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

Determine the order and degree (if defined) of the differential equation $y^{\prime} + y = e^{x}$.

The order and degree of the differential equation $(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^{\frac{7}{5}} =x^3\frac{d^2y}{dx^2}$ are $m$ and $n$ respectively,then $(m + n)$ is equal to-

The differential equation representing the family of curves $y^{2}=2 d(x+\sqrt{d})$, where $d$ is a parameter, is of

The degree of the differential equation $\left( \frac{d^3y}{dx^3} \right)^2 + 4\left( \frac{dy}{dx} \right)^3 = 3\sin \left( \frac{d^2y}{dx^2} \right)$ is:

Determine the order and degree (if defined) of the differential equation $y'' + (y')^2 + 2y = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module