What is the solution for y^{''}+4y=e^{-x}(7-4x+5x^2) ?

The solution for y^{''}+4y=e^{-x}(7-4x+5x^2) is y=c_{1}cos(2x)+c_{2}sin(2x)+e^{-x}x^2+e^{-x}

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y^{''}+4y=e^{-x}(7-4x+5x^2)

y^{^{''}} + 4 y = e^{- x} (7 - 4 x + 5 x^{2})

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

Solution steps

y^{''} (x) + 4 y = e^{- x} (7 - 4 x + 5 x^{2})

Solve linear ODE:

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

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

\frac{d y}{d x} = (y + 1) (y - 2), y (0) = 0

\frac{\partial}{\partial x} (6 ln (6 x^{3} + y^{2}))

g (t) = t^{3} cos (t)

\frac{d}{d x} (ln (2 x^{2} - 1))

x \to - 0.00000001 lim (x^{x})

What is the solution for y^{''}+4y=e^{-x}(7-4x+5x^2) ?
The solution for y^{''}+4y=e^{-x}(7-4x+5x^2) is y=c_{1}cos(2x)+c_{2}sin(2x)+e^{-x}x^2+e^{-x}