What is the solution for 4y^{''}+8y=0 ?

The solution for 4y^{''}+8y=0 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)

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4y^{''}+8y=0

4 y^{^{''}} + 8 y = 0

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

Solution steps

4 y^{''} (t) + 8 y = 0

Solve linear ODE:

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

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

\int_{- 1}^{1} x^{\frac{1}{3}} d x

x \to - 1 lim (sin (π x))

x \to 0 + lim ((tan (3 x))^{x})

y = e^{5 x}, y = e^{6 x}, x = 1

\frac{d}{d x} (2 x y^{3})

What is the solution for 4y^{''}+8y=0 ?
The solution for 4y^{''}+8y=0 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)