What is the solution for y^{''}+3y^'-1=2x^2+x ?

The solution for y^{''}+3y^'-1=2x^2+x is y=c_{1}+c_{2}e^{-3x}+(2x^3)/9-(x^2)/(18)+(10x)/(27)

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y^{''}+3y^'-1=2x^2+x

y^{^{''}} + 3 y^{^{'}} - 1 = 2 x^{2} + x

y = c_{1} + c_{2} e^{- 3 x} + \frac{2 x ^{3}}{9} - \frac{x ^{2}}{18} + \frac{10 x}{27}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} + \frac{2 x ^{3}}{9} - \frac{x ^{2}}{18} + \frac{10 x}{27}

y = c_{1} + c_{2} e^{- 3 x} + \frac{2 x ^{3}}{9} - \frac{x ^{2}}{18} + \frac{10 x}{27}

y^{^{''}} - 4 y = 24 cos (2 t)

y^{^{''}} + 2 y^{^{'}} + 10 y = 2 cos (x) + cos (3 x)

y^{^{''}} + 100 y^{^{'}} + 2500 y = 20

y^{^{''}} - y^{^{'}} - 2 y = 4

y^{^{''}} - 4 y^{^{'}} - y - 2 = 0

What is the solution for y^{''}+3y^'-1=2x^2+x ?
The solution for y^{''}+3y^'-1=2x^2+x is y=c_{1}+c_{2}e^{-3x}+(2x^3)/9-(x^2)/(18)+(10x)/(27)