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. $\endgroup$– AlanSEMar 23, 2012 at 21:54
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$\begingroup$ Sorry, my mistake. $\endgroup$– JozefMar 23, 2012 at 22:04
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$\begingroup$ Perhaps this. $\endgroup$– David MitraMar 23, 2012 at 22:05
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$\begingroup$ It seams that $y=ax$ could be a solution. $\endgroup$– martiniMar 23, 2012 at 22:07
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$\begingroup$ 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! $\endgroup$– HarshDarjiMar 15, 2022 at 7:40
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4 Answers
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Hint: Try $y=zx$. That will let you separate variables.
answered Mar 23, 2012 at 22:07
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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. $\endgroup$ 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 9'2Mar 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.
answered Mar 23, 2012 at 22:44
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$$ y' = \frac{y}{3x - y} = \frac{1}{\frac{3x}{y} - 1} $$
$$ \frac{y}{x} = v \Rightarrow y = vx \Rightarrow \frac{dy}{dx} = x \frac{dv}{dx} + v $$
$$ x \frac{dv}{dx} + v = \frac{1}{\frac{3}{v} - 1} = \frac{v}{3 - v} $$
$$ x \frac{dv}{dx} = \frac{v}{3-v} - v = \frac{v(v-2)}{3-v} $$
$$ \frac{(3-v)dv}{v(v-2)} = \frac{dx}{x} $$
$$ \Rightarrow \int \frac{(3-v)}{v(v-2)} \ dv = \int \frac{dx}{x} = \ln x + c $$
$$ \frac{(3-v)}{v(v-2)} = \frac{A}{v} + \frac{B}{v-2} $$
$$ \Rightarrow A(v-2) + Bv = 3 - v $$
$$ v = 2 \Rightarrow B = \frac{1}{2} $$
$$ v = 0 \Rightarrow A = -\frac{3}{2} $$
$$ \Rightarrow \int \frac{(3-v)}{v(v-2)} \ dv = -\frac{3}{2} \int \frac{1}{v} \ dv + \frac{1}{2} \int \frac{1}{v-2} \ dv = \frac{1}{2} \ln(v-2) - \frac{3}{2} \ln v = \ln x + c $$
$$ v = \frac{y}{x} \Rightarrow \frac{1}{2} \ln \left(\frac{y}{x}-2 \right) - \frac{3}{2} (\ln y - \ln x) = \ln x + c $$
