Differential Equation - Solve by substitution

[JJBladester]

Homework Statement

Solve the following differential equation by using substitution.

Homework Equations

(x + 2y - 1)dx + (2x + 4y - 3)dy = 0

Let u = x + 2y

The Attempt at a Solution

u = x + 2y
du/dx = 1 + 2(dy/dx)

(x + 2y - 1)dx + (2x + 4y - 3)dy = 0

(u - 1)dx + (2u - 3)dy = 0

The solution the teacher gave is as follows but I didn't know how to go from my line above to the answer below. I believe it has something to do with getting the dx and dy out of the new equation involving u and then rearranging to solve by separation variables.

K = x + (x + 2y)² - 3(x + 2y)

 

[quantumdude]

Use this...

du/dx = 1 + 2(dy/dx)

...to eliminate dy in this

(u - 1)dx + (2u - 3)dy = 0

 

[JJBladester]

If u = x + 2y

then du/dx = 1 + 2(dy/dx)

and dy/dx = (du/dx - 1)/2

So...

dy/dx + (x + 2y -1)/(2x + 4y - 2) = 0

(du/dx - 1)/2 + (u - 1)/(2u - 3) = 0

du/dx - 1 + (2u - 2)/(2u - 3) = 0

If I divide (2u - 2) by (2u - 3), I get:

1 + 1/(2u - 3)

Then I have:

du/dx - 1 + 1 + 1/(2u - 3) = 0

du/dx + 1/(2u - 3) = 0

(2u - 3)du + dx = 0

The integration gives me:

u² - 3u + x = c

So, my answer (after putting (x + 2y) back in for u) is:

(x + 2y)² -3(x + 2y) = c

Got it, Tom. Thanks for the push in the right direction!