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

The solution for y^{''}-3y^'=x^2-1 is y=c_{1}e^{3x}+c_{2}-(x^3)/9-(x^2)/9+(7x)/(27)

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

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

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{3}}{9} - \frac{x ^{2}}{9} + \frac{7 x}{27}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{3}}{9} - \frac{x ^{2}}{9} + \frac{7 x}{27}

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{3}}{9} - \frac{x ^{2}}{9} + \frac{7 x}{27}

9 y^{^{''}} + 24 y^{^{'}} + 16 y = 8 e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 25 cos (x)

x^{^{''}} - x = e^{- t} sin (e^{- t}) + cos (e^{- t})

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e x - 10

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} - 5 y = - 5 x

What is the solution for y^{''}-3y^'=x^2-1 ?
The solution for y^{''}-3y^'=x^2-1 is y=c_{1}e^{3x}+c_{2}-(x^3)/9-(x^2)/9+(7x)/(27)