pablos28spainzd

2022-07-01

I have this simple differential equation: ${y}^{\prime }=\left(\mathrm{tan}x\right)y.$
after integrating $\frac{{y}^{\prime }}{y\left(x\right)}=\mathrm{tan}x$
i came up with $\mathrm{log}y\left(x\right)=-\mathrm{log}\mathrm{cos}\left(x\right)+1.$
now my question is this one: why ${e}^{-\mathrm{log}\mathrm{cos}\left(x\right)}=\mathrm{sec}\left(x\right)?$

Jamarcus Shields

Expert

I suppose that $\mathrm{log}\left(x\right)$ is for $\mathrm{ln}\left(x\right)$
${e}^{-\mathrm{log}\mathrm{cos}\left(x\right)}=\left({e}^{\mathrm{log}\mathrm{cos}\left(x\right)}{\right)}^{-1}=\left(\mathrm{cos}\left(x\right){\right)}^{-1}=\frac{1}{\mathrm{cos}\left(x\right)}=\mathrm{sec}\left(x\right)$

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