What is the solution for y^{''}+4y^'+5y=-10x+3e^{-x} ?

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

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y^{''}+4y^'+5y=-10x+3e^{-x}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 10 x + 3 e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - 2 x + \frac{8}{5} + \frac{3}{2} e^{- x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = - 10 x + 3 e^{- x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - 2 x + \frac{8}{5} + \frac{3}{2} e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - 2 x + \frac{8}{5} + \frac{3}{2} e^{- x}

x \to 3 lim (x - 2)

\int \frac{sec ^{2} ( x )}{x} d x

f (x) = \frac{x + 5}{x - 5}

x \to - 2 lim (x^{2} - \frac{4}{x + 2})

y = cos (x), y = 0.5, 0 \leq x \leq π

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