What is the solution for y^{''}-2y=18(cot(t))^3 ?

The solution for y^{''}-2y=18(cot(t))^3 is

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y^{''}-2y=18(cot(t))^3

y^{^{''}} - 2 y = 18 (cot (t))^{3}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + e^{2 t} \cdot \int \frac{9 2 e ^{- 2 t} cot ^{3} ( t )}{2} d t + e^{- 2 t} \cdot \int - \frac{9 2 e ^{2 t} cot ^{3} ( t )}{2} d t

Solution steps

y^{''} (t) - 2 y = 18 (cot (t))^{3}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + e^{2 t} \cdot \int \frac{9 2 e ^{- 2 t} cot ^{3} ( t )}{2} d t + e^{- 2 t} \cdot \int - \frac{9 2 e ^{2 t} cot ^{3} ( t )}{2} d t

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + e^{2 t} \cdot \int \frac{9 2 e ^{- 2 t} cot ^{3} ( t )}{2} d t + e^{- 2 t} \cdot \int - \frac{9 2 e ^{2 t} cot ^{3} ( t )}{2} d t

y^{^{''}} - 2 y^{^{'}} + 2 y = e x sin (x)

- 4 y^{^{''}} - 4 y - y = - (t^{2} + t + 6 e^{- 2} t)

y^{^{''}} + 6 y^{^{'}} + 5 y = e^{x}

y^{^{''}} + 5 y^{^{'}} + 6 = sin (x) - cos (2 x)

y^{^{'''}} - y^{^{'}} = e^{(x)} + e^{(2 x)}

What is the solution for y^{''}-2y=18(cot(t))^3 ?
The solution for y^{''}-2y=18(cot(t))^3 is