What is the solution for y^{''}+4y^'+5y=15x+3e^{-x} ?

The solution for y^{''}+4y^'+5y=15x+3e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+3x-12/5+3/2 e^{-x}

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y^{''}+4y^'+5y=15x+3e^{-x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + 3 e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 3 x - \frac{12}{5} + \frac{3}{2} e^{- x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 15 x + 3 e^{- x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 3 x - \frac{12}{5} + \frac{3}{2} e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 3 x - \frac{12}{5} + \frac{3}{2} e^{- x}

y^{^{''}} + 8 y^{^{'}} + 7 y = - 8 - x + 24 x^{2} + 7 x^{3}

\int x^{2} \cdot 1 - x^{2} d x

y^{^{''}} + 121 y = cos (6 t)

\int \frac{3 ^{x}}{x} d x

3 y^{^{''}} + 8 y^{^{'}} - 3 y = 0

What is the solution for y^{''}+4y^'+5y=15x+3e^{-x} ?
The solution for y^{''}+4y^'+5y=15x+3e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+3x-12/5+3/2 e^{-x}