What is the solution for y^{'''}-2y^{''}+2y^'=0 ?

The solution for y^{'''}-2y^{''}+2y^'=0 is y=c_{1}+e^t(c_{2}cos(t)+c_{3}sin(t))

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y^{'''}-2y^{''}+2y^'=0

y^{^{'''}} - 2 y^{^{''}} + 2 y^{^{'}} = 0

y = c_{1} + e^{t} (c_{2} cos (t) + c_{3} sin (t))

Solution steps

y^{'''} (t) - 2 y^{''} (t) + 2 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + e^{t} (c_{2} cos (t) + c_{3} sin (t))

y = c_{1} + e^{t} (c_{2} cos (t) + c_{3} sin (t))

\frac{d}{d x} (7 x^{2} - 4 x)

f (x) = \frac{x ^{5} - 5 x ^{2}}{2 x ^{5} + 5 x}

\frac{\partial}{\partial x} (3 x z^{3} e^{y^{2}})

\int (ln (2 x))^{2} d x

x \to - 3 lim (\frac{2}{x + 2})

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