The degree of the differential equation $y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$ is:

1

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

2

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

3

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

none of these

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

Open in App

Solution

The correct option is B $2$
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$
$\Rightarrow y_{3}^{2 / 3} = - (2 + 3 y_{2} + y_{1})$
$\Rightarrow y_{3}^{2} = - (2 + 3 y_{2} + y_{1})^{2}$ cubing both sides,
Hence degree of given differential equation is $2$ (power on $y_{3}$ )

Suggest Corrections

Similar questions

Q. The degree of the differential equation

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

is:

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + s i n (\frac{d y}{d x}) + 1 = 0$ is
(a) 3
(b) 2
(c) 1
(d) None of these

Q. Let

y = y (x)

be the solution of the differential equation

e^{x} \sqrt{1 - y^{2}} d x + (\frac{y}{x}) d y = 0,

y (1) = - 1.

Then the value of

(y (3))^{2}

is equal to

Q. Find the order and degree of the differential equation,

y_{2}^{\frac{3}{2}} + y_{1} = 1

Q. The degree of the differential equation

\frac{d^{2} x}{d t^{2}} + 2 x^{3} = 0

Join BYJU'S Learning Program

The degree of the differential equation y2/33+2+3y2+y1=0 is:

The correct option is B 2y2/33+2+3y2+y1=0⇒y2/33=−(2+3y2+y1)⇒y23=−(2+3y2+y1)2 cubing both sides,Hence degree of given differential equation is 2 (power on y3)

The degree of the differential equation $y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$ is:

The correct option is B $2$
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$
$\Rightarrow y_{3}^{2 / 3} = - (2 + 3 y_{2} + y_{1})$
$\Rightarrow y_{3}^{2} = - (2 + 3 y_{2} + y_{1})^{2}$ cubing both sides,
Hence degree of given differential equation is $2$ (power on $y_{3}$ )