What is the solution for x^{''}+2^2*3^2x=0 ?

The solution for x^{''}+2^2*3^2x=0 is x=c_{1}cos(6t)+c_{2}sin(6t)

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x^{''}+2^2*3^2x=0

x^{^{''}} + 2^{2} \cdot 3^{2} x = 0

x = c_{1} cos (6 t) + c_{2} sin (6 t)

Solution steps

x^{''} (t) + 2^{2} \cdot 3^{2} x = 0

Solve linear ODE:

x = c_{1} cos (6 t) + c_{2} sin (6 t)

x = c_{1} cos (6 t) + c_{2} sin (6 t)

y^{^{''}} + 4 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 3

y^{^{'''}} - y^{^{''}} - 12 y^{^{'}} - 40 y = 0, y (0) = e^{5 x}

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

x^{^{''}} + 7 x^{^{'}} + 16 x = 0

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

What is the solution for x^{''}+2^2*3^2x=0 ?
The solution for x^{''}+2^2*3^2x=0 is x=c_{1}cos(6t)+c_{2}sin(6t)