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Solve $$\frac{dy}{dx}=\frac{2x+y}{y}$$

Let $$\frac{dx}{dt}=y \\ \frac{dy}{dt}=2x+y$$

Plugin $\frac{dx}{dt}=y$ into $\frac{dy}{dt}=2x+y$ I get $$2x+\frac{dx}{dt}=\frac{dy}{dt}$$ Using the fact that $\frac{d^2x}{dt^2}=\frac{dy}{dt}$ I get the 2nd order system:

$$\frac{d^2x}{dt^2}-\frac{dx}{dt}-2x=0$$

Solving this I get: $(r-2)(r+1)=0$ , so $r=2,=1$

$$x(t)=Ae^{2t}+Be^{-1t}$$

How do I get my solution in terms of $y,x$? I can't seem to integrate $\frac{dy}{dx}$ by separating variables directly.

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  • $\begingroup$ "given that $\frac{dx}{dy}=y$"... What? How do you know that's true? $\endgroup$ – 5xum Nov 3 '15 at 18:56
  • $\begingroup$ It's $dx/dy=y$. It is given by the question. Then I computed $dy/dx$ but I got nowhere by trying to use separation of variables and integrating. $\endgroup$ – GRS Nov 3 '15 at 19:00
  • $\begingroup$ Have you tried using the chain rule on $dy/dt$? If y is a function of x, the using the chain rule, you have $\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}$ $\endgroup$ – Simon Nov 3 '15 at 19:07
  • $\begingroup$ I don't see where would this bring me. I don't need any t's in my answer, and the answer should be a function of $x$ and $y$ only. $\endgroup$ – GRS Nov 3 '15 at 19:12
  • $\begingroup$ You're right, that would just give you your given. $\endgroup$ – Simon Nov 3 '15 at 19:14
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I'm a bit rusty with differential equations, so I'll just assume that we have the required number of constants of integration...

$$\begin{align} x & = Ae^{2t} + Be^{−t}\\ y & = \frac{dx}{dt}\\ y & = 2Ae^{2t} - Be^{−t}\\ x+y & = 3Ae^{2t}\\ 2x-y & = 3Be^{−t}\\ e^{2t} = \frac{x+y}{3A} & = \left(\frac{3B}{2x-y}\right)^2\\ (2x-y)^2(x+y) & = 27AB^2\\ 4x^3 - 3xy^2 + y^3 & = 27AB^2 \end{align}$$

Now to check that by differentiating:

$$\begin{align} 12x^2 - 3y^2 - 6xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} & = 0\\ \frac{dy}{dx}(3y^2 - 6xy) & = 3y^2 - 12x^2\\ \frac{dy}{dx} & = \frac{y^2 - 4x^2}{y^2 - 2xy}\\ & = \frac{(y - 2x)(y + 2x)}{y(y - 2x)}\\ & = \frac{y + 2x}{y}, \text{when } y \ne 2x\\ \end{align}$$

which agrees with the original equation.


Here's an alternative solution that doesn't use those parametric equations in $t$. Instead, we use a simple $y = vx$ substitution and separation of variables.

$$\begin{align} \frac{dy}{dx}&=\frac{2x+y}{y}\\ ydy &= (2x+y)dx\\ \\ \text{Let } y &= vx\\ dy &= vdx + xdv\\ \\ \text{Substituting, }\\ vx(vdx + xdv) &= (2x+vx)dx\\ v^2dx + vxdv - (2+v)dx &= 0\\ (v^2-v-2)dx + vxdv &= 0\\ \\ \text{Separating variables, }\\ \frac{dx}{x} + \frac{vdv}{v^2-v-2} &= 0\\ \\ \text{Splitting the right term with partial fractions, }\\ \frac{dx}{x} + \frac{1}{3}\left(\frac{(v-2)+2(v+1)}{(v-2)(v+1)}\right)dv &= 0\\ \frac{dx}{x} + \frac{1}{3}\left(\frac{dv}{v+1} + 2 \frac{dv}{v-2}\right) &= 0\\ \\ \text{Integrating, }\\ \int\frac{dx}{x} + \frac{1}{3}\left(\int\frac{dv}{v+1} + 2 \int\frac{dv}{v-2}\right) &= k\\ \log(x) + \frac{1}{3}(\log(v+1) + 2 \log(v-2)) &= k\\ 3\log(x) + \log(v+1) + 2 \log(v-2) &= 3k\\ x^3(v+1)(v-2)^2 &= e^{3k} = C\\ x^3(v+1)(v^2-4v+4) &= C\\ x^3(v^3-4v^2+4v + v^2-4v+4) &= C\\ x^3(v^3-3v^2+4) &= C\\ (xv)^3-3x(vx)^2+4x^3 &= C\\ \\ \text{Replacing $vx$ with $y$, }\\ y^3-3xy^2+4x^3 &= C\\ \end{align}$$

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