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)

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)