# Solution of Exact and Homogeneous differential equation

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.

• If you substitute each back into the ODE, do they both work? – Moo Mar 1 '18 at 16:58
• 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 – Naeem Ivy Mar 1 '18 at 17:05
• @Isham: Looks like by checking, s/he found his issues. – Moo Mar 1 '18 at 17:10
• Oh I see now @mpp – Isham Mar 1 '18 at 17:21

## 1 Answer

$$(5y - 2x) (\dfrac{dy}{dx}) - 2y = 0$$ $$y' =\frac {2y}{(5y - 2x)}$$ 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 $$\boxed{ y^2-\frac {4}5xy=K}$$

• Ohhhh thanks . . Thumbs up 👍👍👍👍 – Naeem Ivy Mar 2 '18 at 17:05