The degree of the differential equation d2xdt2-2x3-0 is

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Linear Differential Equations of First Order

The degree of...

Question

The degree of the differential equation $\frac{d^{2} x}{d t^{2}} + 2 x^{3} = 0$ is

zero

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{[\frac{d^{2} x}{d t^{2}}]}^{3} + [\frac{d x}{d t}] - x t = 0

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

\frac{d^{2} x}{d t^{2}} + 2 \frac{d x}{d t} + x = 0

{(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0

\frac{d^{2} x}{d t^{2}} = - 9 x

x (0) = 1

{\begin{matrix} \frac{d x}{d t} \end{matrix} ∣ ∣ ∣}_{t = 0} = 1,

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