Question

Find order and degree (if defined) of the differential equation $\dfrac{{{d}^{4}}y}{d{{x}^{4}}}+\sin \left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)=0$.

Answer 1

The term order of a differential equation is equal to the value of the highest power/order of the derivative in the equation and degree is equal to the value of the highest power upto which the derivative is increased. A common differential equation to specify the order and degree can be written as:
${{\left( \dfrac{{{d}^{n}}y}{d{{x}^{n}}} \right)}^{m}}$
where $n$ is the value of the order and $m$ is the value of degree.

Complete step-by-step answer:
The equation is given as $\dfrac{{{d}^{4}}y}{d{{x}^{4}}}+\sin \left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)=0$, and the value of order is the highest power of the derivative in the equation, there are two powers one is $\dfrac{{{d}^{4}}y}{d{{x}^{4}}},n=4$ and other is $\dfrac{{{d}^{3}}y}{d{{x}^{3}}},n=3$. Now, $4>3$ thereby, the order of the equation is $4$.
As for the degree, the first derivative $\dfrac{{{d}^{4}}y}{d{{x}^{4}}}$ is equal to $1$ and the second derivative can’t be comprehended as the derivative is attached to sin making it a sine value and not a derivative value.
Hence, degree is undefined.
Therefore, order is $4$ and degree is undefined.

Note: Students may go wrong, while finding the value of the degree as the first degree is raised to power $1$ but the second is undefined in second if we rewrite it as ${{\left( \sin \left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right) \right)}^{1}}$ then we can see that the power is not of that of the derivative but that of the value of sine hence, the degree of the second derivative is undefined.