What is the solution for y^{''}+4y^'+5y=-15x+e-x ?

The solution for y^{''}+4y^'+5y=-15x+e-x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-(16x)/5+(5e+64)/(25)

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y^{''}+4y^'+5y=-15x+e-x

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + e - x

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{16 x}{5} + \frac{5 e + 64}{25}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = - 15 x + e - x

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{16 x}{5} + \frac{5 e + 64}{25}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{16 x}{5} + \frac{5 e + 64}{25}

y^{^{''}} + 2 y = t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 8

y^{^{''}} - y = - 15 t + 4

y^{^{''}} - y = - 15 t + 5

y^{^{''}} + 6 y^{^{'}} + 5 y = 12 e^{x}, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} - 2 y^{^{'}} + y = 3 t + 1 e^{- t}

What is the solution for y^{''}+4y^'+5y=-15x+e-x ?
The solution for y^{''}+4y^'+5y=-15x+e-x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-(16x)/5+(5e+64)/(25)