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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{- 2}{x x})

\frac{d}{d x} ((3 x)^{\frac{1}{2}})

\frac{d}{d x} (14 e^{2 x})

(cos (6 x)) y^{^{'}} - 6 (tan (6 x)) y = - 2 sec (6 x)

\frac{d}{d x} ((x^{6} - 6) x)

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