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

4 y^{^{''}} - y = e^{(\frac{x}{2})} + 7

- y^{^{''}} + y = e^{- 1 \cdot x}, y (0) = 0, y (1) = 0

x^{^{''}} + 64 x = sin (5 t)

0.5 y^{^{''}} + 50 y = cos (t)

x^{^{''}} + x = 1, x (0) = 2, x^{^{'}} (0) = - 1

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