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3. Determine order and degree (if defined) of differential equation

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

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Given function is
\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0
We can rewrite it as
(s^{'})^4+3s.s^{''} =0
Now, it is clear from the above that, the highest order derivative present in differential equation is  s^{''}

Therefore, the order of the given differential equation \left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0  is  2
Now, the given differential equation is a polynomial equation in its derivatives and power raised to s '' is 1
Therefore, it's a degree is 1

Posted by

Gautam harsolia

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