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

The solution for 4y^{''}-y=e^x is y=c_{1}e^{x/2}+c_{2}e^{-x/2}+1/3 e^x

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4y^{''}-y=e^x

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} + \frac{1}{3} e^{x}

Solution steps

4 y^{''} (x) - y = e^{x}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} + \frac{1}{3} e^{x}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} + \frac{1}{3} e^{x}

What is the solution for 4y^{''}-y=e^x ?
The solution for 4y^{''}-y=e^x is y=c_{1}e^{x/2}+c_{2}e^{-x/2}+1/3 e^x