What is the solution for f^{''}(x)-f^'(x)-2f(x)=0,f(0)=2 ?

The solution for f^{''}(x)-f^'(x)-2f(x)=0,f(0)=2 is f(x)=(2-c_{2})e^{2x}+c_{2}e^{-x}

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f^{''}(x)-f^'(x)-2f(x)=0,f(0)=2

f^{^{''}} (x) - f^{^{'}} (x) - 2 f (x) = 0, f (0) = 2

f (x) = (2 - c_{2}) e^{2 x} + c_{2} e^{- x}

Solution steps

f^{''} (x) - f^{'} (x) - 2 f (x) = 0

Solve linear ODE:

f (x) = (2 - c_{2}) e^{2 x} + c_{2} e^{- x}

f (x) = (2 - c_{2}) e^{2 x} + c_{2} e^{- x}

100 (1 0^{- 6}) y^{^{''}} + \frac{1}{500} y^{^{'}} + y = 0

y^{^{''}} + 3 y = 0, y^{^{'}} (1) = 1

16 y^{^{''}} + 40 y^{^{'}} + 50 y = 0

y^{^{''}} - 16 y = 0, y (0) = e^{4 x}, y (0) = 9, y^{^{'}} (0) = 3

y^{^{''}} + y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

What is the solution for f^{''}(x)-f^'(x)-2f(x)=0,f(0)=2 ?
The solution for f^{''}(x)-f^'(x)-2f(x)=0,f(0)=2 is f(x)=(2-c_{2})e^{2x}+c_{2}e^{-x}