What is the solution for y^{''}+3y^'+2y=e^{-2x}+x^2 ?

The solution for y^{''}+3y^'+2y=e^{-2x}+x^2 is y=c_{1}e^{-x}+c_{2}e^{-2x}-e^{-2x}x+(x^2-3x)/2+7/4

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y^{''}+3y^'+2y=e^{-2x}+x^2

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 2 x} + x^{2}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - e^{- 2 x} x + \frac{x ^{2} - 3 x}{2} + \frac{7}{4}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = e^{- 2 x} + x^{2}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - e^{- 2 x} x + \frac{x ^{2} - 3 x}{2} + \frac{7}{4}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - e^{- 2 x} x + \frac{x ^{2} - 3 x}{2} + \frac{7}{4}

y^{^{''}} - 7 y = t^{2}

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

y^{^{'''}} + y^{^{'}} = x + sin (x) + cos (x)

y^{^{''}} + y^{^{'}} - 2 y = 3 e^{- 2 x} - 2 x

y^{^{''}} + y^{^{'}} - 2 y = 2 t, y (0) = 0

What is the solution for y^{''}+3y^'+2y=e^{-2x}+x^2 ?
The solution for y^{''}+3y^'+2y=e^{-2x}+x^2 is y=c_{1}e^{-x}+c_{2}e^{-2x}-e^{-2x}x+(x^2-3x)/2+7/4