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

The solution for y^{''}-4y^'-5y=x^2 is y=c_{1}e^{5x}+c_{2}e^{-x}-(x^2)/5+(8x)/(25)-42/125

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y^{''}-4y^'-5y=x^2

y^{^{''}} - 4 y^{^{'}} - 5 y = x^{2}

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{x ^{2}}{5} + \frac{8 x}{25} - \frac{42}{125}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{x ^{2}}{5} + \frac{8 x}{25} - \frac{42}{125}

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{x ^{2}}{5} + \frac{8 x}{25} - \frac{42}{125}

What is the solution for y^{''}-4y^'-5y=x^2 ?
The solution for y^{''}-4y^'-5y=x^2 is y=c_{1}e^{5x}+c_{2}e^{-x}-(x^2)/5+(8x)/(25)-42/125