What is the solution for 25y^{''}-y=xe^{x/5},y(0)=1,y^'(0)=0 ?

The solution for 25y^{''}-y=xe^{x/5},y(0)=1,y^'(0)=0 is

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25y^{''}-y=xe^{x/5},y(0)=1,y^'(0)=0

25 y^{^{''}} - y = x e^{\frac{x}{5}}, y (0) = 1, y^{^{'}} (0) = 0

y = (\frac{1}{8} + \frac{3}{8}) e^{\frac{x}{5}} - \frac{1}{8} e^{- \frac{x}{5}} + \frac{e ^{\frac{x}{5}} x ^{2}}{20} - \frac{1}{8} e^{- \frac{x}{5}} (2 e^{\frac{2 x}{5}} x - 5 e^{\frac{2 x}{5}})

Solution steps

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

Solve linear ODE:

y = (\frac{1}{8} + \frac{3}{8}) e^{\frac{x}{5}} - \frac{1}{8} e^{- \frac{x}{5}} + \frac{e ^{\frac{x}{5}} x ^{2}}{20} - \frac{1}{8} e^{- \frac{x}{5}} (2 e^{\frac{2 x}{5}} x - 5 e^{\frac{2 x}{5}})

y = (\frac{1}{8} + \frac{3}{8}) e^{\frac{x}{5}} - \frac{1}{8} e^{- \frac{x}{5}} + \frac{e ^{\frac{x}{5}} x ^{2}}{20} - \frac{1}{8} e^{- \frac{x}{5}} (2 e^{\frac{2 x}{5}} x - 5 e^{\frac{2 x}{5}})

x \to \infty lim (\frac{1}{x ( ln ( x ) ) ^{4}})

\int_{1}^{3} 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

\frac{\partial}{\partial v} (uv)

\frac{\partial}{\partial z} (2 x^{4} y z^{3})

\frac{d}{d x} (\frac{3}{4 x ^{8}})

What is the solution for 25y^{''}-y=xe^{x/5},y(0)=1,y^'(0)=0 ?
The solution for 25y^{''}-y=xe^{x/5},y(0)=1,y^'(0)=0 is