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Consider the equation $(5y - 2x) (\dfrac{dy}{dx}) - 2y = 0$ This equation is Exact and Homgeneous differential equation. When i use the exact method then i get the following solution $2xy -(5/2) y^2 = c$ And when I use the Homogeneous method of solution that is putting $y = vx$ in Homogeneous differential equation then I get the following solution $ln (y) + \dfrac{2x}{5y} = c$. Question is that are both right solutions. Im confused about different solution. Explain it please.

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    $\begingroup$ If you substitute each back into the ODE, do they both work? $\endgroup$ – Moo Mar 1 '18 at 16:58
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    $\begingroup$ Actually i just recheck my solution and i found that i did wrong cancellation in the solution. So after correction i got the right answer that is 2xy - (5/2) y^2 = c . And thank you for reminding me that i can verify the solution by plugging it in the given differential equation. Thank you $\endgroup$ – Naeem Ivy Mar 1 '18 at 17:05
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    $\begingroup$ @Isham: Looks like by checking, s/he found his issues. $\endgroup$ – Moo Mar 1 '18 at 17:10
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    $\begingroup$ Oh I see now @mpp $\endgroup$ – Isham Mar 1 '18 at 17:21
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$$(5y - 2x) (\dfrac{dy}{dx}) - 2y = 0$$ $$ y' =\frac {2y}{(5y - 2x)}$$ Since the equation is homogenous then substitute $y=tx$ $$ t'x+t =\frac {2t}{(5t - 2)}$$ $$ t'x =\frac {4t-5t^2}{(5t - 2)}$$ $$ \int \frac{(5t - 2)} {4t-5t^2}dt=\ln|x|+K$$ $$ -\frac 12\int \frac 1tdt+\frac 52\int \frac{ dt} {4-5t}=\ln|x|+K$$ $$ -\frac 12\ln|t|-\frac 12\int \frac{ dt} {t-4/5}=\ln|x|+K$$ $$ -\frac 12\ln|t|-\frac 12\ln|t-4/5|=\ln|x|+K$$ $$ \ln|y|+\ln|\frac yx-4/5|+\ln|x|=K$$ $$ xy(\frac yx-\frac 45)=K$$ Exactly what you had with exact differential method $$\boxed{ y^2-\frac {4}5xy=K}$$

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    $\begingroup$ Ohhhh thanks . . Thumbs up 👍👍👍👍 $\endgroup$ – Naeem Ivy Mar 2 '18 at 17:05

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