suspenc3]

$$\frac{du}{dt}=2+2u+t+tu$$

I manipulated it to:$$-tdt= \frac{2+2u-du}{1+u}$$

Should it be in this form? or try something else?

 

[d_leet]

$$\frac{du}{dt}=2+2u+t+tu$$

Note that 2 + 2u + t + tu, can be factored.

First factor the 2 from the first two terms leaving
2(1+u) + t + tu
Then factor the t from the last two terms and you have
2(1+u) + t(1+u)
Now you should notice that you have a common term of (1+u) which can be factored out leaving
(2+t)(1+u)
And now you can easily separate this equation.

 

[suspenc3]

Riight, I see it now, Thanks

 

[d_leet]

Riight, I see it now, Thanks

No problem glad I could help.