What is the solution for 7y^{''}+26y=0 ?

The solution for 7y^{''}+26y=0 is y=c_{1}cos(sqrt(26/7)t)+c_{2}sin(sqrt(26/7)t)

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7y^{''}+26y=0

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

y = c_{1} cos (\frac{26}{7} t) + c_{2} sin (\frac{26}{7} t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{26}{7} t) + c_{2} sin (\frac{26}{7} t)

y = c_{1} cos (\frac{26}{7} t) + c_{2} sin (\frac{26}{7} t)

\frac{- 5}{6 x ^{6}}

\frac{d}{d x} (\frac{0.3 ( 9 ^{x} )}{x ^{3}})

f (x) = 5 - 2 x^{3}

f (x) = \frac{2}{1 + x ^{2}}

\int \frac{1}{p ^{2}} d p

What is the solution for 7y^{''}+26y=0 ?
The solution for 7y^{''}+26y=0 is y=c_{1}cos(sqrt(26/7)t)+c_{2}sin(sqrt(26/7)t)