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

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

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7y^{''}+y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

\int e^{4 x} \frac{1}{5} d x

\frac{d y}{d t} = (5 + sin (t)) \cdot (\frac{y}{5})

\int \frac{x + 11}{( x - 4 ) ( x + 1 )} d x

y = a x + b

x^{2} = y, x^{2} = - (y - 2)

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