What is the solution for y^{''}=2y+18tan^3(x) ?

The solution for y^{''}=2y+18tan^3(x) is

Popular >

y^{''}=2y+18tan^3(x)

y^{^{''}} = 2 y + 18 tan^{3} (x)

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

Solution steps

y^{''} (x) = 2 y + 18 tan^{3} (x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 2 y = e^{- 2 x} (3 x^{2} + 5 x + 2)

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (x) + e^{x}

5 y^{^{''}} + 25 y^{^{'}} + \frac{1}{1 0 ^{- 2}} y = 10 e^{- \frac{x}{4}}

y^{^{''}} + 2 y^{^{'}} - 35 y = 13 sin (x) - e^{3 x} + 1

y^{^{''}} + a y^{^{'}} + y = e^{- 4 t}

What is the solution for y^{''}=2y+18tan^3(x) ?
The solution for y^{''}=2y+18tan^3(x) is