What is the solution for 2y^{''}+4y=0 ?

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

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2y^{''}+4y=0

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

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

Solution steps

2 y^{''} (t) + 4 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)

\frac{d}{d x} (\frac{1}{x ( 1 + ln ( x ) )})

\int x (x + 3)^{2} d x

(3 x^{2} + 3 x - 1)^{4}

10 \frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 6 y = 0

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