What is the solution for x^{''}+32/2 x=0,x(0)=-1,x^'(0)=-2 ?

The solution for x^{''}+32/2 x=0,x(0)=-1,x^'(0)=-2 is x=-cos(4t)-1/2 sin(4t)

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x^{''}+32/2 x=0,x(0)=-1,x^'(0)=-2

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 2

x = - cos (4 t) - \frac{1}{2} sin (4 t)

Solution steps

x^{''} (t) + \frac{32}{2} x = 0

Solve linear ODE:

x = - cos (4 t) - \frac{1}{2} sin (4 t)

x = - cos (4 t) - \frac{1}{2} sin (4 t)

What is the solution for x^{''}+32/2 x=0,x(0)=-1,x^'(0)=-2 ?
The solution for x^{''}+32/2 x=0,x(0)=-1,x^'(0)=-2 is x=-cos(4t)-1/2 sin(4t)