What is the solution for y^{''''}-y=0,y(0)=2 ?

The solution for y^{''''}-y=0,y(0)=2 is y=(2-c_{2}-c_{3})e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)

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y^{''''}-y=0,y(0)=2

y^{^{''''}} - y = 0, y (0) = 2

y = (2 - c_{2} - c_{3}) e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

Solution steps

y^{(4)} (t) - y = 0

Solve linear ODE:

y = (2 - c_{2} - c_{3}) e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

y = (2 - c_{2} - c_{3}) e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

0.0650400000000 \dots f (x) = \frac{d ^{2} f}{d x ^{2}}

y^{^{''}} + 6 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = 42

2 y^{^{''}} + 35 y = 0, y (0) = 7, y^{^{'}} (0) = 9

4 y^{^{''}} - 0 y^{^{'}} - 9 y = 0

y^{^{''}} + 4 y + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''''}-y=0,y(0)=2 ?
The solution for y^{''''}-y=0,y(0)=2 is y=(2-c_{2}-c_{3})e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)