Homogeneous Differential Equations

$\frac{dy}{dx} = \frac{x^2+y^2}{x^2+xy}$. Homogeneous deg 2.
Put $y=vx \implies v+x\frac{dv}{dx} = \frac{x^2+v^2x^2}{x^2+x(vx)} = \frac{1+v^2}{1+v}$.
$x\frac{dv}{dx} = \frac{1+v^2}{1+v} – v = \frac{1+v^2-v-v^2}{1+v} = \frac{1-v}{1+v}$.
$\int \frac{1+v}{1-v} dv = \int \frac{dx}{x} \implies \int (\frac{2}{1-v} – 1) dv = \log|x|$.
$-2\log|1-v| – v = \log|x| + C \implies (x-y)^2 = C x e^{y/x}$.

$\frac{dy}{dx} = 1 + \frac{y}{x}$. Put $y=vx$.
$v + x\frac{dv}{dx} = 1 + v \implies x\frac{dv}{dx} = 1$.
$\int dv = \int \frac{dx}{x} \implies v = \log|x| + C$.
Ans: $y = x(\log|x| + C)$.

$\frac{dy}{dx} = \frac{x+y}{x-y}$. Put $y=vx$.
$x\frac{dv}{dx} = \frac{1+v}{1-v} – v = \frac{1+v^2}{1-v}$.
$\int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x} \implies \tan^{-1}v – \frac{1}{2}\log(1+v^2) = \log|x| + C$.
Ans: $\tan^{-1}(\frac{y}{x}) = \log\sqrt{x^2+y^2} + C$.

$\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$. Put $y=vx$.
$x\frac{dv}{dx} = \frac{v^2-1}{2v} – v = -\frac{1+v^2}{2v}$.
$\int \frac{2v}{1+v^2} dv = -\int \frac{dx}{x} \implies \log(1+v^2) = -\log|x| + \log C$.
$1+v^2 = \frac{C}{x} \implies x^2+y^2 = Cx$.

$\frac{dy}{dx} = 1 – 2(\frac{y}{x})^2 + \frac{y}{x}$. Put $y=vx$.
$x\frac{dv}{dx} = 1-2v^2+v – v = 1-2v^2$.
$\int \frac{dv}{1-2v^2} = \int \frac{dx}{x}$.
$\frac{1}{2\sqrt{2}} \log|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}| = \log|x| + C$.

$x dy = (y+\sqrt{x^2+y^2}) dx \implies \frac{dy}{dx} = \frac{y}{x} + \sqrt{1+(\frac{y}{x})^2}$.
Put $y=vx \implies x\frac{dv}{dx} = \sqrt{1+v^2}$.
$\int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x} \implies \log|v+\sqrt{1+v^2}| = \log|x| + \log C$.
$v+\sqrt{1+v^2} = Cx \implies y + \sqrt{x^2+y^2} = Cx^2$.

$\frac{dy}{dx} = \frac{y \cos(y/x) + x \sin(y/x) \cdot (y/x)}{x \cos(y/x) \dots}$ (Simplified: $x\frac{dv}{dx} = \dots$)
Substitution reduces to $\int \cot v dv + \int \frac{1}{v} dv$.
Ans: $xy \cos(\frac{y}{x}) = C$.

This is form $\frac{dx}{dy} = F(x/y)$. Put $x=vy$.
$\frac{dx}{dy} = \frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} \implies v + y\frac{dv}{dy} = \frac{e^v(v-1)}{1+e^v}$.
$y\frac{dv}{dy} = \frac{ve^v – e^v – v – ve^v}{1+e^v} = \frac{-(v+e^v)}{1+e^v}$.
$\int \frac{1+e^v}{v+e^v} dv = -\int \frac{dy}{y} \implies \log|v+e^v| = -\log|y| + \log C$.
$y(v+e^v) = C \implies x + y e^{x/y} = C$.

$\frac{dy}{dx} = \frac{y-x}{y+x}$. Same as Q3 form.
$\frac{1}{2}\log(x^2+y^2) + \tan^{-1}(y/x) = C$.
At $x=1, y=1: \frac{1}{2}\log 2 + \frac{\pi}{4} = C$.
Ans: $\log(x^2+y^2) + 2\tan^{-1}(y/x) = \frac{\pi}{2} + \log 2$.

$\frac{dy}{dx} = -\frac{xy+y^2}{x^2}$. Put $y=vx$.
$x\frac{dv}{dx} = -(v+v^2) – v = -(2v+v^2)$.
Solving gives $\frac{1}{2}\log|\frac{x^2 y}{2x+y}| = C$.
At $(1,1)$, find C.
Ans: $x^2 y = 2x+y$. (Simplified form).

$\frac{dy}{dx} = \frac{2xy+y^2}{2x^2}$. Put $y=vx$.
$x\frac{dv}{dx} = v + \frac{v^2}{2} – v = \frac{v^2}{2}$.
$\int 2v^{-2} dv = \int \frac{dx}{x} \implies \frac{-2}{v} = \log|x| + C$.
$-\frac{2x}{y} = \log|x| + C$. Put $x=1, y=2 \implies -1 = 0 + C \implies C=-1$.
Ans: $y = \frac{2x}{1-\log|x|}$.

Check degrees of each term.
(A) Terms like $4x$ (deg 1) and $5$ (deg 0) mixed. No.
(B) $xy$ (deg 2) vs $x^3$ (deg 3). No.
(C) $x^3$ (deg 3), $2y^2$ (deg 2). No.
(D) $y^2$ (deg 2), $x^2$ (deg 2), $xy$ (deg 2), $y^2$ (deg 2). Yes.