What is the solution for y^{''}+28y=0 ?

The solution for y^{''}+28y=0 is y=c_{1}cos(2sqrt(7)t)+c_{2}sin(2sqrt(7)t)

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y^{''}+28y=0

y^{^{''}} + 28 y = 0

y = c_{1} cos (27 t) + c_{2} sin (27 t)

Solution steps

y^{''} (t) + 28 y = 0

Solve linear ODE:

y = c_{1} cos (27 t) + c_{2} sin (27 t)

y = c_{1} cos (27 t) + c_{2} sin (27 t)

\int \frac{y ^{2}}{8} d y

x sin (\frac{3}{x})

\int e^{16 x^{2}} d x

\int_{- 1}^{x^{2}} (- 2 t + 2) d t

\frac{5}{x}

What is the solution for y^{''}+28y=0 ?
The solution for y^{''}+28y=0 is y=c_{1}cos(2sqrt(7)t)+c_{2}sin(2sqrt(7)t)