What is the solution for 4y^{''}-y=(ex)/(2+7) ?

The solution for 4y^{''}-y=(ex)/(2+7) is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-(ex)/9

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4y^{''}-y=(ex)/(2+7)

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{e x}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{e x}{9}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{e x}{9}

4 y^{^{''}} - 4 y^{^{'}} + y = (- 3 x + 4) sin (x)

y^{^{''}} + 4 y = 2 sec^{3} (2 x)

y^{^{''}} + y^{^{'}} - 6 = 0

y^{^{''}} - 2 y^{^{'}} + 65 y = e^{x} cos (8 x)

x^{^{''}} + 8 x = 16 cos (4 t)

What is the solution for 4y^{''}-y=(ex)/(2+7) ?
The solution for 4y^{''}-y=(ex)/(2+7) is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-(ex)/9