What is the solution for 4y^{''}-y=(ex)/(2+5) ?

The solution for 4y^{''}-y=(ex)/(2+5) is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-(ex)/7

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4y^{''}-y=(ex)/(2+5)

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{e x}{7}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{e x}{7}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{e x}{7}

What is the solution for 4y^{''}-y=(ex)/(2+5) ?
The solution for 4y^{''}-y=(ex)/(2+5) is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-(ex)/7