The order and degree of the differential equation -1 - dydx 2 - -d2ydx2 23 is

The correct option is A 2,3 Given, √(1+ ( dydx )^2)= ( d^2ydx^2 )^3/2 squaring on both sides, we get 1+ ( dydx )^2= ( d^2ydx^2 )^3 Therefore, order ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

The order and...

Question

The order and degree of the differential equation $\sqrt{1 + {(\frac{d y}{d x})}^{2}} = {(] \frac{d^{2} y}{d x^{2}})}^{\frac{2}{3}}$ is

2, 3

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

3, 2

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

1, 2

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

2, 2

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

Open in App

Solution

The correct option is A $2, 3$
Given,

$\sqrt{1 + {(\frac{d y}{d x})}^{2}} = {(\frac{d^{2} y}{d x^{2}})}^{\frac{3}{2}}$

squaring on both sides, we get

$1 + {(\frac{d y}{d x})}^{2} = {(\frac{d^{2} y}{d x^{2}})}^{3}$

Therefore, order 2, degree 3.

Suggest Corrections

Similar questions

Q. The degree of the differential equation

{(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x sin (\frac{d^{2} y}{d x^{2}})

Q. The order and degree of the differential equation

\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 - {(\frac{d y}{d x})}^{4}}

are respectively.

The order and degree of differential equation $[1 + {(\frac{d y}{d x})}^{2}] = \frac{d^{2} y}{d x^{2}}$ are
a) $2, \frac{3}{2}$
b) 2, 3
c) 2, 1
d) 3, 4

The order and degree of the differential equation ${[4 + {(\frac{d y}{d x})}^{2}]}^{2 / 3} = \frac{d^{2} y}{d x^{2}}$ are

Q. The order and degree of

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

is:

Join BYJU'S Learning Program

The order and degree of the differential equation √1+(dydx)2=(]d2ydx2)23 is

The correct option is A 2,3Given,√1+(dydx)2=(d2ydx2)32squaring on both sides, we get1+(dydx)2=(d2ydx2)3Therefore, order 2, degree 3.

The order and degree of the differential equation $\sqrt{1 + {(\frac{d y}{d x})}^{2}} = {(] \frac{d^{2} y}{d x^{2}})}^{\frac{2}{3}}$ is

The correct option is A $2, 3$
Given,

$\sqrt{1 + {(\frac{d y}{d x})}^{2}} = {(\frac{d^{2} y}{d x^{2}})}^{\frac{3}{2}}$

squaring on both sides, we get

$1 + {(\frac{d y}{d x})}^{2} = {(\frac{d^{2} y}{d x^{2}})}^{3}$

Therefore, order 2, degree 3.