Separable differential equation

[jLinder]

Hi, I'm new here. I really need help with this equation. I have found a solution to it at cramster.com although I don't know how they have done one of the steps. (Chapter 10.3, exercise 9 in James Stewarts Calculus 6E.)

From cramster.com:
(1) du/dt = 2 + 2u + t + tu
(2) -> du/dt = 2(u + 1) + t(u + 1) - I get it so long..
(3) -> du/dt = (1 + u)(2 + t) - Here's where I'm lost, how do I get from (2) to (3)? Is there a systematic method for this? I think I can manage the rest by myself.

Thanks!

 

[srmichael]

Hi, I'm new here. I really need help with this equation. I have found a solution to it at cramster.com although I don't know how they have done one of the steps. (Chapter 10.3, exercise 9 in James Stewarts Calculus 6E.)

From cramster.com:
(1) du/dt = 2 + 2u + t + tu
(2) -> du/dt = 2(u + 1) + t(u + 1) - I get it so long..
(3) -> du/dt = (1 + u)(2 + t) - Here's where I'm lost, how do I get from (2) to (3)? Is there a systematic method for this? I think I can manage the rest by myself.

Thanks!

They are factoring (2). If you were to distribute (3) you would see that you get (1 + u)(2) + (1 + u)(t). Think of it this way...Let w = 1 + u so now you have:

w(2 + t) which when distributed becomes (w)(2) + (w)(t) or, when substituting back in for w becomes: (1 + u)(2) + (1 + u)(t)

 

[jLinder]

Thank You!

So it's really like getting from 3t+6u to 3(t+2u). I've never thought of it that way, thanks for opening my eyes a little more! Another tool in the toolbox! =)

What I've noticed with math so far is that it's very much about recognizing (hidden) patterns to simplify the problems. My teacher told us so in the beginning of my education but it's first when one experince it that really one notice that.