What is the solution for y^{'''}+6y=7y^'+4e^x ?

The solution for y^{'''}+6y=7y^'+4e^x is y=c_{1}e^x+c_{2}e^{2x}+c_{3}e^{-3x}-e^xx

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y^{'''}+6y=7y^'+4e^x

y^{^{'''}} + 6 y = 7 y^{^{'}} + 4 e^{x}

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

Solution steps

y^{'''} (x) + 6 y = 7 y^{'} (x) + 4 e^{x}

Solve linear ODE:

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

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

y^{^{''}} + y - 12 = 0

x^{^{''}} - 3 x^{^{'}} + 2 x = 20 sin (2 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 8 e^{6 x}

x^{^{''}} - 5 x^{^{'}} + 6 x = 12, x (0) = 2, x^{^{'}} (0) = 0

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

What is the solution for y^{'''}+6y=7y^'+4e^x ?
The solution for y^{'''}+6y=7y^'+4e^x is y=c_{1}e^x+c_{2}e^{2x}+c_{3}e^{-3x}-e^xx