What is the solution for 36y^{''}+25y=0 ?

The solution for 36y^{''}+25y=0 is y=c_{1}cos((5t)/6)+c_{2}sin((5t)/6)

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36y^{''}+25y=0

36 y^{^{''}} + 25 y = 0

y = c_{1} cos (\frac{5 t}{6}) + c_{2} sin (\frac{5 t}{6})

Solution steps

36 y^{''} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{5 t}{6}) + c_{2} sin (\frac{5 t}{6})

y = c_{1} cos (\frac{5 t}{6}) + c_{2} sin (\frac{5 t}{6})

2 y^{^{''}} - 17 y^{^{'}} - 9 y = 0, y (0) = 4, y^{^{'}} (0) = - 7

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - 5

u^{^{''}} = H (t)

0.25 \cdot 0.04 x^{^{''}} + 2 \cdot 0.04 x^{^{'}} + x = 0

What is the solution for 36y^{''}+25y=0 ?
The solution for 36y^{''}+25y=0 is y=c_{1}cos((5t)/6)+c_{2}sin((5t)/6)