What is the solution for 3y^{''}+12y=0,y(0)=5,y^'(0)=12 ?

The solution for 3y^{''}+12y=0,y(0)=5,y^'(0)=12 is y=5cos(2t)+6sin(2t)

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3y^{''}+12y=0,y(0)=5,y^'(0)=12

3 y^{^{''}} + 12 y = 0, y (0) = 5, y^{^{'}} (0) = 12

y = 5 cos (2 t) + 6 sin (2 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (2 t) + 6 sin (2 t)

y = 5 cos (2 t) + 6 sin (2 t)

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

36 y^{^{''}} - 81 y = 0, y (0) = 0, y^{^{'}} (0) = 1

4 x^{^{''}} + 20 x^{^{'}} + 169 x = 0, x (0) = 3, x^{^{'}} (0) = 12

\frac{4}{32} y^{^{''}} + 3 y = 0

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y^{^{'}} (\frac{π}{2}) = 12

What is the solution for 3y^{''}+12y=0,y(0)=5,y^'(0)=12 ?
The solution for 3y^{''}+12y=0,y(0)=5,y^'(0)=12 is y=5cos(2t)+6sin(2t)