What is the solution for y^{''}+6y^'+8y=x^2 ?

The solution for y^{''}+6y^'+8y=x^2 is y=c_{1}e^{-2x}+c_{2}e^{-4x}+(x^2)/8-(3x)/(16)+7/64

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y^{''}+6y^'+8y=x^2

y^{^{''}} + 6 y^{^{'}} + 8 y = x^{2}

y = c_{1} e^{- 2 x} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{3 x}{16} + \frac{7}{64}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 8 y = x^{2}

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{3 x}{16} + \frac{7}{64}

y = c_{1} e^{- 2 x} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{3 x}{16} + \frac{7}{64}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = e^{4 x}

y^{^{''}} - 4 y^{^{'}} - 3 y = cos (2 x)

y^{^{''}} + 4 y^{^{'}} = - 32 (x + 2), y (0) = 0, y^{^{'}} (0) = - 10

3 y^{^{''}} - 4 y^{^{'}} - 5 y = (x^{2} + 1) e^{- x}

y^{^{''}} + k^{2} y = k sin (k x)

What is the solution for y^{''}+6y^'+8y=x^2 ?
The solution for y^{''}+6y^'+8y=x^2 is y=c_{1}e^{-2x}+c_{2}e^{-4x}+(x^2)/8-(3x)/(16)+7/64