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

The solution for y^{''}+4y^'=x^2+3x+4 is y=c_{1}+c_{2}e^{-4x}+(x^3)/(12)+(5x^2)/(16)+(27x)/(32)

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

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

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{3}}{12} + \frac{5 x ^{2}}{16} + \frac{27 x}{32}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{3}}{12} + \frac{5 x ^{2}}{16} + \frac{27 x}{32}

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{3}}{12} + \frac{5 x ^{2}}{16} + \frac{27 x}{32}

\frac{d ^{2} y}{d x ^{2}} - \frac{1}{5} \frac{d y}{d x} + \frac{3}{400} = 0

(D^{2} - 3 D + 2) y = 6 e^{3 x} + sin (2 x)

y^{^{''}} + 18 y^{^{'}} + 81 y = x

y^{^{''}} + 81 y = 2 sin (a t)

y^{^{''}} + 19 y = 4 e^{x}

What is the solution for y^{''}+4y^'=x^2+3x+4 ?
The solution for y^{''}+4y^'=x^2+3x+4 is y=c_{1}+c_{2}e^{-4x}+(x^3)/(12)+(5x^2)/(16)+(27x)/(32)