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^{^{''}} + 4 y^{^{'}} + 13 y = e^{- t}

y^{^{''}} + 5 y^{^{'}} - 6 y = 2 e^{- x}

y^{^{''}} + 3 y^{^{'}} - y = t

y^{^{''}} - 5 y^{^{'}} + 4 y = 3 e^{x} sin (x)

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} (12 x^{2} - 6 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}