What is the solution for 2y^{''}-0y^'+18y=0 ?

The solution for 2y^{''}-0y^'+18y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)

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2y^{''}-0y^'+18y=0

2 y^{^{''}} - 0 y^{^{'}} + 18 y = 0

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

2 y^{''} (t) - 0 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

x^{^{''}} + 4 x^{^{'}} + 4 x = 0

u^{^{''}} + 49 H (t) = 0, H (0) = 3, u^{^{'}} (0) = 7

24 y^{^{''}} + 14 y^{^{'}} - 3 y = 0

y^{^{''}} + 5 \cdot y^{^{'}} + a \cdot y = 0, y (0) = 1, y^{^{'}} (0) = 7

5 y^{^{''}} + c y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

What is the solution for 2y^{''}-0y^'+18y=0 ?
The solution for 2y^{''}-0y^'+18y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)