What is the solution for y^{'''}-y^{''}-y^'+y=0 ?

The solution for y^{'''}-y^{''}-y^'+y=0 is y=c_{1}e^{-t}+c_{2}e^t+c_{3}te^t

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

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

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

Solution steps

y^{'''} (t) - y^{''} (t) - y^{'} (t) + y = 0

Solve linear ODE:

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

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

\frac{d}{d x} ((1 + e^{- x})^{x})

\int 3 x^{3} e^{3 x} d x

f (x) = x^{2} + x, 4, at x = 2

\int \frac{18}{1 - cos ( 6 x )} d x

sin (t - \frac{( 2 \cdot π )}{3})

What is the solution for y^{'''}-y^{''}-y^'+y=0 ?
The solution for y^{'''}-y^{''}-y^'+y=0 is y=c_{1}e^{-t}+c_{2}e^t+c_{3}te^t