Steps in solving $(x + 2y + 3)dx + (2x + 4y - 1)dy = 0$.

Question

For this problem we are referring to this part of a theorem in the book:

$\textbf{Case 2.}$ $ \ $ If $a_2/b_1 = b_2/b_1 = k,$ then the transformation $z = a_{1}x + b_{1}y$ reduces the equation to a separable equation in the variables x and z.

So we consider the equation: $(x + 2y + 3)dx + (2x + 4y - 1)dy = 0$.

We get that $k = 2.$ Also, we let $z = x + 2y.$ The equation then transforms into

$$(z + 3)dx + (2z-1)\Big(\frac{dz - dx}{2}\Big) = 0.$$

This next step is what lost me completely. The book says that this equation could also be written as

$$7dx + (2z - 1)dz = 0,$$

but I can't see how. Why is $z$ four? How does $\Big(\frac{dz - dx}{2}\Big)$ get reduced to just $dz$?

Answer 1

Let $x+2y=v \implies 1+2y'=v'$, then $$\frac{dy}{dx}=\frac{-(x+2y+3)}{2x+4y-1)} \implies \frac{1}{2} (v'-1)=-\frac{v+3}{2v-1}$$ $$\implies v'=1-\frac{2v+6}{2v-1} \implies v'=\frac{-7}{2v-1}$$ $$\implies \int (2v-1) dv= -7\int dx+C$$ $$\implies v^2-v=-7x+c, v=x+2y$$

Answer 2

Multiply with 2 and rearrange the terms $$ ((2z+6)-(2z-1))\,dx+(2z-1)\,dz=0 $$ This then cancels to exactly the claimed formula, $$ 7\,dx+(2z-1)\,dz=0 $$

Answer 3

$$(z + 3)dx + (2z-1)\Big(\frac{dz - dx}{2}\Big) = 0.$$

$$(z + 3)dx + \dfrac {2z-1}{2} dz -\dfrac {2z-1}{2} dx = 0.$$ Collect all the $dx$ terms toegether: $$(z + 3-z+\dfrac 12)dx + \dfrac 12(2z-1)dz = 0.$$ $$\dfrac 72dx + \dfrac 12(2z-1)dz = 0.$$ $$7dx + (2z-1)dz = 0.$$