What is the solution for x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0 ?

The solution for x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0 is x=e^{-3t}(1/2 cos(4t)+3/8 sin(4t))

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x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = \frac{1}{2}, x^{^{'}} (0) = 0

x = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

Solution steps

x^{''} (t) + 6 x^{'} (t) + 25 x = 0

Solve linear ODE:

x = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

x = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

What is the solution for x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0 ?
The solution for x^{''}+6x^'+25x=0,x(0)= 1/2 ,x^'(0)=0 is x=e^{-3t}(1/2 cos(4t)+3/8 sin(4t))