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)

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)