What is the solution for 4y^{''}+7y=0 ?

The solution for 4y^{''}+7y=0 is y=c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2)

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4y^{''}+7y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (- 3 y)

\frac{\partial}{\partial x} (e^{2 x} - x cos (t))

5 \frac{d y}{d x} + 30 y = 6

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

\frac{d}{d x} (x^{3} - 3 x^{2} + 4)

What is the solution for 4y^{''}+7y=0 ?
The solution for 4y^{''}+7y=0 is y=c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2)