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

Answered by abdullahhhhh last updated on 01/Jan/22

$$\mathbf{y}={\mathbf{e}}^{\mathbf{x}+\mathbf{y}}$$$$\frac{\mathbf{dy}}{\mathbf{dx}}=(1+\frac{\mathbf{dy}}{\mathbf{dx}}){\mathbf{e}}^{\mathbf{x}+\mathbf{y}}$$$$(1-{\mathit{e}}^{\mathit{x}+\mathit{y}})\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}={\mathit{e}}^{\mathit{x}+\mathit{y}}$$$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{{\mathit{e}}^{\mathit{x}+\mathit{y}}}{1-{\mathit{e}}^{\mathit{x}+\mathit{y}}}=\frac{\mathit{y}}{1-\mathit{y}}$$$$$$

Commented by abdullahhhhh last updated on 01/Jan/22

$$\mathbf{look}\mathbf{lny}=\mathbf{ln}({\mathbf{e}}^{\mathbf{x}+{\mathbf{e}}^{\mathbf{x}+{\mathbf{e}}^{\mathbf{x}+...}})=\mathbf{x}+{\mathbf{e}}^{\mathbf{x}+{\mathbf{e}}^{\mathbf{x}+...}}}$$$$\mathbf{lny}=\mathbf{x}+\mathbf{y}$$$$$$

Commented by mkam last updated on 01/Jan/22

$$whereareyoufromabdullah$$

Commented by mkam last updated on 01/Jan/22

$$howy=e^{x+y}?$$

Commented by abdullahhhhh last updated on 01/Jan/22

$$\mathbf{from}\mathbf{egypt}$$

Commented by mkam last updated on 01/Jan/22

$$youarewelcomiamfromiraq$$

Commented by mkam last updated on 01/Jan/22

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{r}$$

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