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

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

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

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

y = - e^{- t} + 3

Solution steps

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

Solve linear ODE:

y = - e^{- t} + 3

y = - e^{- t} + 3

y^{^{''}} - 6 y^{^{'}} - 2 y = 0

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

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

y^{^{'''}} - 4 y^{^{''}} - 16 y^{^{'}} + 64 y = 0

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

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