What is the solution for y^{''}+4y^'+5y=50x+13e^{3x} ?

The solution for y^{''}+4y^'+5y=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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y^{''}+4y^'+5y=50x+13e^{3x}

y^{^{''}} + 4 y^{^{'}} + 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

y^{''} (x) + 4 y^{'} (x) + 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^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{x ^{2}}

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

y^{^{'''}} - y^{^{'}} = 10 sin (3 x)

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

y^{^{''}} + y^{^{'}} = x + 1

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