What is the solution for y^{'''}-y^{''}-y^'+y=9e^{-t}+3 ?

The solution for y^{'''}-y^{''}-y^'+y=9e^{-t}+3 is y=c_{1}e^{-t}+c_{2}e^t+c_{3}te^t+(9e^{-t}t)/4+3

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y^{'''}-y^{''}-y^'+y=9e^{-t}+3

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 9 e^{- t} + 3

y = c_{1} e^{- t} + c_{2} e^{t} + c_{3} t e^{t} + \frac{9 e ^{- t} t}{4} + 3

Solution steps

y^{'''} (t) - y^{''} (t) - y^{'} (t) + y = 9 e^{- t} + 3

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{t} + c_{3} t e^{t} + \frac{9 e ^{- t} t}{4} + 3

y = c_{1} e^{- t} + c_{2} e^{t} + c_{3} t e^{t} + \frac{9 e ^{- t} t}{4} + 3

y^{^{''}} + 10 y^{^{'}} + 25 y = 14 \cdot e^{- 5 x}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 9 e^{- t} + 6

y^{^{''}} = - 4 y^{^{'}} - 5 y + 2

16 \frac{d ^{4} y}{d x ^{4}} + 24 \frac{d ^{2} y}{d x ^{2}} + 9 = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 sin (4 x)

What is the solution for y^{'''}-y^{''}-y^'+y=9e^{-t}+3 ?
The solution for y^{'''}-y^{''}-y^'+y=9e^{-t}+3 is y=c_{1}e^{-t}+c_{2}e^t+c_{3}te^t+(9e^{-t}t)/4+3