The order and degree of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given differential equation is $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$.Squaring both sides,we get $1+\frac{1}{(\frac{dy

Vedclass · Accepted Answer

$2, 3$

Vedclass · Answer

(A) Given differential equation is $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$.Squaring both sides,we get $1+\frac{1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$.Multiplying both sides by $(\frac{dy}{dx})^2$,we get $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$.The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.The power of the highest order derivative after making the equation free from radicals and fractions is $3$,so the degree is $3$.Thus,the order and degree are $2$ and $3$ respectively.

The order and degree of the differential equation $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$ are respectively:

Similar Questions

The order and degree of the differential equation $(\frac{d^3y}{dx^3})^{\frac{2}{3}} - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 4 = 0$ are respectively:

The order of the differential equation,whose general solution is given by $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$,where $c_1, c_2, c_3, c_4$ and $c_5$ are arbitrary constants,is

The differential equation of all parabolas having their axis of symmetry coinciding with the $x$-axis has its order and degree respectively:

The sum of the degree and order of the differential equation $(1+y_1^2)^{2/3} = y_2$ is

The differential equation $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + \sin y + x^2 = 0$ is of the following type:

Vedclass Test Series

Exam Paper Generator

Online Exam Module