Find the general solution of y''sin^3 x = 4ysin 3x , given…

Find the general solution of $y''\sin^3 x = 4y\sin 3x$, given that $y_1=\sin^4x$ is a particular solution.

Step 1: Standard form

Using $\sin 3x = 3\sin x-4\sin^3x$: \[\frac{4\sin3x}{\sin^3x} = \frac{4(3\sin x-4\sin^3x)}{\sin^3x} = \frac{12}{\sin^2x}-16 = 12\csc^2x-16,\]

\[ y'' - (12\csc^2x-16)y = 0. \]

Step 2: Verify $y_1$

\[ y_1' = 4\sin^3x\cos x, \qquad y_1'' = 12\sin^2x\cos^2x-4\sin^4x = 12\sin^2x-16\sin^4x. \]

\[ y_1''-(12\csc^2x-16)y_1 = (12\sin^2x-16\sin^4x)-(12\sin^2x-16\sin^4x)=0.\ \checkmark \]

Step 3: Substitution $y=y_1v$

$y' = 4\sin^3x\cos x\,v+\sin^4x\,v'$, $y'' = (12\sin^2x-16\sin^4x)v + 8\sin^3x\cos x\,v' + \sin^4x\,v''$.

Step 4: Reduce the order

Substituting into $y''-(12\csc^2x-16)y=0$, the $v$ terms reproduce Step 2's identity and cancel, leaving

\[ \sin^4x\,v'' + 8\sin^3x\cos x\,v' = 0 \implies \sin x\,v''+8\cos x\,v'=0 \quad(\text{divide by }\sin^3x). \]

Step 5: Solve for $w=v'$

\[ \frac{dw}{w} = -8\cot x\,dx \implies \ln|w| = -8\ln|\sin x|+\mathrm{const} \implies w = \frac{A}{\sin^8x}. \]

Step 6: Integrate for $v$: a cosecant reduction chain

We need $v = A\displaystyle\int\csc^8x\,dx+B$. The reduction formula

\[ I_n = \int\csc^nx\,dx = -\frac{\csc^{n-2}x\cot x}{n-1}+\frac{n-2}{n-1}I_{n-2} \]

(from $\frac{d}{dx}\bigl[\csc^{n-2}x\cot x\bigr]=-(n-1)\csc^nx+(n-2)\csc^{n-2}x$), starting from $I_2=-\cot x$, gives

\[ I_4 = -\frac{\csc^2x\cot x}{3}+\frac23I_2 = -\frac{\cot x(\csc^2x+2)}{3}, \]

\[ I_6 = -\frac{\csc^4x\cot x}{5}+\frac45I_4 = -\frac{\csc^4x\cot x}{5}-\frac{4\cot x(\csc^2x+2)}{15}, \]

\[ I_8 = -\frac{\csc^6x\cot x}{7}+\frac67I_6 = -\frac{\csc^6x\cot x}{7}-\frac{6\csc^4x\cot x}{35}-\frac{8\cot x(\csc^2x+2)}{35}. \]

Combining over the common denominator 35: \[I_8 = -\frac{\cot x}{35}\left(5\csc^6x+6\csc^4x+8\csc^2x+16\right), \qquad v = -\frac{A\cot x}{35}\left(5\csc^6x+6\csc^4x+8\csc^2x+16\right)+B.\]

Step 7: General solution

Multiplying by $y_1=\sin^4x$ and using $\sin^4x\cot x\csc^{2k}x = \cos x\sin^{3-2k}x$: \[y = B\sin^4x - \frac{A\cos x}{35}\left(\frac{5}{\sin^3x}+\frac{6}{\sin x}+8\sin x+16\sin^3x\right).\]

Renaming $C_1=B$, $C_2=-A/35$:

Step 8: Linear independence

The equation has no $y'$ term ($P=0$), so Abel's formula gives a constant Wronskian. With $y_2$ built from $w=v'=\csc^8x$ (taking $A=1$): \[W(y_1,y_2) = y_1^2w = \sin^8x\cdot\csc^8x = 1 \neq 0,\]

confirming that $y_1,y_2$ form a fundamental system.