What is the solution for y^{''}+4y^'+20x=0 ?

The solution for y^{''}+4y^'+20x=0 is y=c_{1}+c_{2}e^{-4x}-(5x^2)/2+(5x)/4

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y^{''}+4y^'+20x=0

y^{^{''}} + 4 y^{^{'}} + 20 x = 0

y = c_{1} + c_{2} e^{- 4 x} - \frac{5 x ^{2}}{2} + \frac{5 x}{4}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 20 x = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{5 x ^{2}}{2} + \frac{5 x}{4}

y = c_{1} + c_{2} e^{- 4 x} - \frac{5 x ^{2}}{2} + \frac{5 x}{4}

What is the solution for y^{''}+4y^'+20x=0 ?
The solution for y^{''}+4y^'+20x=0 is y=c_{1}+c_{2}e^{-4x}-(5x^2)/2+(5x)/4