I want to find a general solution of the following homogeneous equation
$$\frac{dy}{dt}=\frac{3t+12y}{t+14y}$$
I have tried using the substitution $z = y/t$ to make the equation separable, but then it gets a little bit tedious to solve.
$$\frac{dy}{dt}=\frac{3+12z}{1+14z}=z+\frac{dz}{dt}t$$
$$\int \frac{1+14z}{-14z^2+11z+3} dz = \ln(t)+c_1$$
So I'm wondering if there's another way to solve it?
My teacher in ODEs show me this "rarely unknown"substitution method, it Works with homogeneous and almost homogeneous ODEs.
1° Let $y=p^S$ and $t=q^R$, respectively $dy=Sp^{S-1}dS$, $dt=Rq^{R-1}dR$; R and S are constants. Then substitute in the ODE:
$$\frac{dy}{dx}=\frac{3t+12y}{t+14y}\rightarrow\frac{Sp^{S-1}dS}{Rq^{R-1}dR}=\frac{3q^R+12p^S}{q^R+14p^S}$$ $$\frac{S\ dS}{R\ dR}=\frac{q^{R-1}(3q^R+12p^S)}{p^{S-1}(q^R+14p^S)}$$ $$\frac{S\ dS}{R\ dR}=\frac{3q^{2R-1}+12p^Sq^{R-1}}{p^{S-1}q^R+14p^{2S-1}}$$ Match the all the upper exponents with all the lowers
$$2R-1+S+R-1+=S-1+R+2S-1\\3R+S=3S+R\\R=S$$
The only restriction is $(R,S)>0\quad$ I'll do it easier with $R=1$ we obtain: $$\frac{dS}{dR}=\frac{3q+12p}{q+14p}$$ we got the same initial ODE, so we're right.
$\star\quad $ 2° Method we want $F(\lambda^\alpha t,\lambda^\beta y)=\lambda^{\beta-\alpha}F(t,y)\ $ to be homogeneous, and the substitution must be $y=\sigma t^{\frac{\alpha}{\beta}}$ $$\frac{3\lambda t+12\lambda y}{\lambda t+14\lambda y}\\\lambda^0\left(\frac{3t+12y}{t+14y}\right)$$ it works with bigger exponents but in this case $\alpha=\beta=1$ then $y=\sigma t$ the easiest substitution .
$\star\quad $ 3° Another method could be; from $dy/dt=3(t-4y)/(t+14y)\ $Let $u=t+4y\ $ then $du=dt+4dy$ $$\frac{\frac14(du-dt)}{dt}=\frac{3u}{t+\frac72(u-t)}\\\frac14\frac{du}{dt}-\frac14=\frac{6u}{7u-5t}\\\frac{du}{dt}=\frac{31u-5t}{7u-5t}$$ but its unnecessary cause its the same traditional method, in another form.Also you can do it letting $\omega=t+14y$
All these methods are long and again unnecessary, your integral is easier enough, the first method its for different exponents for a ODE like this: $y'=(x^2y+yx^3)/(y+y^2x)$