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)

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 5

\frac{d ^{2} y}{d x ^{2}} + 9 \frac{d y}{d x} + 8 y = 0

2 y^{^{''}} + 12 y^{^{'}} + 68 y = 0, y (0) = 3, y^{^{'}} (0) = 0

x^{^{'''}} + 7 x^{^{''}} - 5 x^{^{'}} - 75 x = 0

y^{^{''}} - 14 y^{^{'}} + 5 y = 0

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)