What is the solution for 4y^{''}-y=e^{x/2}+5 ?

The solution for 4y^{''}-y=e^{x/2}+5 is y=c_{1}e^{x/2}+c_{2}e^{-x/2}+(e^{x/2}x)/4-5

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4y^{''}-y=e^{x/2}+5

4 y^{^{''}} - y = e^{\frac{x}{2}} + 5

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} + \frac{e ^{\frac{x}{2}} x}{4} - 5

Solution steps

4 y^{''} (x) - y = e^{\frac{x}{2}} + 5

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} + \frac{e ^{\frac{x}{2}} x}{4} - 5

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} + \frac{e ^{\frac{x}{2}} x}{4} - 5

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

y^{^{'''}} - 6 y^{^{''}} = 2 - cos (x)

y^{^{''}} + 7 y^{^{'}} - 12 y = e^{3 x} - 2 e^{x} - 5 x

y^{^{''}} - 6 y^{^{'}} + 9 y = t

\frac{d ^{2} y}{d x ^{2}} + 4 y = e^{4 x} sin (3 x)

What is the solution for 4y^{''}-y=e^{x/2}+5 ?
The solution for 4y^{''}-y=e^{x/2}+5 is y=c_{1}e^{x/2}+c_{2}e^{-x/2}+(e^{x/2}x)/4-5