The proof of d^2y/dx^2 - 2 cot x dy/dx +3y = 0 is determined as 6cos²xsinx - 3sin³x - 6cos²xsinx + 3sin³x = 0
The equation is given as;
[tex]\frac{d^2y}{dx^2} - 2\ cotx \ \frac{dy}{dx} \ + 3y = 0 \ \ ---(1)[/tex]
y = sin³x
y = sin³x
let u = sinx (du/dx = cosx)
y = u³
dy/du = 3u²
[tex]\frac{dy}{dx} = \frac{dy}{du} \ .\ \frac{du}{dx} \\\\[/tex]
dy/dx = 3u² x cosx
dy/dy = 3sin²xcosx
y' = 3sin²xcosx
y' = 3sin²xcosx
let u = sin²x (du/dx = 2sinxcosx)
v = cosx (dv/dx = -sinx)
[tex]y'' = \frac{d^2y}{dx^2} = V\frac{du}{dx} + U\frac{dv}{dx}[/tex]
y'' = 3[cosx(2sinxcosx) + sin²x(-sinx)]
y'' = 6cos²xsinx - 3sin³x
Substitute first derivative and second derivative of y in equation (1)
[tex]\frac{d^2y}{dx^2} - 2\ cotx \ \frac{dy}{dx} \ + 3y \\\\(6cos^2xsinx \ - \ 3sin^3x) \ - \ 2(\frac{cosx}{sinx} )(3sin^2xcosx) \ + 3(sin^3x) \\\\6cos^2xsinx \ - \ 3sin^3x \ - 6cos^2xsinx \ + \ 3sin^3x = 0 \ \ proved[/tex]
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