What is the solution for y^{''}+12y=4e^{9x} ?

The solution for y^{''}+12y=4e^{9x} is y=c_{1}cos(2sqrt(3)x)+c_{2}sin(2sqrt(3)x)+4/93 e^{9x}

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y^{''}+12y=4e^{9x}

y^{^{''}} + 12 y = 4 e^{9 x}

y = c_{1} cos (23 x) + c_{2} sin (23 x) + \frac{4}{93} e^{9 x}

Solution steps

y^{''} (x) + 12 y = 4 e^{9 x}

Solve linear ODE:

y = c_{1} cos (23 x) + c_{2} sin (23 x) + \frac{4}{93} e^{9 x}

y = c_{1} cos (23 x) + c_{2} sin (23 x) + \frac{4}{93} e^{9 x}

0.01 y^{^{''}} - 0.06 y^{^{'}} + 0.08 y = 10 x

y^{^{''}} - y^{^{'}} - 2 y = 10, y (- 1) = 0, y^{^{'}} (- 1) = 1

y^{^{''}} - 3 y^{^{'}} + 2 y = 1

y^{^{''}} + 36 y = 288 sin (6 x) + 96 cos (6 x)

y^{^{''}} - y^{^{'}} - 2 y = 1, + t y (0) = 2, y^{^{'}} (0) = 0

What is the solution for y^{''}+12y=4e^{9x} ?
The solution for y^{''}+12y=4e^{9x} is y=c_{1}cos(2sqrt(3)x)+c_{2}sin(2sqrt(3)x)+4/93 e^{9x}