$(x+2y-4)dx - (2x-4y)dy = 0$
$(\bar x + 2 \bar y) d \bar x - (2 \bar x - \bar y) d \bar y = 0 $
$\bar y = V \bar x $
$ d \bar y = V d\bar x + dV \bar x $
Plugged it in exactly like this and could not figure out how to separate it
$ ( \bar x + 2V \bar x )d \bar x - (2 \bar x - V \bar x) (V + dV \bar x) $
$ \bar x (1 + 2V) d \bar x - 2 \bar x (1 - V) (V + dV \bar x) = 0$
$\endgroup$ 21 Answer
$\begingroup$$$ (x + 2y - 4)dx - (2x -4y)dy = 0 $$
Let $X = x + a$ and $Y= y+ b$ and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 4)dX - (2X-2a - 4Y +4b)dY =0$$
It's easy to find what the "suitable" values should be, since we just want them to cancel out the constants! So we solve the following system of equations:
$$\begin{align*} -4 &= a + 2b\\ 0 &= 2a - 4b \end{align*}$$I'll let you verify that the correct values are $a=-2$ and $b=-1$. This leaves us with$$(X + 2Y)dX - (2X-4Y)dY=0$$
Now let $Y=VX$, and hence $dY=VdX+XdV$. Then
\begin{align} X(1 + 2V)dX - X(2-4V)(VdX+XdV) &=0\\ X(1+4V^2)dX - X^2(2-4V)dV &=0\\ X(1+4V^2)dX &= X^2(2-4V)dV\\ \dfrac{1}{X}dX &= \dfrac{2-4V}{1+4V^2}dV \\ \log \left|X \right|+C_1&=\tan^{-1}(2V)-\frac{1}{2}\log(4V^2+1)\\ 2\log\left|X\right|+C_2&=2\tan^{-1}(2V)-\log(4V^2+1)\\ C_2&=2\tan^{-1}\left(\frac{2Y}{X}\right)-\log\left(\frac{4Y^2}{X^2}+1\right)\\&\quad-2\log\left|X\right|\\ C_2&=2\tan^{-1}\left(\frac{2Y}{X}\right)-\log\left(4Y^2+X^2\right)\\ C_2&=2\tan^{-1}\left(\frac{2y-2}{x-2}\right)\\&\quad-\log\Big(4(y-1)^2+(x-2)^2\Big) \end{align}which may also be written as
$$C_3=\log\Big(4(y-1)^2+(x-2)^2\Big)-2\tan^{-1}\left(\frac{2y-2}{x-2}\right)$$
$\endgroup$ 8
