How can I show that an ODE has other solutions than the trivial ones when it is not analytically solvable? I have the following first-order differential equation:
$$
(a x - b y) y'(x) - c y =0,
$$
where $a,b,c>0$ and $x>=0$.
There are two obvious solutions: $y(x)\equiv 0$ and $y(x)=\frac{a-c}{b} x$. The numerical analysis suggests that there are infinitely many solutions between these two solutions. For example, the following is a streamline plot for $a=3/4,b=7/4,c=1/4$ generated by Mathematica. The red line is the linear solution, and the yellow line is the zero function solution.
My question is, how can I rigorously show that there are infinitely many solutions between these two trivial solutions? Or, at the very least, can I formally show that there is one solution that is one of the above solutions? Since an explicit construction seems not feasible, I don't know where to start.

a = 3/4; b = 7/4; c = 1/4;
gr1 = StreamPlot[{1, (c y)/(a x - b y)}, {x, 0, 2}, {y, -.05, 1}];
gr2 = ContourPlot[y == (a - c)/b x, {x, 0, 2}, {y, 0, 3}, ContourStyle -> Red];
sol3 = DSolve[{(a x - b y[x]) y'[x] - c y[x] == 0, y[0] == 0}, y, {x, 0, 2}][[1]];
gr3 = Plot[Evaluate[y[x] /. sol3], {x, 0, 2}, PlotStyle -> Yellow];
sol4 = NDSolve[{(a x - b y[x]) y'[x] - c y[x] == 0, y[1] == .2}, y, {x, 0, 2}][[1]];
gr4 = Plot[Evaluate[y[x] /. sol4], {x, 0, 2}, PlotStyle -> Green];
Show[gr1, gr2, gr3, gr4]

 A: In fact, we can find exact solutions for this ODE.
For $x>0$, division by $x$ shows that the equation is equivalent to
$$\left(a-b\frac{y(x)}{x}\right)y'(x)-c\frac{y(x)}{x}=0$$
or, provided we stipulate that $a-b\frac{y(x)}{x}$ never be zero,
$$y'(x)=\frac{c\frac{y(x)}{x}}{a-b\frac{y(x)}{x}}$$
Writing $F(t):=ct/(a-bt)$, it can be seen that our transformed equation is
$$y'(x)=F\left(\frac{y(x)}{x}\right)$$
which is a first-order homogeneous differential equation. We can solve such equations by introducing the substitution $v(x):=y(x)/x$ and transforming the equation into a separable one in terms of $v$:
$$y'(x)=F\left(\frac{y(x)}{x}\right)\rightarrow \frac{v'(x)}{F(v(x))-v(x)}=\frac{1}{x}$$
After integrating and substituting $u=v(x)$, we are left with evaluating $\int \frac{1}{F(u)-u} du$, which is amenable to elementary techniques since $F(u)=cu/(a-bu)$ is a rational function.
A: $$(a x - b y) y'(x) - c y =0$$
$$(a x - b y)  - cx' y =0$$
$$a x   - cx' y =by$$$$
x'-\dfrac a{cy}x=-\dfrac bc$$
Should  be integrable and at least lead to an implicit solution.
$$x(y)=f(y)$$
