What is the solution for y^{''}=2y+tan^3(x),y^1(x)= 1/2 tan(x) ?

The solution for y^{''}=2y+tan^3(x),y^1(x)= 1/2 tan(x) is

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y^{''}=2y+tan^3(x),y^1(x)= 1/2 tan(x)

y^{^{''}} = 2 y + tan^{3} (x), y^{1} (x) = \frac{1}{2} tan (x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{2 e ^{- 2 x} tan ^{3} ( x )}{4} d x + e^{- 2 x} \cdot \int - \frac{2 e ^{2 x} tan ^{3} ( x )}{4} d x

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{2 e ^{- 2 x} tan ^{3} ( x )}{4} d x + e^{- 2 x} \cdot \int - \frac{2 e ^{2 x} tan ^{3} ( x )}{4} d x

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{2 e ^{- 2 x} tan ^{3} ( x )}{4} d x + e^{- 2 x} \cdot \int - \frac{2 e ^{2 x} tan ^{3} ( x )}{4} d x

What is the solution for y^{''}=2y+tan^3(x),y^1(x)= 1/2 tan(x) ?
The solution for y^{''}=2y+tan^3(x),y^1(x)= 1/2 tan(x) is