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Question Number 55939 by MJS last updated on 06/Mar/19
$$\frac{{y}'}{{y}}=\mathrm{1}−\mathrm{cot}\:{x} \\ $$$${y}=? \\ $$
Answered by kaivan.ahmadi last updated on 06/Mar/19
$$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}=\mathrm{1}−{cotx}\Rightarrow \\ $$$$\frac{\mathrm{1}}{{y}}{dy}=\left(\mathrm{1}−{cotx}\right){dx}\Rightarrow \\ $$$$\int\frac{\mathrm{1}}{{y}}{dy}=\int\left(\mathrm{1}−{cotx}\right){dx}\Rightarrow \\ $$$${lny}={x}−{ln}\mid{sinx}\mid\Rightarrow{y}={e}^{{x}−{ln}\mid{sinx}\mid} \Rightarrow \\ $$$${y}=\frac{{e}^{{x}} }{\mid{sinx}\mid}+{C} \\ $$
Commented by MJS last updated on 06/Mar/19
$$\mathrm{thank}\:\mathrm{you} \\ $$