Find the order degree of differential equation dsdt4-3sd2sdt2-0-

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Degree of a Differential Equations

Find the orde...

Question

Find the order degree of differential equation ${(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$ .

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Determine order and degree(if defined) of differential equation

Determine order and degree (when defined) of differential equations.
${(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0.$

\frac{d^{4} y}{d x^{4}} + sin (y^{'''}) = 0

y^{'} + 5 y = 0

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

{(\frac{d^{2} y}{d x^{2}})}^{2} - cos (\frac{d y}{d x}) = 0

\frac{d^{2} y}{d x^{2}} = cos 3 x + sin 3 x

(y^{'''})^{2} + (y)^{3} + (y)^{4} + y^{5} = 0

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

y^{'} + y = e^{x}

y^{''} + (y) 2 + 2 y = 0

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

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

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

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