What is the solution for y^{'''}+y^'=0,y(0)=1,y^'(0)=1,y^{''}(0)=2 ?

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

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y^{'''}+y^'=0,y(0)=1,y^'(0)=1,y^{''}(0)=2

y^{^{'''}} + y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1, y^{^{''}} (0) = 2

y = sin (t) - 2 cos (t) + 3

Solution steps

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

Solve linear ODE:

y = sin (t) - 2 cos (t) + 3

y = sin (t) - 2 cos (t) + 3

(D^{3} - 6 D^{2} + 11 D - 6) y = 0

20 y^{^{''}} + 240 y^{^{'}} + 800 y = 0

a y^{^{''}} + b y^{^{'}} - cy = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 3, y (0) = 5

y^{^{''}} + 1 y = 0, y (\frac{π}{2}) = 2, y^{^{'}} (\frac{π}{2}) = 3

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