What is the solution for y^{''}+36y=0,y(0)=5,y^'(0)=12 ?

The solution for y^{''}+36y=0,y(0)=5,y^'(0)=12 is y=5cos(6t)+2sin(6t)

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y^{''}+36y=0,y(0)=5,y^'(0)=12

y^{^{''}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = 12

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

Solution steps

y^{''} (t) + 36 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1

y^{^{''''}} - 3 y^{^{''}} - 4 y = 0

y^{^{''}} + 0.5 y^{^{'}} + y = 0

y^{^{''}} + 64 π^{2} y = 0

What is the solution for y^{''}+36y=0,y(0)=5,y^'(0)=12 ?
The solution for y^{''}+36y=0,y(0)=5,y^'(0)=12 is y=5cos(6t)+2sin(6t)