Solve the differential equation y''' sin^4 x = sin 2x

Solve the differential equation $y''' \sin^4 x = \sin 2x$.

Step 1: Simplify the right-hand side

Using $\sin 2x = 2\sin x\cos x$, \[y''' = \frac{\sin 2x}{\sin^4 x} = \frac{2\sin x\cos x}{\sin^4 x} = \frac{2\cos x}{\sin^3 x} = 2\cot x\csc^2 x.\]

Step 2: First integration

\[y'' = \int 2\cot x\csc^2 x\,dx.\]

With $u = \cot x$, so $du = -\csc^2 x\,dx$, \[y'' = \int 2u\cdot(-du) = -u^2 + C_1 = -\cot^2 x + C_1.\]

Using $\cot^2 x = \csc^2 x - 1$ and renaming $1+C_1$ as $C_1$, $y'' = -\csc^2 x + C_1.$

Step 3: Second integration

\[y' = \int(-\csc^2 x + C_1)\,dx = \cot x + C_1 x + C_2.\]

Step 4: Third integration

\[y = \int(\cot x + C_1 x + C_2)\,dx = \ln|\sin x| + \frac{C_1}{2}x^2 + C_2 x + C_3.\]

Step 5: General solution

Absorbing $C_1/2$ into $C_1$,

\[y = \ln|\sin x| + C_1 x^2 + C_2 x + C_3,\] valid for $x \neq k\pi$, $k\in\mathbb{Z}$.

Step 6: Verification

Differentiating the absorbed form three times: $y' = \cot x + 2C_1 x + C_2$, $y'' = -\csc^2 x + 2C_1$, $y''' = 2\csc^2 x\cot x = \frac{2\cos x}{\sin^3 x}$.

Then \[y'''\sin^4 x = \frac{2\cos x}{\sin^3 x}\cdot\sin^4 x = 2\sin x\cos x = \sin 2x.\]