What is the solution for 2500y^{''}+81y=0 ?

The solution for 2500y^{''}+81y=0 is y=c_{1}cos((9t)/(50))+c_{2}sin((9t)/(50))

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2500y^{''}+81y=0

2500 y^{^{''}} + 81 y = 0

y = c_{1} cos (\frac{9 t}{50}) + c_{2} sin (\frac{9 t}{50})

Solution steps

2500 y^{''} (t) + 81 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{9 t}{50}) + c_{2} sin (\frac{9 t}{50})

y = c_{1} cos (\frac{9 t}{50}) + c_{2} sin (\frac{9 t}{50})

y^{^{''}} + 5 y^{^{'}} - 24 y = 0, y (0) = x, y^{^{'}} (0) = 56

y^{^{''}} - 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

\frac{d ^{2} y}{d t ^{2}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 0

- 2 y^{^{''}} + 5 y^{^{'}} + 10 y = 0

What is the solution for 2500y^{''}+81y=0 ?
The solution for 2500y^{''}+81y=0 is y=c_{1}cos((9t)/(50))+c_{2}sin((9t)/(50))