What is the solution for y^{''}+9y=9sec^2(3x) ?

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

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y^{''}+9y=9sec^2(3x)

y^{^{''}} + 9 y = 9 sec^{2} (3 x)

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

Solution steps

y^{''} (x) + 9 y = 9 sec^{2} (3 x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 8 e^{- t} cos (2 t)

y^{^{''}} + 4 y = 5 t^{2} e^{t}

y^{^{''}} + y = cos (2 t)

D^{2} (D - 2)^{3} y = 48 e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}

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