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

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

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

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

y = c_{1} - e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{1} - e^{- t}

y = c_{1} - e^{- t}

3 y^{^{''}} + 2 y^{^{'}} = 0, y (0) = - 6, y^{^{'}} (0) = - 18

\frac{3}{16} y^{^{''}} + 12 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - 3

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + (a^{2} - 2) y = 0

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

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