What is the solution for x^{''}+2x^'+50x=0,x(0)=-3/2 ,x^'(0)=-2 ?

The solution for x^{''}+2x^'+50x=0,x(0)=-3/2 ,x^'(0)=-2 is x=e^{-t}(-3/2 cos(7t)-1/2 sin(7t))

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x^{''}+2x^'+50x=0,x(0)=-3/2 ,x^'(0)=-2

x^{^{''}} + 2 x^{^{'}} + 50 x = 0, x (0) = - \frac{3}{2}, x^{^{'}} (0) = - 2

x = e^{- t} (- \frac{3}{2} cos (7 t) - \frac{1}{2} sin (7 t))

Solution steps

x^{''} (t) + 2 x^{'} (t) + 50 x = 0

Solve linear ODE:

x = e^{- t} (- \frac{3}{2} cos (7 t) - \frac{1}{2} sin (7 t))

x = e^{- t} (- \frac{3}{2} cos (7 t) - \frac{1}{2} sin (7 t))

What is the solution for x^{''}+2x^'+50x=0,x(0)=-3/2 ,x^'(0)=-2 ?
The solution for x^{''}+2x^'+50x=0,x(0)=-3/2 ,x^'(0)=-2 is x=e^{-t}(-3/2 cos(7t)-1/2 sin(7t))