What is the solution for 3y^{''}+12y=0,y(0)=8,y^'(0)=6 ?

The solution for 3y^{''}+12y=0,y(0)=8,y^'(0)=6 is y=8cos(2t)+3sin(2t)

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3y^{''}+12y=0,y(0)=8,y^'(0)=6

3 y^{^{''}} + 12 y = 0, y (0) = 8, y^{^{'}} (0) = 6

y = 8 cos (2 t) + 3 sin (2 t)

Solution steps

3 y^{''} (t) + 12 y = 0

Solve linear ODE:

y = 8 cos (2 t) + 3 sin (2 t)

y = 8 cos (2 t) + 3 sin (2 t)

y^{^{''}} + 12 y^{^{'}} + 38 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = - 1

y^{^{''}} - r^{2} y = 0

\frac{5}{32} y^{^{''}} + \frac{1}{4} y = 0

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

What is the solution for 3y^{''}+12y=0,y(0)=8,y^'(0)=6 ?
The solution for 3y^{''}+12y=0,y(0)=8,y^'(0)=6 is y=8cos(2t)+3sin(2t)