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})

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

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

\frac{\partial}{\partial x} (x z + yz)

\frac{d}{d x} (2 x^{2} + 8 x)

y^{^{'}} = - \frac{1}{x} y^{2} - \frac{1}{x} y

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)