What is the solution for y^{''}+9y=9sec^2(3t) ?

The solution for y^{''}+9y=9sec^2(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)-1+ln(sec(3t)+tan(3t))sin(3t)

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y^{''}+9y=9sec^2(3t)

y^{^{''}} + 9 y = 9 sec^{2} (3 t)

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

Solution steps

y^{''} (t) + 9 y = 9 sec^{2} (3 t)

Solve linear ODE:

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

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

y^{^{''}} + 25 y = 40 cos (5 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}, y (4) = 1, y^{^{'}} (4) = 3

y^{^{''}} + 25 y = 2 x cos (5 x)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 12 y = 3 cos (x)

4 y^{^{''}} - 16 y = \frac{1}{e ^{x}}

What is the solution for y^{''}+9y=9sec^2(3t) ?
The solution for y^{''}+9y=9sec^2(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)-1+ln(sec(3t)+tan(3t))sin(3t)