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

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

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

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

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 10 x - 8 + \frac{13}{10} e^{x}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 10 x - 8 + \frac{13}{10} e^{x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 10 x - 8 + \frac{13}{10} e^{x}

(d^{2} - 4) y (d, t) = 5 e^{3 t}, y (0) = 1

y^{^{''}} - 3 y^{^{'}} = 2 e^{x} sin (x)

y^{^{''}} - y^{^{'}} + y = sin (t)

y^{^{''}} + 2 y^{^{'}} + 3 y = 12 e^{x} + 3 x

y^{^{''}} - y = (1 - e^{2 x})^{- \frac{3}{2}}

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