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I need to integrate an ODE of the form $$\frac{1}{x}\frac{d}{dx}\left(x\frac{dy}{dx}+\frac{yx^2}{2}\right)=0.$$ I know I need to integrate w.r.t $x$ to create a first order ODE but I'm not sure how as the $dx$ is in the denominator.

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  • $\begingroup$ You have that $\left(x\frac{dy}{dx}+\frac{yx^2}{2}\right)=C$ $\endgroup$ May 23, 2020 at 16:41

2 Answers 2

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This is a bit of a "hack" solution, but we can multiply both sides by $x$ to find $$\frac{\mathrm{d}}{\mathrm{d}x}\left(x\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{yx^2}{2}\right)=0$$ Which tells us that $\text{deg}\left(x\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{yx^2}{2}\right)=0$, i.e, it is a constant. So now our equation is $$x\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{yx^2}{2}=C$$ $$y'+\frac{x}{2}y=\frac{C}{x}$$ This is a first order linear ordinary differential equation, so we can find solutions via the method of Integrating Factors.

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$$\frac{1}{x}\frac{d}{dx}\left(x\frac{dy}{dx}+\frac{yx^2}{2}\right)=0$$ Integration gives us: $$xy'+\frac{yx^2}{2}=C$$ It's a first order linear DE.


Whenever you see this $$\dfrac {df}{dx}=0$$ You can deduce that $$f=c$$ Because the only function that gives zero once you differentiate it is the constant function.

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  • $\begingroup$ Ah, you beat me to it. Well done. $\endgroup$
    – K.defaoite
    May 23, 2020 at 16:46
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    $\begingroup$ Thanks Oh will upvote your answer. @K.defaoite $\endgroup$ May 23, 2020 at 16:46
  • $\begingroup$ I feel like im being daft but which rule did you use to obtain that answer? I just can't see it $\endgroup$
    – PokerF4ce
    May 23, 2020 at 17:22
  • $\begingroup$ @PokerF4ce You have a derivative that's equal to zero ...so you integrate and get a constant on RHS because the derivative of a constant is zero. $\endgroup$ May 23, 2020 at 17:23
  • $\begingroup$ @PokerF4ce I added some lines I hope it's more clear. $\endgroup$ May 23, 2020 at 17:25

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