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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!

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  • $\begingroup$ You lost me here $y'=\frac {y} {3x-y} \Rightarrow 3x-y^2=y$. Maybe this is using techniques I'm not familiar with, but this seems wrong. $\endgroup$
    – AlanSE
    Commented Mar 23, 2012 at 21:54
  • $\begingroup$ Sorry, my mistake. $\endgroup$
    – Jozef
    Commented Mar 23, 2012 at 22:04
  • $\begingroup$ Perhaps this. $\endgroup$ Commented Mar 23, 2012 at 22:05
  • $\begingroup$ It seams that $y=ax$ could be a solution. $\endgroup$
    – martini
    Commented Mar 23, 2012 at 22:07
  • $\begingroup$ See $y'$ is essentially ${dy \over dx}$. So just do ${dx \over dy}$ and then try to solve it. It'll be finished in a matter of seconds! $\endgroup$
    – HarshDarji
    Commented Mar 15, 2022 at 7:40

4 Answers 4

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

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  • $\begingroup$ You mean the denominator is $3x-y^2$? I don't know, nonlinear equations are difficult. One might be able to come up with a special trick reducing to a Riccati, but a little scribbling got me nowhere. In the original problem, taking advantage of homogeneity was an automatic reflex. $\endgroup$ Commented Mar 24, 2012 at 16:06
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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 $x=cy^{3}+\frac{y}{2}$.

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  • $\begingroup$ If you care to, you can explicitly solve for $y$ in terms of $x$ when $c\geq0$. If $c=0$, then of course $y=2x$. If $c>0$, then the cubic formula yields $y=\sqrt[3]{ \frac{x}{2c}+\sqrt{\frac{x^2}{4c^2}+\frac{1}{216c^3}}}+\sqrt[3]{ \frac{x}{2c}-\sqrt{\frac{x^2}{4c^2}+\frac{1}{216c^3}}}$. $\endgroup$
    – 2'5 9'2
    Commented Mar 24, 2012 at 0:03
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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