Question

dy/dx - 3ycotx = sin2x

Answer 1

The general solution to the differential equation is: $y = -2\sin^2 x + C\sin^3 x$

Answer 2

The given differential equation is: $\frac{dy}{dx} - 3y\cot x = \sin 2x$ This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Here, $P(x) = -3\cot x$ and $Q(x) = \sin 2x$. The integrating factor (I.F.) is $e^{\int P(x) dx}$. $\int P(x) dx = \int -3\cot x dx = -3\ln|\sin x| = \ln(\sin^{-3} x)$ $I.F. = e^{\ln(\sin^{-3} x)} = \sin^{-3} x = \csc^3 x$ The general solution is $y \times (I.F.) = \int Q(x) \times (I.F.) dx + C$. $y \csc^3 x = \int \sin 2x \csc^3 x dx + C$ The integral is evaluated as: $\int \sin 2x \csc^3 x dx = \int (2\sin x \cos x) \csc^3 x dx = \int \frac{2\cos x}{\sin^2 x} dx$ Using substitution $u = \sin x$, $du = \cos x dx$: $\int 2u^{-2} du = -2u^{-1} = -2\csc x$ Thus, the solution becomes: $y \csc^3 x = -2\csc x + C$ Solving for $y$: $y = (-2\csc x + C)\sin^3 x = -2\sin^2 x + C\sin^3 x$