# Homework Help: A simple DE.

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1. Jun 4, 2017

### Buffu

1. The problem statement, all variables and given/known data

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

$y(-2) = 2$

2. Relevant equations

3. 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 ?

2. Jun 4, 2017

@Buffu I looked over your solution=I don't see any mistakes.

3. Jun 4, 2017

### Buzz Bloom

Hi Buffu:

I think your equation where dz/dx = f(x)is wrong.
You want to do two things first.
1. Substitute z=2x+3y into the given equation.
2. Substitute an expression for dy=g(dz) into the result of step 1.

Hope that helps.

Regards,
Buzz

4. Jun 5, 2017

### Buffu

$\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0 \iff \dfrac{(3z + 3) - 4z - 2}{2z + 1} + {dz \over dx} = 0\iff \dfrac{ 1 - z}{2z + 1} + {dz \over dx} = 0$