- #1

Buffu

- 849

- 146

## Homework Statement

##(2x + 3y + 1)dx + (4x + 6y + 1) dy = 0##

##y(-2) = 2##

## Homework Equations

## The Attempt at a Solution

Let ##z = 2x + 3y##

then ##z^\prime = 2 + 3y^prime##

##\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0##

##\dfrac{dz}{dx} = \dfrac{z- 1}{2z + 1}##

##x = 2z + 3\log(1 - z) + C = 2(2x + 3y) + 3\log|(1 - 2x - 3y)| + C##

Plugging the values, I get ##C = -6##

So, ##0 = x + 2y + log|3y + 2x - 1| - 2##

But the answer given is ##0 = x + 2y \color{red}{-} log|3y + 2x - 1| - 2##.

Is the answer in the book wrong ?