Find the general solution of y'' + (tan x - 2cot x) y' +…

Find the general solution of $y'' + (\tan x - 2\cot x)\,y' + 2\cot^2 x\cdot y = 0$, given that $y_1 = \sin x$ is a particular solution.

Step 1: Verify $y_1$

\[ y_1' = \cos x, \qquad y_1'' = -\sin x. \]

Using $\tan x\cos x=\sin x$, $\cot x\cos x=\dfrac{\cos^2x}{\sin x}$, and $\cot^2x\sin x=\dfrac{\cos^2x}{\sin x}$: $-\sin x + \sin x - \frac{2\cos^2x}{\sin x} + \frac{2\cos^2x}{\sin x} = 0.$ $\checkmark$

Step 2: Substitution $y=y_1v$

\[ y' = \cos x\,v+\sin x\,v', \qquad y'' = -\sin x\,v+2\cos x\,v'+\sin x\,v''. \]

Step 3: Reduce the order

Substitute into the equation and expand $(\tan x-2\cot x)(\cos x\,v+\sin x\,v')$: $\sin x\,v-\frac{2\cos^2x}{\sin x}v \;+\; \frac{\sin^2x}{\cos x}v'-2\cos x\,v'.$

The coefficient of $v$ reproduces Step 1's identity (times $v$) and vanishes. Collecting $v'$ and $v''$: \[\sin x\,v'' + \left(2\cos x+\frac{\sin^2x}{\cos x}-2\cos x\right)v' = \sin x\,v''+\frac{\sin^2x}{\cos x}v' = 0.\]

Dividing by $\sin x$: $v'' + \tan x\,v' = 0.$

Step 4: Solve for $w=v'$

\[ \frac{dw}{w} = -\tan x\,dx \implies \ln|w| = \ln|\cos x|+\mathrm{const} \implies w = A\cos x. \]

Step 5: Integrate for $v$

\[ v' = A\cos x \implies v = A\sin x + B. \]

Step 6: General solution

Then $y=\sin x\,(A\sin x+B) = A\sin^2x+B\sin x$. Renaming $B=C_1$, $A=C_2$:

Step 7: Verify $y_2=\sin^2x$

\[ y_2' = \sin 2x, \qquad y_2'' = 2\cos 2x = 2-4\sin^2x. \]

\[ (2-4\sin^2x) + (\tan x-2\cot x)\sin2x + 2\cot^2x\sin^2x = (2-4\sin^2x)+(2\sin^2x-4\cos^2x)+2\cos^2x \]

\[ = 2-2\sin^2x-2\cos^2x = 2-2(\sin^2x+\cos^2x) = 0.\ \checkmark \]

Step 8: Linear independence

\[ W(y_1,y_2) = \sin x\cdot2\sin x\cos x - \sin^2x\cdot\cos x = \sin^2x\cos x \neq 0 \]

for $x\neq 0,\pi/2,\pi,\dots$, so $y_1,y_2$ form a fundamental system.