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

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

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

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

y = - e^{- t} + 2

Solution steps

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

Solve linear ODE:

y = - e^{- t} + 2

y = - e^{- t} + 2

49 y^{^{''}} + 28 y^{^{'}} + 3 y = 0, y (2) = 9, y^{^{'}} (2) = 3

u^{^{''}} - a H (t) = 0

y^{^{''}} - 8 y + 15 y = 0

(3 + 2) y^{^{''}} - 6 (3 + 3 + 3 + 1) y^{^{'}} + 2 (0 + 7) y = 0

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

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