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I have

$$\left(2x+ 1-\frac{y^2}{x^2}\right)dx+ \frac{2y}{x}dy= 0$$

which is not exact. However, it could be made exact by using the formula for integrating factors. I have two ways,

With $$M=\left(2x+ 1-\frac{y^2}{x^2}\right)$$ and $$N=\frac{2y}{x}$$ we can form the following integrating factors:

$$\phi(x)=\frac{N_x-M_y}{M}=-\frac{\frac{4y}{x^2}}{2x+1-\frac{y^2}{x^2}}$$

or

$$\psi(x)=\frac{M_y-N_x}{N}=\frac{ (x^3 - x^2 + y^2)}{yx^2}$$

Then multiplying these in, should give an exact form of the ODE. But neither of the two make the ODE exact. Are there other formulas one can use?

Thanks

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    $\begingroup$ Isn't $F(x,y) = x^2 + x + \frac{y^2}{x}$ a potential or am I missing something? $\endgroup$
    – Klaus
    Oct 11, 2022 at 8:54
  • $\begingroup$ see can i post an alternative solution $\endgroup$
    – Vanessa
    Oct 11, 2022 at 8:56
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    $\begingroup$ Note that the x-derivative of $2y/x$ is $-2y/x^2$, you seem to have missed the sign. $\endgroup$ Oct 11, 2022 at 10:48

3 Answers 3

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$$\left(2x+ 1-\frac{y^2}{x^2}\right)dx+ \frac{2y}{x}dy= 0$$ $$\left(2x+1\right)dx+\dfrac {(-y^2dx +{2xy}dy)}{x^2}= 0$$ $$\left(2x+1\right)dx+\dfrac {(-y^2dx +{x}dy^2)}{x^2}= 0$$ $$\left(2x+1\right)dx+d \left (\dfrac {y^2}{x}\right)= 0$$ Integrate.

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Combine the two differentials into $y'=dy/dx$: $$2x+1-(y/x)^2=-2yy'/x$$ Set $z=y/x,y'=z'x+z$: $$2x+1-z^2=-2zx(z'x+z)/x$$ $$z^2+2zz'x=-2x-1$$ $$(z^2x)'=-2x-1\implies z^2x=-x^2-x+K$$ $$\frac{y^2}x+x^2+x=K$$

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Ok, this differential form is indeed exact. The integrating factor is 1, and the first integral is $$ I=\dfrac{x^3+x^2+y^2}{x} $$

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  • $\begingroup$ indeed this is what i was gonna write $\endgroup$
    – Vanessa
    Oct 11, 2022 at 8:56
  • $\begingroup$ Once you check it is exact you apply direct integration $\endgroup$ Oct 11, 2022 at 8:58
  • $\begingroup$ Haha, that was a weird exercise. It says "The ODE is not exact, find an IF to make it exact". So confusing the student is part of good old fashion calculus. $\endgroup$ Oct 11, 2022 at 9:00

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