What is the solution for 25y^{''}+11y=0 ?

The solution for 25y^{''}+11y=0 is y=c_{1}cos((sqrt(11)t)/5)+c_{2}sin((sqrt(11)t)/5)

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25y^{''}+11y=0

25 y^{^{''}} + 11 y = 0

y = c_{1} cos (\frac{11 t}{5}) + c_{2} sin (\frac{11 t}{5})

Solution steps

25 y^{''} (t) + 11 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{11 t}{5}) + c_{2} sin (\frac{11 t}{5})

y = c_{1} cos (\frac{11 t}{5}) + c_{2} sin (\frac{11 t}{5})

x \to 2 - lim (\frac{1}{x ^{2} - 9})

\int \frac{- 2 x ^{3} + x}{- 4 x ^{4} + 4 x ^{2}} d x

\int \frac{x}{1 - x ^{3}} d x

\int (3 x + 1) e^{2 x} d x

n = 1 \sum \infty n (n - 1) x^{n}

What is the solution for 25y^{''}+11y=0 ?
The solution for 25y^{''}+11y=0 is y=c_{1}cos((sqrt(11)t)/5)+c_{2}sin((sqrt(11)t)/5)