Solve differential equation y'+y\cot(x)=sin(2x)

Question

Solve differential equation

y^{'} + y \cot (x) = \sin (2 x)

Answer 1

y^{'} + P (x) y = Q (x)

I . F . = e^{\int P d x}

y (I . F .) = \int (I . F .) Q (x) d x + c

P (x) = \cot (x), Q (x) d x + c

I . F . = e^{\int \cot (x) d x} = e^{\log (\sin (x))} = \sin (x)

y (\sin (x)) = \int \sin (x) \sin (2 x) d x + c

y (\sin (x)) = \int \sin (x) 2 \sin (x) \cos (x) d x + c

y (\sin (x)) = 2 \int \sin^{2} (x) \cos (x) d x + c

Fisrt solve the integral
Let

\sin (x) = t \Rightarrow \cos (x) d x = d t

\int \sin^{2} (x) \cos (x) d x = \int t^{2} d t = \frac{t^{3}}{3} + c_{1} = \frac{\sin^{3} (x)}{3} + c_{1}

y (\sin (x)) = 2 \frac{\sin^{3} (x)}{3} + c_{1} + c

y (\sin (x)) = \frac{2}{3} \sin^{3} (x) + C, [c + c_{1} = C]

Expert Answer