What is the solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0 ?

The solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0 is y=c_{2}sin(4t)

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4y^{''}+(16)/(0.25)y=0,y(0)=0

4 y^{^{''}} + \frac{16}{0.25} y = 0, y (0) = 0

y = c_{2} sin (4 t)

Solution steps

4 y^{''} (t) + \frac{16}{0.25} y = 0

Solve linear ODE:

y = c_{2} sin (4 t)

y = c_{2} sin (4 t)

y^{^{''}} + 3 y^{^{'}} - 4 y = 0, y (0) = 3, y^{^{'}} (0) = 2

2 y^{^{''}} - y^{^{'}} - 4 y = 0, y (2) = 5, y^{^{'}} (2) = 0

y^{^{''}} - 6 y + 10 y = 0

y^{^{''}} - 7 y^{^{'}} = - 12 y

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

What is the solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0 ?
The solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0 is y=c_{2}sin(4t)