What is the solution for y^{'''}-3y^{''}=e^{2x} ?

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

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y^{'''}-3y^{''}=e^{2x}

y^{^{'''}} - 3 y^{^{''}} = e^{2 x}

y = c_{1} e^{3 x} + c_{2} + c_{3} x - \frac{1}{4} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} + c_{3} x - \frac{1}{4} e^{2 x}

y = c_{1} e^{3 x} + c_{2} + c_{3} x - \frac{1}{4} e^{2 x}

y^{^{''}} - y = 8 e^{9 x} + x e^{9 x} + 5 x^{2} e^{9 x}

y^{^{''}} + y = e^{x} + x^{3}, y (0) = 2, y^{^{'}} (0) = 0

(D^{3} + 4 D) y = sin (2 x)

y^{^{''}} - 4 y = x^{- 1} e^{2 x}

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

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