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) Step $1$: Write the given equation: $\sqrt{1 + \frac{1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.Step $2$: Simplify the expression inside the square root:

Vedclass · Accepted Answer

Order $2$, Degree $3$

Vedclass · Answer

(A) Step $1$: Write the given equation: $\sqrt{1 + \frac{1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.Step $2$: Simplify the expression inside the square root: $\sqrt{\frac{(dy/dx)^2 + 1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.Step $3$: This becomes $\frac{\sqrt{(dy/dx)^2 + 1}}{|dy/dx|} = (d^2y/dx^2)^{3/2}$.Step $4$: Square both sides to remove the square root on the right: $\frac{(dy/dx)^2 + 1}{(dy/dx)^2} = (d^2y/dx^2)^3$.Step $5$: Multiply by $(dy/dx)^2$ to clear the fraction: $(dy/dx)^2 + 1 = (d^2y/dx^2)^3 \cdot (dy/dx)^2$.Step $6$: The highest order derivative is $d^2y/dx^2$, so the order is $2$. The power of the highest order derivative is $3$, so the degree is $3$.

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

Similar Questions

The order and degree of the differential equation $x\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + y^2 = 0$ are respectively:

The degree of the differential equation $(1 + \frac{dy}{dx})^2 = (\frac{d^3y}{dx^3})^{1/3}$ is . . . . . . .

If $m$ and $n$ are the degree and order of the differential equation $\left(1+y_{1}^{2}\right)^{2 / 3}=y_{2}$,then the value of $\frac{m+n}{m-n}$ is

The order and degree of the differential equation $ y = x \frac{dy}{dx} + \frac{2}{dy/dx} $ are

The number of arbitrary constants in the general solution of a differential equation of fourth order is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module