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

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

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

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

y = 1 - e^{- t}

Solution steps

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

Solve linear ODE:

y = 1 - e^{- t}

y = 1 - e^{- t}

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 26 y = 0

\frac{d ^{2} y}{d x ^{2}} + y = 0, y = cos (x) + sin (x)

64 y^{^{''}} + 96 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 8

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + 16 y = 0

y^{^{''}} + \frac{32}{13} y^{^{'}} + 32 y = 0

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