What is the solution for 3y^{''}+48y=0,y(0)=8,y^'(0)=-33 ?

The solution for 3y^{''}+48y=0,y(0)=8,y^'(0)=-33 is y=8cos(4t)-33/4 sin(4t)

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3y^{''}+48y=0,y(0)=8,y^'(0)=-33

3 y^{^{''}} + 48 y = 0, y (0) = 8, y^{^{'}} (0) = - 33

y = 8 cos (4 t) - \frac{33}{4} sin (4 t)

Solution steps

3 y^{''} (t) + 48 y = 0

Solve linear ODE:

y = 8 cos (4 t) - \frac{33}{4} sin (4 t)

y = 8 cos (4 t) - \frac{33}{4} sin (4 t)

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

2 y^{^{''}} + 4 y^{^{'}} - 10 y = 0

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

y^{^{''}} + a y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{1}{4} x^{^{''}} - x = 0, x (0) = 1, x^{^{'}} (0) = 5

What is the solution for 3y^{''}+48y=0,y(0)=8,y^'(0)=-33 ?
The solution for 3y^{''}+48y=0,y(0)=8,y^'(0)=-33 is y=8cos(4t)-33/4 sin(4t)