What is the solution for 25y^{''}+2y=0 ?

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

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25y^{''}+2y=0

25 y^{^{''}} + 2 y = 0

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = - x^{3}, (2, - 8)

x \to 0 lim (tan (x))

\int_{- \infty}^{- 1} e^{- 30 t} d t

\frac{d}{d x} (\frac{x}{y ^{2} + 1})

\frac{d}{d x} (lo g_{e} (2))

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