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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 = 0

y^{^{'''}} - 6 y^{^{''}} = 4 - sin (x)

y^{^{''}} + 3 y = 5, y (0) = 1, y^{^{'}} (0) = 1

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

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