What is the solution for y^{''}+4y^'=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y^'=0,y(0)=0,y^'(0)=1 is y= 1/4-1/4 e^{-4t}

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y^{''}+4y^'=0,y(0)=0,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{4} - \frac{1}{4} e^{- 4 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) = 0

Solve linear ODE:

y = \frac{1}{4} - \frac{1}{4} e^{- 4 t}

y = \frac{1}{4} - \frac{1}{4} e^{- 4 t}

y^{^{''}} + 2 y^{^{'}} + 65 y = 0, y (0) = - 1, y^{^{'}} (0) = 33

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 1, y (0) = 0

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 8

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y^{0} = 1

What is the solution for y^{''}+4y^'=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y^'=0,y(0)=0,y^'(0)=1 is y= 1/4-1/4 e^{-4t}