What is the solution for y^{''}=2y+20tan^3(x),y(0)=10tan(x) ?

The solution for y^{''}=2y+20tan^3(x),y(0)=10tan(x) is

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y^{''}=2y+20tan^3(x),y(0)=10tan(x)

y^{^{''}} = 2 y + 20 tan^{3} (x), y (0) = 10 tan (x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int 52 e^{- 2 x} tan^{3} (x) d x + e^{- 2 x} \cdot \int - 52 e^{2 x} tan^{3} (x) d x

Solution steps

y^{''} (x) = 2 y + 20 tan^{3} (x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int 52 e^{- 2 x} tan^{3} (x) d x + e^{- 2 x} \cdot \int - 52 e^{2 x} tan^{3} (x) d x

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int 52 e^{- 2 x} tan^{3} (x) d x + e^{- 2 x} \cdot \int - 52 e^{2 x} tan^{3} (x) d x

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{4} e^{3 t}

y^{^{''}} + 4 y = 11 t^{3} sin (2 t)

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 4 x

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8, y (0) = 2, y^{^{'}} (0) = 2

What is the solution for y^{''}=2y+20tan^3(x),y(0)=10tan(x) ?
The solution for y^{''}=2y+20tan^3(x),y(0)=10tan(x) is