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Question Number 181615 by Mastermind last updated on 27/Nov/22

Determine whether this is Homogenous  or not  (dy/dx)=(y/(y−2x))    .

$$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{this}\:\mathrm{is}\:\mathrm{Homogenous} \\ $$$$\mathrm{or}\:\mathrm{not} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{y}−\mathrm{2x}} \\ $$$$ \\ $$$$. \\ $$

Answered by hmr last updated on 27/Nov/22

(dy/dx) = ((((y/x)))/(((y/x))−2))    • z = ((y/x)) → (dy/dx) = z + x (dz/dx)    z + x (dz/dx) = (z/(z − 2))  x (dz/dx) = (z/(z − 2)) − z   x (dz/dx) =  ((3z − z^2 )/(z − 2))  −(1/x) + ((z − 2)/(3z − z^2 )) (dz/dx) = 0  ∫−(1/x)dx + ∫((z − 2)/(3z − z^2 )) dz = c    • ((z − 2)/(3z − z^2 )) = (a/(3−z)) + (b/z)  z − 2 = az + b(3 − z)  z − 2 = (a − b)z + 3b   { ((a − b = 1)),((3b = −2 → b = ((−2)/3) → a = (1/3))) :}    −ln∣x∣ + (1/3)ln∣3 − z∣ −(2/3)ln∣z∣= c  −ln∣x∣ + (1/3)ln∣3 − (y/x)∣ −(2/3)ln∣(y/x)∣= c

$$\frac{{dy}}{{dx}}\:=\:\frac{\left(\frac{{y}}{{x}}\right)}{\left(\frac{{y}}{{x}}\right)−\mathrm{2}} \\ $$$$ \\ $$$$\bullet\:{z}\:=\:\left(\frac{{y}}{{x}}\right)\:\rightarrow\:\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$$ \\ $$$${z}\:+\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{z}}{{z}\:−\:\mathrm{2}} \\ $$$${x}\:\frac{{dz}}{{dx}}\:=\:\frac{{z}}{{z}\:−\:\mathrm{2}}\:−\:{z}\: \\ $$$${x}\:\frac{{dz}}{{dx}}\:=\:\:\frac{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }{{z}\:−\:\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{{x}}\:+\:\frac{{z}\:−\:\mathrm{2}}{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }\:\frac{{dz}}{{dx}}\:=\:\mathrm{0} \\ $$$$\int−\frac{\mathrm{1}}{{x}}{dx}\:+\:\int\frac{{z}\:−\:\mathrm{2}}{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }\:{dz}\:=\:{c} \\ $$$$ \\ $$$$\bullet\:\frac{{z}\:−\:\mathrm{2}}{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }\:=\:\frac{{a}}{\mathrm{3}−{z}}\:+\:\frac{{b}}{{z}} \\ $$$${z}\:−\:\mathrm{2}\:=\:{az}\:+\:{b}\left(\mathrm{3}\:−\:{z}\right) \\ $$$${z}\:−\:\mathrm{2}\:=\:\left({a}\:−\:{b}\right){z}\:+\:\mathrm{3}{b} \\ $$$$\begin{cases}{{a}\:−\:{b}\:=\:\mathrm{1}}\\{\mathrm{3}{b}\:=\:−\mathrm{2}\:\rightarrow\:{b}\:=\:\frac{−\mathrm{2}}{\mathrm{3}}\:\rightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$ \\ $$$$−{ln}\mid{x}\mid\:+\:\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{3}\:−\:{z}\mid\:−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{z}\mid=\:{c} \\ $$$$−{ln}\mid{x}\mid\:+\:\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{3}\:−\:\frac{{y}}{{x}}\mid\:−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\frac{{y}}{{x}}\mid=\:{c} \\ $$

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