What is the solution for f^{''}(t)-f^'(t)-2f(t)=0,f(0)=1,f^'(0)=0 ?

The solution for f^{''}(t)-f^'(t)-2f(t)=0,f(0)=1,f^'(0)=0 is f(t)=(1-2/3)e^{2t}+2/3 e^{-t}

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f^{''}(t)-f^'(t)-2f(t)=0,f(0)=1,f^'(0)=0

f^{^{''}} (t) - f^{^{'}} (t) - 2 f (t) = 0, f (0) = 1, f^{^{'}} (0) = 0

f (t) = (1 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

Solution steps

f^{''} (t) - f^{'} (t) - 2 f (t) = 0

Solve linear ODE:

f (t) = (1 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

f (t) = (1 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

y^{^{''''}} + 5 y^{^{''}} = 0

y^{^{''}} + 25 y = 0, y (π) = 0, y (0) = 0

y^{^{''}} = a y^{^{'}}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = x e^{5 x}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

What is the solution for f^{''}(t)-f^'(t)-2f(t)=0,f(0)=1,f^'(0)=0 ?
The solution for f^{''}(t)-f^'(t)-2f(t)=0,f(0)=1,f^'(0)=0 is f(t)=(1-2/3)e^{2t}+2/3 e^{-t}