How does one solve the differential equation $y'=\frac {y} {3x-y} $ ?

Question

I need to find a solution to the differential following equation:

$y'=\frac {y} {3x-y} $.

I tried to use use some kind of substitution, but I didn't manage to solve it.

Any suggestion\help?

Thanks a lot!

Answer 1

Hint: Try $y=zx$. That will let you separate variables.

Answer 2

We can transform this equation to linear differential equation.

$ \frac{dx}{dy}= \frac{3x}{y}-1 $

And integrating factor $\lambda=e^{{\int{\frac{-3}{y}dy}}}=y^{-3}$.

Then $ x=\frac{\int(-1)y^{-3}dy}{y^{-3}}$. Therefore $x=cy^{3}+\frac{y}{2}$.

Answer 3

I followed $y=z.x$ transform and I got easily the general solution : $cy^3+y=2x$

$3cy^2y'+y'=2$

$(3cy^3+y)y'=2y$

$(3.(2x-y)+y)y'=2y$

$(6x-2y)y'=2y$

$(3x-y)y'=y$

$y'=\frac{y}{3x-y}$

If you cannot manage the y=z.x transform ,let me know.

Answer 4

$$ y' = \frac{y}{3x - y} = \frac{1}{\frac{3x}{y} - 1} $$

$$ \frac{y}{x} = v \Rightarrow y = vx \Rightarrow \frac{dy}{dx} = x \frac{dv}{dx} + v $$

$$ x \frac{dv}{dx} + v = \frac{1}{\frac{3}{v} - 1} = \frac{v}{3 - v} $$

$$ x \frac{dv}{dx} = \frac{v}{3-v} - v = \frac{v(v-2)}{3-v} $$

$$ \frac{(3-v)dv}{v(v-2)} = \frac{dx}{x} $$

$$ \Rightarrow \int \frac{(3-v)}{v(v-2)} \ dv = \int \frac{dx}{x} = \ln x + c $$

$$ \frac{(3-v)}{v(v-2)} = \frac{A}{v} + \frac{B}{v-2} $$

$$ \Rightarrow A(v-2) + Bv = 3 - v $$

$$ v = 2 \Rightarrow B = \frac{1}{2} $$

$$ v = 0 \Rightarrow A = -\frac{3}{2} $$

$$ \Rightarrow \int \frac{(3-v)}{v(v-2)} \ dv = -\frac{3}{2} \int \frac{1}{v} \ dv + \frac{1}{2} \int \frac{1}{v-2} \ dv = \frac{1}{2} \ln(v-2) - \frac{3}{2} \ln v = \ln x + c $$

$$ v = \frac{y}{x} \Rightarrow \frac{1}{2} \ln \left(\frac{y}{x}-2 \right) - \frac{3}{2} (\ln y - \ln x) = \ln x + c $$