The degree of the differential equation left( | vedclass

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

Question

(B) Given the differential equation: $\left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} + 4 - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} = 0$. To find the degre

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the differential equation: $\left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} + 4 - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} = 0$. To find the degree,we must first eliminate the fractional exponent of the highest order derivative. Rearrange the equation: $\left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} = 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4$. Cube both sides to remove the fractional power: $\left( \left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} \right)^3 = \left( 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4 \right)^3$. This simplifies to: $\left( \frac{d^3y}{dx^3} \right)^2 = \left( 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4 \right)^3$. The highest order derivative present is $\frac{d^3y}{dx^3}$,which is of order $3$. The power (exponent) of the highest order derivative after making the equation a polynomial in derivatives is $2$. Therefore,the degree of the differential equation is $2$.

The degree of the differential equation $\left( \frac{d^3y}{dx^3} \right)^{\frac{2}{3}} + 4 - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} = 0$ is

Similar Questions

The order and degree of the differential equation $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$ are

The sum of the order and degree of the differential equation $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2}}=\left(1+\frac{d y}{d x}\right)^{\frac{4}{3}}$ is

The sum of the degree and order of the differential equation $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[5]{\frac{dy}{d x}-5}$ is

The order of the differential equation whose general solution is $y = a_1(a_2 + a_3) \cdot \cos(x + a_4) - a_5 e^{x + a_6}$ is . . . . . . .

Which of the following equations is non-linear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module