Question

dy/dx=ytanx-2sinx

Answer 1

Hey raj, \(\Large \color{royalblue}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

Moving some stuff around gives us:\[\Large y'-(\tan x)y=-2\sin x\]

Answer 3

Our integrating factor will be:\[\Large \mu=e^{\int\limits(-\tan x)dx}, \qquad \mu=e^{\ln(\cos x)}, \qquad \mu=\cos x\]

Answer 4

Multiplying through by our integrating factor:\[\Large (\cos x) y'-(\sin x)y=-2\sin x \cos x\]Left side simplifies down, applying product rule in reverse:\[\Large \left(y \cos x\right)'\quad=\quad -2\sin x \cos x\]Integrating:\[\Large y \cos x\quad=\quad -2\int\limits \sin x \cos x\;dx\]

Answer 5

Shouldn't be too bad from there :)