Consider the following differential equations | vedclass

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

Question

(C) For $D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$. Multiplying by $\frac{dy}{dx}$, we get $y \frac{dy}{dx}=4(\frac{dy}{dx})^2+3x$. The highest derivat

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

(C) For $D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$. Multiplying by $\frac{dy}{dx}$, we get $y \frac{dy}{dx}=4(\frac{dy}{dx})^2+3x$. The highest derivative is $\frac{dy}{dx}$, so order is $1$. The power of the highest derivative is $2$, so degree is $2$.For $D_2: \frac{d^2y}{dx^2}=(3+(\frac{dy}{dx})^2)^{\frac{4}{3}}$. Cubing both sides, we get $(\frac{d^2y}{dx^2})^3=(3+(\frac{dy}{dx})^2)^4$. The highest derivative is $\frac{d^2y}{dx^2}$, so order is $2$. The power of the highest derivative is $3$, so degree is $3$.For $D_3: [1+(\frac{dy}{dx})]^2=(\frac{dy}{dx})^2$. Expanding, $1+(\frac{dy}{dx})^2+2\frac{dy}{dx}=(\frac{dy}{dx})^2$, which simplifies to $1+2\frac{dy}{dx}=0$. The highest derivative is $\frac{dy}{dx}$, so order is $1$. The power of the highest derivative is $1$, so degree is $1$.Sum of orders $= 1+2+1 = 4$.Sum of degrees $= 2+3+1 = 6$.Required ratio $= \frac{4}{6} = \frac{2}{3}$.

Similar Questions

The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$ is

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:

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

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

The order and the degree of the differential equation $\frac{d^4y}{dx^4} - 4\frac{d^3y}{dx^3} + 8\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 4y = 0$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module