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 and degree of the differential equation $y=px+\sqrt{a^{2}p^{2}+b^{2}}$,where $p=\frac{dy}{dx}$,are respectively:

If $y=a^3 e^{b^2 x+c}$ is the general solution of a differential equation,where $a$ and $c$ are arbitrary constants and $b$ is a fixed constant,then the order of the differential equation is

For the differential equation given below,determine its order and degree (if defined):
$\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$

Degree of the differential equation $e^{\frac{dy}{dx}} + (\frac{dy}{dx})^3 = x$ is

The order and degree of the differential equation $3 - (\frac{d^3y}{dx^3})^{7/3} = (\frac{dy}{dx})^5$ are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module