What is the solution for y^{''}+9y=18(sec(3x))^3 ?

The solution for y^{''}+9y=18(sec(3x))^3 is y=c_{1}cos(3x)+c_{2}sin(3x)+tan(3x)sin(3x)

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y^{''}+9y=18(sec(3x))^3

y^{^{''}} + 9 y = 18 (sec (3 x))^{3}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + tan (3 x) sin (3 x)

Solution steps

y^{''} (x) + 9 y = 18 (sec (3 x))^{3}

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + tan (3 x) sin (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + tan (3 x) sin (3 x)

y^{^{''}} - 3 y^{^{'}} - 4 y = 6

3 y^{^{''}} + 192.08 y = 48.02 cos (8 t)

y^{^{''}} - 2 y^{^{'}} - 24 y = 16 - (x + 2) e^{4 x}

y^{^{''}} - 7 y^{^{'}} + 12 y = x

3 \frac{d ^{2} x}{d t ^{2}} + 10 x = 2 sin (3 t)

What is the solution for y^{''}+9y=18(sec(3x))^3 ?
The solution for y^{''}+9y=18(sec(3x))^3 is y=c_{1}cos(3x)+c_{2}sin(3x)+tan(3x)sin(3x)