What is the solution for y^{''}+4y^'+5y=5x-1 ?

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

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y^{''}+4y^'+5y=5x-1

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x - 1

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 5 x - 1

Solve linear ODE:

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

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

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

\frac{\partial}{\partial y} (16 - x^{2} - 4 y)

y^{^{''}} - y = sin (2 t)

f (x) = \frac{e ^{x}}{x + 3}

\frac{( s + 5 )}{( s ^{2} + 4 s + 13 )}

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