What is the solution for 3/4 x^{''}+192x=0,x(0)=-1/4 ,x^'(0)=0 ?

The solution for 3/4 x^{''}+192x=0,x(0)=-1/4 ,x^'(0)=0 is x=-1/4 cos(16t)

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3/4 x^{''}+192x=0,x(0)=-1/4 ,x^'(0)=0

\frac{3}{4} x^{^{''}} + 192 x = 0, x (0) = - \frac{1}{4}, x^{^{'}} (0) = 0

x = - \frac{1}{4} cos (16 t)

Solution steps

\frac{3}{4} x^{''} (t) + 192 x = 0

Solve linear ODE:

x = - \frac{1}{4} cos (16 t)

x = - \frac{1}{4} cos (16 t)

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

\frac{1}{2} x^{^{''}} + 25 x = 0

3 y^{^{''}} - 24 y^{^{'}} = - 48 y

3 y^{^{''}} + 48 y = 0, y (0) = 3, y^{^{'}} (0) = 20

5 \frac{d ^{2} x}{d t ^{2}} - 3 x = 0

What is the solution for 3/4 x^{''}+192x=0,x(0)=-1/4 ,x^'(0)=0 ?
The solution for 3/4 x^{''}+192x=0,x(0)=-1/4 ,x^'(0)=0 is x=-1/4 cos(16t)