What is the solution for x^{''}+48x=0 ?

The solution for x^{''}+48x=0 is x=c_{1}cos(4sqrt(3)t)+c_{2}sin(4sqrt(3)t)

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x^{''}+48x=0

x^{^{''}} + 48 x = 0

x = c_{1} cos (43 t) + c_{2} sin (43 t)

Solution steps

x^{''} (t) + 48 x = 0

Solve linear ODE:

x = c_{1} cos (43 t) + c_{2} sin (43 t)

x = c_{1} cos (43 t) + c_{2} sin (43 t)

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

y^{^{'''}} + 3 y^{^{''}} - 6 y^{^{'}} - 8 y = 0

(D^{2} + 16) y = 0

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 15 y = 0

77 y^{^{''''}} + 338 y^{^{''}} + 225 y = 0

What is the solution for x^{''}+48x=0 ?
The solution for x^{''}+48x=0 is x=c_{1}cos(4sqrt(3)t)+c_{2}sin(4sqrt(3)t)