${\left( {\cfrac{{ds}}{{dt}}} \right)^4} + 3s\cfrac{{{d^2}s}}{{d{t^2}}} = 0$
${\left( {\cfrac{{ds}}{{dt}}} \right)^4} + 3s\cfrac{{{d^2}s}}{{d{t^2}}} = 0$
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
.: The highest order derivative is
$\cfrac{{{d^2}s}}{{d{t^2}}}$ and the degree of
its highest order is one.
Therefore, the given
differential equation is of order 2 and degree 1.
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