What is the solution for (D^2+4D+5)y=50x+13e^{3x} ?

The solution for (D^2+4D+5)y=50x+13e^{3x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+10x-8+1/2 e^{3x}

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(D^2+4D+5)y=50x+13e^{3x}

(D^{2} + 4 D + 5) y = 50 x + 13 e^{3 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 10 x - 8 + \frac{1}{2} e^{3 x}

Solution steps

(D^{2} + 4 D + 5) y = 50 x + 13 e^{3 x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 10 x - 8 + \frac{1}{2} e^{3 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 10 x - 8 + \frac{1}{2} e^{3 x}

y^{^{''}} + y = 4 x cos (x)

(D^{2} - 1) y = \frac{2}{1 + e ^{x}}

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 x^{2} + x + 9

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{9 + e ^{x}}

What is the solution for (D^2+4D+5)y=50x+13e^{3x} ?
The solution for (D^2+4D+5)y=50x+13e^{3x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+10x-8+1/2 e^{3x}