What is the solution for y^{''}+3y^'+2y=x^3+4x^2+2x+1 ?

The solution for y^{''}+3y^'+2y=x^3+4x^2+2x+1 is y=c_{1}e^{-x}+c_{2}e^{-2x}+(x^3)/2-(x^2)/4+x/4+3/8

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y^{''}+3y^'+2y=x^3+4x^2+2x+1

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{3} + 4 x^{2} + 2 x + 1

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x ^{3}}{2} - \frac{x ^{2}}{4} + \frac{x}{4} + \frac{3}{8}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = x^{3} + 4 x^{2} + 2 x + 1

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x ^{3}}{2} - \frac{x ^{2}}{4} + \frac{x}{4} + \frac{3}{8}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x ^{3}}{2} - \frac{x ^{2}}{4} + \frac{x}{4} + \frac{3}{8}

What is the solution for y^{''}+3y^'+2y=x^3+4x^2+2x+1 ?
The solution for y^{''}+3y^'+2y=x^3+4x^2+2x+1 is y=c_{1}e^{-x}+c_{2}e^{-2x}+(x^3)/2-(x^2)/4+x/4+3/8