What is the solution for y^{''}-y^'+1/4 y=8+e^{x/2} ?

The solution for y^{''}-y^'+1/4 y=8+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+32+(e^{x/2}x^2)/2

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y^{''}-y^'+1/4 y=8+e^{x/2}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 8 + e^{\frac{x}{2}}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 32 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

Solution steps

y^{''} (x) - y^{'} (x) + \frac{1}{4} y = 8 + e^{\frac{x}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 32 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 32 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x} + 4 x - 12

y^{^{''}} + y - 5 = e^{2 x} - 3 + cos (3 x)

y^{^{''}} - 4 y = 2 e^{3 x}

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

y^{^{''}} + 2 y^{^{'}} + 4 y = 3 sin (5 t)

What is the solution for y^{''}-y^'+1/4 y=8+e^{x/2} ?
The solution for y^{''}-y^'+1/4 y=8+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+32+(e^{x/2}x^2)/2