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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{24}{32} y^{^{''}} + 72 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{1} = x e^{- x}

6 y^{^{''''}} + 5 y^{^{'''}} + 25 y^{^{''}} + 20 y^{^{'}} + 4 y = 0

4 \frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} + 5 y = 0

4 y^{^{''}} + y^{^{'}} + \frac{1}{16} y = 0

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