What is the solution for y^{''}-y=e^{(2x)} ?

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

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y^{''}-y=e^{(2x)}

y^{^{''}} - y = e^{(2 x)}

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

Solution steps

y^{''} (x) - y = e^{(2 x)}

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 7 sin (x)

y^{^{'''}} - y^{^{''}} - 9 y^{^{'}} + 9 y = 7 - e^{x} + e^{3 x}

3 y^{^{''}} - y = e^{x} - 5

y^{^{''}} + 4 y^{^{'}} + 4 y = 375 \cdot t

y^{^{''}} - 2 y^{^{'}} - 15 y = 4 e^{t}

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