Question

Solve differential equation \(\displaystyle{y}'+{y}{\cot{{\left({x}\right)}}}={\sin{{\left({2}{x}\right)}}}\)

Answer 1

\(\displaystyle{y}'+{P}{\left({x}\right)}{y}={Q}{\left({x}\right)}\)
\(\displaystyle{I}.{F}.={e}^{{\int{P}{\left.{d}{x}\right.}}}\)
\(\displaystyle{y}{\left({I}.{F}.\right)}=\int{\left({I}.{F}.\right)}{Q}{\left({x}\right)}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{P}{\left({x}\right)}={\cot{{\left({x}\right)}}},\ {Q}{\left({x}\right)}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{I}.{F}.={e}^{{\int{\cot{{\left({x}\right)}}}{\left.{d}{x}\right.}}}={e}^{{{\log{{\left({\sin{{\left({x}\right)}}}\right)}}}}}={\sin{{\left({x}\right)}}}\)
\(\displaystyle{y}{\left({\sin{{\left({x}\right)}}}\right)}=\int{\sin{{\left({x}\right)}}}{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{y}{\left({\sin{{\left({x}\right)}}}\right)}=\int{\sin{{\left({x}\right)}}}{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{y}{\left({\sin{{\left({x}\right)}}}\right)}={2}\int{{\sin}^{{2}}{\left({x}\right)}}{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}+{c}\)
Fisrt solve the integral
Let
\(\displaystyle{\sin{{\left({x}\right)}}}={t}\Rightarrow{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\)
\(\displaystyle\int{{\sin}^{{2}}{\left({x}\right)}}{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}=\int{t}^{{2}}{\left.{d}{t}\right.}={\frac{{{t}^{{3}}}}{{{3}}}}+{c}_{{1}}={\frac{{{{\sin}^{{3}}{\left({x}\right)}}}}{{{3}}}}+{c}_{{1}}\)
\(\displaystyle{y}{\left({\sin{{\left({x}\right)}}}\right)}={2}{\frac{{{{\sin}^{{3}}{\left({x}\right)}}}}{{{3}}}}+{c}_{{1}}+{c}\)
\(\displaystyle{y}{\left({\sin{{\left({x}\right)}}}\right)}={\frac{{{2}}}{{{3}}}}{{\sin}^{{3}}{\left({x}\right)}}+{C},\ {\left[{c}+{c}_{{1}}={C}\right]}\)