What is the solution for y^{'''}-y^{''}-y^'+y=3e^{-t}+2 ?

The solution for y^{'''}-y^{''}-y^'+y=3e^{-t}+2 is y=c_{1}e^{-t}+c_{2}e^t+c_{3}te^t+(3e^{-t}t)/4+2

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

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

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

Solution steps

y^{'''} (t) - y^{''} (t) - y^{'} (t) + y = 3 e^{- t} + 2

Solve linear ODE:

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

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x^{2} + 6 e^{x}

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

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

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{'''}-y^{''}-y^'+y=3e^{-t}+2 ?
The solution for y^{'''}-y^{''}-y^'+y=3e^{-t}+2 is y=c_{1}e^{-t}+c_{2}e^t+c_{3}te^t+(3e^{-t}t)/4+2