What is the solution for (D^2-4D+13)y=24e^{2x}cos(3x) ?

The solution for (D^2-4D+13)y=24e^{2x}cos(3x) is y=e^{2x}(c_{1}cos(3x)+c_{2}sin(3x))+4e^{2x}xsin(3x)

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(D^2-4D+13)y=24e^{2x}cos(3x)

(D^{2} - 4 D + 13) y = 24 e^{2 x} cos (3 x)

y = e^{2 x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + 4 e^{2 x} x sin (3 x)

Solution steps

(D^{2} - 4 D + 13) y = 24 e^{2 x} cos (3 x)

Solve linear ODE:

y = e^{2 x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + 4 e^{2 x} x sin (3 x)

y = e^{2 x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + 4 e^{2 x} x sin (3 x)

y^{^{''}} - 2 y^{^{'}} + 10 y = 3 e^{x} tan (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 7 e^{- 2 t}

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

y^{^{''}} - 5 y = x^{2} - 2 x

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 e^{- t}

What is the solution for (D^2-4D+13)y=24e^{2x}cos(3x) ?
The solution for (D^2-4D+13)y=24e^{2x}cos(3x) is y=e^{2x}(c_{1}cos(3x)+c_{2}sin(3x))+4e^{2x}xsin(3x)