What is the solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-2/3 ?

The solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-2/3 is x=-2/3 cos(4sqrt(3)t)

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3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-2/3

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - \frac{2}{3}

x = - \frac{2}{3} cos (43 t)

Solution steps

\frac{3}{4} \cdot x^{''} (t) + 36 x = 0

Solve linear ODE:

x = - \frac{2}{3} cos (43 t)

x = - \frac{2}{3} cos (43 t)

y^{^{''}} - 6 y^{^{'}} - 12 y = 0

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

y^{^{''}} - 8 y^{^{'}} - 105 y = 0

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 0

y^{^{''''}} - 2 y^{^{'''}} - 6 y^{^{''}} = 0

What is the solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-2/3 ?
The solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-2/3 is x=-2/3 cos(4sqrt(3)t)