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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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