What is the solution for y^{''}+5y^'+4y=3x^2 ?

The solution for y^{''}+5y^'+4y=3x^2 is y=c_{1}e^{-x}+c_{2}e^{-4x}+(3x^2)/4-(15x)/8+63/32

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y^{''}+5y^'+4y=3x^2

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 x^{2}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + \frac{3 x ^{2}}{4} - \frac{15 x}{8} + \frac{63}{32}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 3 x^{2}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + \frac{3 x ^{2}}{4} - \frac{15 x}{8} + \frac{63}{32}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + \frac{3 x ^{2}}{4} - \frac{15 x}{8} + \frac{63}{32}

y^{^{''}} + 4 y^{^{'}} + 3 y = 7 e^{- x}

y^{^{''}} + 8 y^{^{'}} + 16 y = e^{- 4 t}

y^{^{''}} + 9 y = sin (3 x) + e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 x)

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

What is the solution for y^{''}+5y^'+4y=3x^2 ?
The solution for y^{''}+5y^'+4y=3x^2 is y=c_{1}e^{-x}+c_{2}e^{-4x}+(3x^2)/4-(15x)/8+63/32