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

The solution for 2y^{''}+3y^'+y=x^2 is y=c_{1}e^{-x/2}+c_{2}e^{-x}+x^2-6x+14

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

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

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- x} + x^{2} - 6 x + 14

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- x} + x^{2} - 6 x + 14

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- x} + x^{2} - 6 x + 14

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{2 x}}{( e ^{x} + 1 ) ^{2}}

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

y^{^{''}} + y = 6 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = t e^{t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = x - 6

What is the solution for 2y^{''}+3y^'+y=x^2 ?
The solution for 2y^{''}+3y^'+y=x^2 is y=c_{1}e^{-x/2}+c_{2}e^{-x}+x^2-6x+14