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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{t} sin (t)

y^{^{''}} + 7 y + 12 y = t^{2} e^{- 3 t}

5 y^{^{''}} + 88 y^{^{'}} + 240 y = 82 sin (6 x)

(d^{2} - 4) y = e^{- 3 x} sin (e^{- x})

(D^{2} - 6 D + 9) y = 1 + x + x^{2}

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