Find the general solution of y'' + y'tan x - sin 2x = 0

Find the general solution of $y'' + y'\tan x - \sin 2x = 0$.

Step 1: Substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[p' + p\tan x = \sin 2x. \tag{1}\]

This is a first-order linear ODE in $p(x)$.

Step 2: Integrating factor

\[ \mu = e^{\displaystyle\int \tan x\,dx} = e^{\ln|\sec x|} = \frac{1}{\cos x}. \]

Step 3: Multiply by the integrating factor

Multiply (1) by $\dfrac{1}{\cos x}$: \[\frac{d}{dx}\left[\frac{p}{\cos x}\right] = \frac{\sin 2x}{\cos x}.\]

Using $\sin 2x = 2\sin x\cos x$: $\frac{d}{dx}\left[\frac{p}{\cos x}\right] = 2\sin x.$

Step 4: First integration

\[ \frac{p}{\cos x} = \int 2\sin x\,dx = -2\cos x + C_1. \]

Thus: \[p = y' = -2\cos^2 x + C_1\cos x. \tag{2}\]

Step 5: Second integration

Integrate (2) using $\cos^2 x = \dfrac{1+\cos 2x}{2}$: $y = \int\left(-(1+\cos 2x) + C_1\cos x\right)dx = -x - \frac{\sin 2x}{2} + C_1\sin x + C_2.$

Step 6: General solution

The solution is recorded in the result below.

Step 7: Verification

Differentiate: $y' = -1 - \cos 2x + C_1\cos x = -2\cos^2 x + C_1\cos x$, $y'' = 4\cos x\sin x - C_1\sin x = 2\sin 2x - C_1\sin x$.

Check $y'' + y'\tan x - \sin 2x$: $2\sin 2x - C_1\sin x + (-2\cos^2 x + C_1\cos x)\tan x - \sin 2x$.

\[ = 2\sin 2x - C_1\sin x - 2\cos x\sin x + C_1\sin x - \sin 2x = 2\sin 2x - \sin 2x - \sin 2x = 0.\ \checkmark \]