What are the order and degree, respectively, of the differential equation
${(\frac{d^{3} y}{d x^{3}})}^{2} = y^{4} + {(\frac{d y}{d x})}^{5}$ ?

4, 5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2, 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3, 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5, 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3, 2$
Order is the highest derivative of the dependent variable with respect to the independent variable and degree is the highest power to which the highest order derivative in the differential equation is raised.

so, $O r d e r = 3$ and $D e g r e e = 2$

Therefore, Answer is $C$

Suggest Corrections

Similar questions

{(\frac{d^{3} y}{d x^{3}})}^{2} + {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{4} + y^{5} = 0

\frac{d^{4} y}{d x^{4}} + (\frac{d^{3} y}{d x^{3}})^{2} + (\frac{d^{2} y}{d x^{2}})^{3} + (\frac{d y}{d x})^{4} = 0

{[1 + {(\frac{d y}{d x})}^{5}]}^{\frac{2}{3}} = \frac{d^{3} y}{d x^{3}}

\frac{d^{3} y}{d x^{3}} + 4 \sqrt{{(\frac{d y}{d x})}^{3} + y^{2}} = 0

{[1 + {(\frac{d y}{d x})}^{2}]}^{5 / 2} = \frac{d^{3} y}{{d x}^{3}}

Join BYJU'S Learning Program