Solve the IVP of the DE of form dy/dt=f(y/t)...

Question

I am trying to solve the following IVP:

$$ \frac{dy}{dt} = \frac{2y-3t}{3y-4t},\ \ \ \ y(1) = 2 $$

Here is my approach. I applied a substitution $ v = \frac{y}{t} $, which yields me the below:

$$ t\frac{dv}{dt} + v = \frac{2v-3}{3v-4} $$

I subtracted $v$ from both sides, and simplified, to get this seperable DE:

$$ \frac{dt}{t} = \frac{3v-4}{-3(v-1)^2} $$

This is the integral that I made, based off the initial value condition, with $s$ and $r$ as placeholder variables for $t$ and $v$ respectively:

$$ \int_{1}^{t} \frac{ds}{s} = \int_{2}^{v} \frac{3r-4}{-3(r-1)^2} dr $$

I can leave the answer implicitly. In the end, I substituted back y/t for v...

$$ t = exp\ \ (\int_{2/1}^{y/t} \frac{3(\frac{m}{n})-4}{-3(\frac{m}{n}-1)^2} \frac{dm}{dn}) $$

...where $m$ and $n$ are sort of placeholder variables. Is this approach (and answer) correct? I am allowed to leave my answer in implicit form. Thank you for helping out!

Answer 1

Substituting $$y(t)=v(t)t$$ we get $$t\frac{dv(t)}{dt}+v(t)=\frac{-3t+2tv(t)}{-4t+3tv(t)}$$ and this is $$\frac{\frac{dv(t)}{dt}(3v(t)-4)}{(v(t)-1)^2}=-\frac{3}{t}$$ and now integrate! This gives $$-3\log(v(t)-1)+\frac{1}{v(t)-1}=-3\log(t)+C_1$$

Answer 2

$$\frac{dy}{dt} = \frac{2y-3t}{3y-4t},\ \ \ \ y(1) = 2$$

$$ (2y-3t)dt+(-3y+4t)dy=0$$

Use the integrating factor $$\mu (t,y)= \frac 1 {(t-y)^2}$$ Then the differential equation is exact and you can easily get an implicit form... $$ \begin{cases} \displaystyle f(t,y)= \int \frac {2y-3t}{(t-y)^2} dt+\phi(y) \\ \displaystyle f(t,y)= \int \frac {4t-3y}{(t-y)^2} dy+\psi(t) \end{cases} $$ $$ \implies 3 \ln|t-y|+ \frac t {y-t}=K \,\,\,;\, y(1)=2 \implies K=1$$ $$\boxed { 3 \ln|t-y|+ \frac t {y-t}=1}$$