What is the solution for y^{''}+0y^'+36y=0,y(0)=1 ?

The solution for y^{''}+0y^'+36y=0,y(0)=1 is y=cos(6t)+c_{2}sin(6t)

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

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

y = cos (6 t) + c_{2} sin (6 t)

Solution steps

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

Solve linear ODE:

y = cos (6 t) + c_{2} sin (6 t)

y = cos (6 t) + c_{2} sin (6 t)

y^{^{''}} + 20 y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = 0.6

4 y^{^{''}} + 577 y = 0, y (0) = 1, y^{^{'}} (0) = 3

2 y^{^{''}} + 4 y^{^{'}} + k y = 0, y (0) = 2, y^{^{'}} (0) = 0

0.1 \cdot y^{^{''}} + y^{^{'}} + y = 0

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

What is the solution for y^{''}+0y^'+36y=0,y(0)=1 ?
The solution for y^{''}+0y^'+36y=0,y(0)=1 is y=cos(6t)+c_{2}sin(6t)