What is the solution for 5/8 x^{''}+40x=0,x(0)= 1/4 ,x^'(0)=0 ?

The solution for 5/8 x^{''}+40x=0,x(0)= 1/4 ,x^'(0)=0 is x= 1/4 cos(8t)

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5/8 x^{''}+40x=0,x(0)= 1/4 ,x^'(0)=0

\frac{5}{8} x^{^{''}} + 40 x = 0, x (0) = \frac{1}{4}, x^{^{'}} (0) = 0

x = \frac{1}{4} cos (8 t)

Solution steps

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

Solve linear ODE:

x = \frac{1}{4} cos (8 t)

x = \frac{1}{4} cos (8 t)

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 6

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 2

4 \frac{d ^{2} y}{d x ^{2}} + 36 y = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2}{3} x}

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

What is the solution for 5/8 x^{''}+40x=0,x(0)= 1/4 ,x^'(0)=0 ?
The solution for 5/8 x^{''}+40x=0,x(0)= 1/4 ,x^'(0)=0 is x= 1/4 cos(8t)