What is the solution for 25y^{''}-y=(xex)/5 ?

The solution for 25y^{''}-y=(xex)/5 is y=c_{1}e^{x/5}+c_{2}e^{-x/5}-(ex^2)/5-10e

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25y^{''}-y=(xex)/5

25 y^{^{''}} - y = \frac{x e x}{5}

y = c_{1} e^{\frac{x}{5}} + c_{2} e^{- \frac{x}{5}} - \frac{e x ^{2}}{5} - 10 e

Solution steps

25 y^{''} (x) - y = \frac{x e x}{5}

Solve linear ODE:

y = c_{1} e^{\frac{x}{5}} + c_{2} e^{- \frac{x}{5}} - \frac{e x ^{2}}{5} - 10 e

y = c_{1} e^{\frac{x}{5}} + c_{2} e^{- \frac{x}{5}} - \frac{e x ^{2}}{5} - 10 e

y^{^{''}} - 2 y^{^{'}} + y = e^{x} + e^{2 x}

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} (0 = (2 x))

y^{^{''}} - 6 y^{^{'}} + 9 y = 18

y^{^{''}} - y = 8 t e^{t} + 2 e^{t}

y^{^{''}} + 18 y = 3 e^{8 x}

What is the solution for 25y^{''}-y=(xex)/5 ?
The solution for 25y^{''}-y=(xex)/5 is y=c_{1}e^{x/5}+c_{2}e^{-x/5}-(ex^2)/5-10e