What is the solution for linear x^{''}+9x=0 ?

The solution for linear x^{''}+9x=0 is x=c_{1}cos(3t)+c_{2}sin(3t)

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linear x^{''}+9x=0

linear

x^{^{''}} + 9 x = 0

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

Solution steps

x^{''} (t) + 9 x = 0

Solve linear ODE:

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

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

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

x, \frac{d ^{2} x}{d y ^{2}} = a^{2} x

\frac{d ^{3} y}{d x ^{3}} - 2 y^{^{''}} + 1 y = 0

\frac{1}{16} y^{^{''}} + 4 y = 0

y^{^{''''}} + 3 y^{^{'''}} - y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

What is the solution for linear x^{''}+9x=0 ?
The solution for linear x^{''}+9x=0 is x=c_{1}cos(3t)+c_{2}sin(3t)