What is the solution for x^{''}+kx=0,x(0)= 2/3 ,x^'(0)=5 ?

The solution for x^{''}+kx=0,x(0)= 2/3 ,x^'(0)=5 is x= 2/3 cos(sqrt(k)t)+(5sin(sqrt(k)t))/(sqrt(k))

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x^{''}+kx=0,x(0)= 2/3 ,x^'(0)=5

x^{^{''}} + k x = 0, x (0) = \frac{2}{3}, x^{^{'}} (0) = 5

x = \frac{2}{3} cos (k t) + \frac{5 sin ( k t )}{k}

Solution steps

x^{''} (t) + k x (t) = 0

Solve linear ODE:

x = \frac{2}{3} cos (k t) + \frac{5 sin ( k t )}{k}

x = \frac{2}{3} cos (k t) + \frac{5 sin ( k t )}{k}

y^{^{''}} - 2 y^{^{'}} - 143 y = 0, y (0) = 12, y^{^{'}} (0) = - 12

2 y^{^{''}} + 7 y^{^{'}} + 3 y = 0, y^{^{'}} (0) = 0, y (0) = 3

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = e^{3} x cos (2 x)

y^{^{''}} = \frac{( a - 2 )}{a} y

16 y^{^{''}} - 32 y^{^{'}} + 65 y = 0

What is the solution for x^{''}+kx=0,x(0)= 2/3 ,x^'(0)=5 ?
The solution for x^{''}+kx=0,x(0)= 2/3 ,x^'(0)=5 is x= 2/3 cos(sqrt(k)t)+(5sin(sqrt(k)t))/(sqrt(k))