What is the solution for y^{''}+2y^'=0,y(0)=1,y^'(0)=1 ?

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

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

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

y = 1 + \frac{1}{2} - \frac{1}{2} e^{- 2 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) = 0

Solve linear ODE:

y = 1 + \frac{1}{2} - \frac{1}{2} e^{- 2 t}

y = 1 + \frac{1}{2} - \frac{1}{2} e^{- 2 t}

x \to 1 - lim (\frac{x ^{n}}{x - n})

\frac{d}{d x} (ln ∣ e^{x y} cos (x) ∣)

2 y^{^{'}} - y = e^{\frac{x}{3}}

7 x + 2, at x = 2

5^{3 x}

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