What is the solution for (D^2-4D+13)y=40sin(3x) ?

The solution for (D^2-4D+13)y=40sin(3x) is y=e^{2x}(c_{1}cos(3x)+c_{2}sin(3x))+sin(3x)+3cos(3x)

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(D^2-4D+13)y=40sin(3x)

(D^{2} - 4 D + 13) y = 40 sin (3 x)

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

Solution steps

(D^{2} - 4 D + 13) y = 40 sin (3 x)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 9 sin (x))

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 5

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 5 e^{- 3 x} - e^{- x}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} = x^{2}

What is the solution for (D^2-4D+13)y=40sin(3x) ?
The solution for (D^2-4D+13)y=40sin(3x) is y=e^{2x}(c_{1}cos(3x)+c_{2}sin(3x))+sin(3x)+3cos(3x)