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Question Number 91643 by jagoll last updated on 02/May/20

(dy/dx) = ((y−x)/x)

$$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}}{{x}}\: \\ $$$$ \\ $$

Commented by john santu last updated on 02/May/20

Commented by Prithwish Sen 1 last updated on 02/May/20

ydx−xdy = xdy  −(xdy−ydx)=xdy  −((xdy−ydx)/x^2 ) = (dy/x)  −∫d((y/x))=∫(dx/x)  −(y/x) = lnCx      C= constant  Cx = e^((−y)/x)

$$\mathrm{ydx}−\mathrm{xdy}\:=\:\mathrm{xdy} \\ $$$$−\left(\mathrm{xdy}−\mathrm{ydx}\right)=\mathrm{xdy} \\ $$$$−\frac{\mathrm{xdy}−\mathrm{ydx}}{\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{dy}}{\mathrm{x}} \\ $$$$−\int\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\int\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$−\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{lnCx}\:\:\:\:\:\:\mathrm{C}=\:\mathrm{constant} \\ $$$$\boldsymbol{\mathrm{Cx}}\:=\:\boldsymbol{\mathrm{e}}^{\frac{−\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}} \\ $$

Commented by jagoll last updated on 02/May/20

thank you both

$${thank}\:{you}\:{both} \\ $$

Commented by mathmax by abdo last updated on 02/May/20

⇒(dy/dx) =(y/x)−1  let (y/x) =z ⇒y =xz ⇒(dy/dx) =z +x (dz/dx) ⇒  z +x z^′  =z−1 ⇒xz^′  =−1 ⇒z^′ =−(1/x) ⇒z =−lnx +k ⇒  y =x(−lnx +k) =kx −xlnx

$$\Rightarrow\frac{{dy}}{{dx}}\:=\frac{{y}}{{x}}−\mathrm{1}\:\:{let}\:\frac{{y}}{{x}}\:={z}\:\Rightarrow{y}\:={xz}\:\Rightarrow\frac{{dy}}{{dx}}\:={z}\:+{x}\:\frac{{dz}}{{dx}}\:\Rightarrow \\ $$$${z}\:+{x}\:{z}^{'} \:={z}−\mathrm{1}\:\Rightarrow{xz}^{'} \:=−\mathrm{1}\:\Rightarrow{z}^{'} =−\frac{\mathrm{1}}{{x}}\:\Rightarrow{z}\:=−{lnx}\:+{k}\:\Rightarrow \\ $$$${y}\:={x}\left(−{lnx}\:+{k}\right)\:={kx}\:−{xlnx} \\ $$$$ \\ $$

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