What is the solution for y^{'''}-y^{''}-y^'+y=3e^{-t}+4 ?

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

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y^{'''}-y^{''}-y^'+y=3e^{-t}+4

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 3 e^{- t} + 4

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

Solution steps

y^{'''} (t) - y^{''} (t) - y^{'} (t) + y = 3 e^{- t} + 4

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 3 e^{- t} + 5

- 4 y^{^{''}} + 9 y = 15

y^{^{''}} - 4 y^{^{'}} + 3 y = t^{2} e^{2 t}

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

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

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