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Solve the following differential equation:

$a(x\frac{dy}{dx}+2y)=xy\frac{dy}{dx}$. --Edited: see edit notes

I am having trouble solving this equation, problems that I run into are outlined below.

First, this is a non-exact differential equation. I will not put the work here, but it can be seen if you put the equation in the form $\frac{dM}{dy}=\frac{dN}{dx}$, and it comes out that $\frac{dM}{dy}\neq\frac{dN}{dx}$. So, integrating factors must be used in order to move forward solving.

But this becomes very difficult. If you choose to multiply your ODE by some function $\mu(x)$ or a $\mu(y)$ or $\mu(x,y)$, attempting to find the integrating factor, nothing cancels and you are left integrating something that cannot be integrated.

I also tried to multiply the ODE by $x^\alpha y^\beta$ (because I originally thought that $N$ and $M$ were sums of products of powers of $x$ and $y$, but this also proved inadequate, as the $a$s in the ODE did not cancel, and you are left with two sides of an equation that you cannot get to equal one another.

Leaving me back where I started- ground zero. Does anyone have any ideas on how to find this integrating factor?

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4 Answers 4

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The equation is separable,

$$2a^2y=x(y-a)\frac{dy}{dx},$$

$$2a^2\frac{dx}{x}=\frac{y-a}{y}dy,$$

$$2a^2\log x+C=y-a\log y.$$

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  • $\begingroup$ Apologies, I miswrote the ODE, I will nonetheless check and see if it will be able to be separated the way that you did $\endgroup$
    – Pascal
    Commented Sep 30, 2018 at 21:42
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It's separable $$a(x\frac{dy}{dx}+2ay)=xy\frac{dy}{dx}$$ $$a(xy'+2ay)=xyy'$$ $$2a^2y=xy'(y-a)$$ $$2a^2\int \frac {dx}x=\int \frac {(y-a)}{y}dy$$


$$a(x\frac{dy}{dx}+2y)=xy\frac{dy}{dx}$$ $$a(xy'+2y)=xyy'$$ $$y'x(a-y)=-2ay$$ This last equation is separable $$\int \frac {a-y}{y}dy=-2a\int \frac {dx}x$$

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  • $\begingroup$ Apologies, I miswrote the ODE, I will nonetheless check and see if it will be able to be separated the way that you did $\endgroup$
    – Pascal
    Commented Sep 30, 2018 at 21:43
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    $\begingroup$ @Pascal no problem edit your question $\endgroup$ Commented Sep 30, 2018 at 21:44
  • $\begingroup$ Great, I think I was thinking too hard and getting ahead of myself with this one. Thank you for your clarity and work. $\endgroup$
    – Pascal
    Commented Sep 30, 2018 at 21:47
  • $\begingroup$ I added some lines @Pascal $\endgroup$ Commented Sep 30, 2018 at 21:49
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Starting from Isham's answer (extracting $y'$ from the equation $$\int \frac {a-y}{y}dy=-2a\int \frac {dx}x$$ that is to say $$a \log(y)-y=-2a\log(x)+c$$ from which $$y=-a\, W\left(-\frac{e^{\frac{c}{a}}}{a x^2}\right)$$ where appears Lambert function.

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$$a\left(x\frac{dy}{dx}+2y\right)=xy\frac{dy}{dx}$$ we can see a common factor of $x\frac{dy}{dx}$ so we can obtain: $$x\frac{dy}{dx}\left(y-a\right)=2ya$$ so: $$\int\frac{y-a}{2ya}dy=\int\frac1xdx$$ so: $$2\ln|x|=\int\left(\frac{1}{a}-\frac{1}{y}\right)dy$$ $$2\ln|x|=\frac{y}{a}-\ln|y|+C$$ $$x^2y=e^{\frac{y}{a}+C}$$ $$x=\sqrt{y^{-1}e^{\frac{y+C_1}{a}}}$$ and this can then be used solving the Lambert W function: https://en.wikipedia.org/wiki/Lambert_W_function

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