What is the solution for 5/8 x^{''}+40x=0,x(0)= 1/3 ,x^'(0)=0 ?

The solution for 5/8 x^{''}+40x=0,x(0)= 1/3 ,x^'(0)=0 is x= 1/3 cos(8t)

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5/8 x^{''}+40x=0,x(0)= 1/3 ,x^'(0)=0

\frac{5}{8} x^{^{''}} + 40 x = 0, x (0) = \frac{1}{3}, x^{^{'}} (0) = 0

x = \frac{1}{3} cos (8 t)

Solution steps

\frac{5}{8} x^{''} (t) + 40 x = 0

Solve linear ODE:

x = \frac{1}{3} cos (8 t)

x = \frac{1}{3} cos (8 t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = a, y^{^{'}} (0) = 10

y^{^{''}} + 2 α y^{^{'}} + α^{2} y = 0

y^{^{''}} - y^{^{'}} - 2 \cdot y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 68 y = 0, y (0) = 7

What is the solution for 5/8 x^{''}+40x=0,x(0)= 1/3 ,x^'(0)=0 ?
The solution for 5/8 x^{''}+40x=0,x(0)= 1/3 ,x^'(0)=0 is x= 1/3 cos(8t)