What is the solution for 8y^{''}+26y=0 ?

The solution for 8y^{''}+26y=0 is y=c_{1}cos((sqrt(13)t)/2)+c_{2}sin((sqrt(13)t)/2)

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8y^{''}+26y=0

8 y^{^{''}} + 26 y = 0

y = c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2})

Solution steps

8 y^{''} (t) + 26 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2})

y = c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2})

\frac{e ^{- s}}{s ^{3}}

\int \frac{1}{x + 1 + x ^{2}} d x

\frac{1}{s ^{3}}

x^{2} y^{^{'}} - 2 x y = 3 y^{4}, y (1) = \frac{1}{2}

\int \frac{1}{e ^{x} + e ^{2 x}} d x

What is the solution for 8y^{''}+26y=0 ?
The solution for 8y^{''}+26y=0 is y=c_{1}cos((sqrt(13)t)/2)+c_{2}sin((sqrt(13)t)/2)