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

(B) Step $1$: Write the given differential equation: $(\frac{d^3y}{dx^3})^{\frac{2}{3}} = 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4$.Step $2$: To find t

Vedclass · Accepted Answer

$3, 2$

Vedclass · Answer

(B) Step $1$: Write the given differential equation: $(\frac{d^3y}{dx^3})^{\frac{2}{3}} = 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4$.Step $2$: To find the degree, the equation must be a polynomial in terms of derivatives. Remove the fractional exponent by raising both sides to the power of $3$: $((\frac{d^3y}{dx^3})^{\frac{2}{3}})^3 = (3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4)^3$.Step $3$: This simplifies to $(\frac{d^3y}{dx^3})^2 = (3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4)^3$.Step $4$: The order is the highest derivative present, which is $3$. The degree is the power of the highest derivative after rationalizing, which is $2$. Thus, the order is $3$ and the degree is $2$.

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:

Similar Questions

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

The order and degree of the differential equation $\frac{d^2 y}{d x^2}=\sqrt{\frac{d y}{d x}}$ are respectively

For the differential equation $\frac{d^3y}{dx^3} + \cos \left( \frac{d^2y}{dx^2} \right) = 0$, which of the following is true?

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

The differential equation $\frac{d^3y}{dx^3}-5y \frac{dy}{dx}+xy=0$ represents :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module