What is the solution for x^{''}+6x^'+25x=0,x(0)=3,x^'(0)=8 ?

The solution for x^{''}+6x^'+25x=0,x(0)=3,x^'(0)=8 is x=e^{-3t}(3cos(4t)+17/4 sin(4t))

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x^{''}+6x^'+25x=0,x(0)=3,x^'(0)=8

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = 3, x^{^{'}} (0) = 8

x = e^{- 3 t} (3 cos (4 t) + \frac{17}{4} sin (4 t))

Solution steps

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

Solve linear ODE:

x = e^{- 3 t} (3 cos (4 t) + \frac{17}{4} sin (4 t))

x = e^{- 3 t} (3 cos (4 t) + \frac{17}{4} sin (4 t))

\frac{d ^{2} y}{d x ^{2}} = k y

y^{^{''}} - 12 y^{^{'}} + 20 y = 0

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 0

y^{^{''}} - y^{^{'}} = 0, y (0) = cos (x)

3 V^{^{''}} + 4 V^{^{'}} + 3 V = 0

What is the solution for x^{''}+6x^'+25x=0,x(0)=3,x^'(0)=8 ?
The solution for x^{''}+6x^'+25x=0,x(0)=3,x^'(0)=8 is x=e^{-3t}(3cos(4t)+17/4 sin(4t))