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Question Number 108498 by mohammad17 last updated on 17/Aug/20

Answered by Aziztisffola last updated on 17/Aug/20

$$1)\mathrm{y}={(\mathrm{x}+2)}^{\mathrm{x}}={\mathrm{e}}^{\mathrm{xln}(\mathrm{x}+2)}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=(\ln (\mathrm{x}+2)+\frac{\mathrm{x}}{\mathrm{x}+2}){(\mathrm{x}+2)}^{\mathrm{x}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}(-1)=-1$$

Commented by mohammad17 last updated on 17/Aug/20

$$thankyousir$$

Answered by Aziztisffola last updated on 17/Aug/20

$$3){\mathrm{xy}}^{\prime }+\mathrm{y}={\mathrm{e}}^{\mathrm{x}}\Rightarrow {\mathrm{y}}^{\prime }+\frac{1}{\mathrm{x}}\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}$$$$\mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \frac{\mathrm{dx}}{\mathrm{x}}}={\mathrm{e}}^{\ln \mid \mathrm{x}\mid }=\mathrm{x}$$$$\mathrm{yx}=\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}\mathrm{xdx}={\mathrm{e}}^{\mathrm{x}}+\mathrm{C}$$$$\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}+\frac{\mathrm{C}}{\mathrm{x}}$$$$4)\mathrm{ch}\left(5\mathrm{x}\right)+\mathrm{sh}\left(5\mathrm{x}\right)={\mathrm{e}}^{5\mathrm{x}}$$$$\int \frac{\mathrm{dx}}{\mathrm{ch}\left(5\mathrm{x}\right)+\mathrm{sh}\left(5\mathrm{x}\right)}=\int \frac{\mathrm{dx}}{{\mathrm{e}}^{5\mathrm{x}}}=\int {\mathrm{e}}^{-5\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{5}{\mathrm{e}}^{-5\mathrm{x}}+\mathrm{C}$$$$5)\overrightarrow{{\mathrm{V}}_1}(1;1;1)\mathrm{and}\overrightarrow{{\mathrm{V}}_2}(1;1;-2)$$$$\overrightarrow{{\mathrm{V}}_1}.\overrightarrow{{\mathrm{V}}_2}=1+1-2=0\Rightarrow \overrightarrow{{\mathrm{V}}_1}\mathrm{\perp }\overrightarrow{{\mathrm{V}}_2}$$

Commented by mohammad17 last updated on 17/Aug/20

$$thankyousir$$

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