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}

y^{^{''}} - 2 y^{^{'}} + 1 = x^{2}

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 t e^{t}, y (0) = 1, y^{^{'}} (0) = - 5

y^{^{''}} - 3 y^{^{'}} - 4 y = 8 e^{x} cos (2 x)

y^{^{''}} + 6 y^{^{'}} + 5 y = - 1020 sin (3 t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 4 a

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