Question-162833

Question Number 162833 by mkam last updated on 01/Jan/22

Answered by abdullahhhhh last updated on 01/Jan/22

y=e^(x+y) (dy/dx)=(1+(dy/dx)) e^(x+y) (1−e^(x+y) )(dy/dx)=e^(x+y) (dy/dx)=(e^(x+y) /(1−e^(x+y) ))=(y/(1−y))

$$\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}} \\ $$$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\left(\mathrm{1}+\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\right)\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}} \\ $$$$\left(\mathrm{1}−\boldsymbol{{e}}^{\boldsymbol{{x}}+\boldsymbol{{y}}} \right)\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\boldsymbol{{e}}^{\boldsymbol{{x}}+\boldsymbol{{y}}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}+\boldsymbol{{y}}} }{\mathrm{1}−\boldsymbol{{e}}^{\boldsymbol{{x}}+\boldsymbol{{y}}} }=\frac{\boldsymbol{{y}}}{\mathrm{1}−\boldsymbol{{y}}} \\ $$$$ \\ $$

Commented by abdullahhhhh last updated on 01/Jan/22

look lny=ln(e^(x+e^(x+e^(x+...) ) )=x+e^(x+e^(x+...) ) ) lny=x+y

$$\boldsymbol{\mathrm{look}}\:\boldsymbol{\mathrm{lny}}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{e}}^{\left.\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}+…} } \right)=\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}+…} } } \right. \\ $$$$\boldsymbol{\mathrm{lny}}=\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:\: \\ $$$$ \\ $$

Commented by mkam last updated on 01/Jan/22