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^{^{'}} + ln (x) y = ln (x)

\int (sin (x) + cos (x))^{2} d x

\frac{d}{d x} (\frac{x - 3}{6 x - 5})

x \to 2 lim (\frac{1}{2 - x})

\frac{\partial}{\partial y} (e^{- x} cos (y))

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