What is the solution for 20y^{''}+60y=0 ?

The solution for 20y^{''}+60y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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20y^{''}+60y=0

20 y^{^{''}} + 60 y = 0

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

20 y^{''} (t) + 60 y = 0

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

\frac{d ^{2} y}{d x ^{2}} - 36 y = 0

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 4

6 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

What is the solution for 20y^{''}+60y=0 ?
The solution for 20y^{''}+60y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)