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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (e^{(x + y)^{2}})

x \to 4 + lim (- 3)

f (x) = \frac{2 x}{x ^{2} + 1}, 1, at x = 1

\frac{d}{d x} (cos^{2} (9 x) + sin^{2} (9 x))

\frac{d}{d t} (4 e^{t})

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