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

The solution for y^{''}-y^'+y/4 =4+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+16+(e^{x/2}x^2)/2

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

y^{^{''}} - y^{^{'}} + \frac{y}{4} = 4 + e^{\frac{x}{2}}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 16 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

Solution steps

y^{''} (x) - y^{'} (x) + \frac{y}{4} = 4 + e^{\frac{x}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 16 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 16 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y^{^{''}} + 5 y = cos (7 t)

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- x} + sin (x)

y^{^{''}} + 2 y^{^{'}} = cos (4 x)

2 x^{^{''}} + 4 x + 20 x = 20 H (t)

2 y^{^{''}} - y^{^{'}} - y = 25 e^{- x}

What is the solution for y^{''}-y^'+y/4 =4+e^{x/2} ?
The solution for y^{''}-y^'+y/4 =4+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+16+(e^{x/2}x^2)/2