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

The solution for y^{'''}-3y^{''}+4y=e^{2x} is y=c_{1}e^{-x}+c_{2}e^{2x}+c_{3}xe^{2x}+(e^{2x}x^2)/6

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y^{'''}-3y^{''}+4y=e^{2x}

y^{^{'''}} - 3 y^{^{''}} + 4 y = e^{2 x}

y = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x} + \frac{e ^{2 x} x ^{2}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x} + \frac{e ^{2 x} x ^{2}}{6}

y = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x} + \frac{e ^{2 x} x ^{2}}{6}

y^{^{''}} + 9 y = 3 cos (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{- 2 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{x}

y^{^{''}} = - 3 y^{^{'}} + 2 y + 2 x + 3, y (0) = 2, y (1) = 1

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

What is the solution for y^{'''}-3y^{''}+4y=e^{2x} ?
The solution for y^{'''}-3y^{''}+4y=e^{2x} is y=c_{1}e^{-x}+c_{2}e^{2x}+c_{3}xe^{2x}+(e^{2x}x^2)/6