What is the solution for 4y^{''}+26y=0 ?

The solution for 4y^{''}+26y=0 is y=c_{1}cos(sqrt(13/2)t)+c_{2}sin(sqrt(13/2)t)

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4y^{''}+26y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

y = 5 x

\int (3 x^{2} - 6 x - 7) d x

\frac{1}{( s ^{2} - 4 s + 5 )}

\frac{d}{d x} (sin (3 x) + 3 x cos (3 x))

\frac{d}{d x} (x - e^{- x})

What is the solution for 4y^{''}+26y=0 ?
The solution for 4y^{''}+26y=0 is y=c_{1}cos(sqrt(13/2)t)+c_{2}sin(sqrt(13/2)t)