What is the solution for y^{''}+2y^'-3y=3x^2-4x ?

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

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y^{''}+2y^'-3y=3x^2-4x

y^{^{''}} + 2 y^{^{'}} - 3 y = 3 x^{2} - 4 x

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 9 y = 18 x + 36 sin (3 x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 sin (x)

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{2}}

x^{^{''}} + 100 x^{^{'}} + \frac{1}{0.0004} x = 30

y^{^{''}} - y = 1 + x^{2} e^{x}

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