What is the solution for y^{''}+5y^'+4x=0 ?

The solution for y^{''}+5y^'+4x=0 is y=c_{1}+c_{2}e^{-5x}-(2x^2)/5+(4x)/(25)

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y^{''}+5y^'+4x=0

y^{^{''}} + 5 y^{^{'}} + 4 x = 0

y = c_{1} + c_{2} e^{- 5 x} - \frac{2 x ^{2}}{5} + \frac{4 x}{25}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 x = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 x} - \frac{2 x ^{2}}{5} + \frac{4 x}{25}

y = c_{1} + c_{2} e^{- 5 x} - \frac{2 x ^{2}}{5} + \frac{4 x}{25}

y^{^{''}} + 25 y = 25, y (0) = 1, y^{^{'}} (0) = - 5

2 y^{^{''}} + 20 y^{^{'}} + 50 y = 10 sin (3 x)

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

y^{^{'''}} - 19 y^{^{'}} + 30 y = x^{2} \cdot e^{2 x}

y^{^{''}} - 4 y^{^{'}} - 5 y = 125, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+5y^'+4x=0 ?
The solution for y^{''}+5y^'+4x=0 is y=c_{1}+c_{2}e^{-5x}-(2x^2)/5+(4x)/(25)