what is the solution of the ordinary differential equation? would have any complete solution to the equation plus complete below?
$$
y'= \frac{4y-{3x}}{2x-y}
$$
a part of the solution:$$
y'=xv'+v
$$
$$ 
y=xv
$$
not complete solution
$$
xv'+v= \frac{4xv-{3x}}{2x-xv}
$$
Continuation
$$
xv'+v= \frac{2v-{3}}{1-v}
$$
Continuation
$$
xv'= \frac{2v-3-v+{v^2}}{1-v}
$$
Continuation
$$
 x\frac{dy}{dx}= \frac{2v-3-v+{v^2}}{1-v}
$$
Continuation
$$
xv'= \frac{1-v}{v^2-v-3}
$$
Answer $$|y-x|=C|y+x|^3$$
 A: That's a homogeneous differential equation. I'd solve it like this:
divide by x $$y' = \frac{4\left(\frac{y}{x}\right) - 3}{2 - \left(\frac{y}{x}\right)}$$
Set $v = \frac{y}{x}$, so $y = xv$ and $y' = v + xv'$
Substitute: $$v + xv' = \frac{4v - 3}{2 - v}$$
$$xv' = \frac{4v - 3}{2 - v} - v$$
$$xv' = \frac{4v - 3 - 2v + v^2}{2 - v}$$
$$xv' = \frac{v^2 + 2v - 3}{2 - v}$$
$$xv' = \frac{(v + 3)(v - 1)}{-(v - 2)}$$
$$v' = \frac{(v + 3)(v - 1)}{v - 2}\left(-\frac{1}{x}\right)$$
Solve by separation of variables, where $B(v) = \frac{(v + 3)(v - 1)}{v - 2}$ and $A(x) = \left(-\frac{1}{x}\right)$. First $B(v) = 0$:
$$(v + 3)(v - 1) = 0$$
Solutions are $v = -3$ and $v = 1$, so $y(x) = -3x$ and $y(x) = x$.
Secondly, you solve $$\int_{}^{}\frac{v - 2}{(v + 3)(v - 1)} \,\mathrm dv = \int_{}^{}\left(-\frac{1}{x}\right) \,\mathrm dx$$
The one on the right is $-ln|x| + C$. The other can be solved using residues: $$\int_{}^{}\left(\frac{A}{v + 3} + \frac{B}{v - 1}\right) \,\mathrm dv$$
$$A = \lim\limits_{v\rightarrow -3} \frac{v - 2}{v - 1} = \frac{5}{4}$$
$$B = \lim\limits_{v\rightarrow 1} \frac{v - 2}{v + 3} = -\frac{1}{4}$$
You put these values and find that the integral is $\frac{5}{4}ln|v + 3| -\frac{1}{4}ln|v - 1| + C$
so: $$\frac{5}{4}ln|v + 3| -\frac{1}{4}ln|v - 1| = -ln|x|+ C /*4$$
$$5ln|v + 3| -ln|v - 1| = -ln|x^4|+ C$$
You go on and find $$\frac{(v + 3)^5}{v - 1} = \frac{C}{x^4}$$ with C $\neq$ 0.
Replace v with $\frac{y}{x}$ and finally find $$(y + 3x)^5 = C(y - x)$$
If C = 0, we find that $y = -3x$
