MCQ
Which of the following equation is non-linear
  • A
    $\frac{{dy}}{{dx}} + \frac{y}{x} = \log x$
  • $y\frac{{dy}}{{dx}} + 4x = 0$
  • C
    $dx + dy = 0$
  • D
    $\frac{{dy}}{{dx}} = \cos x$

Answer

Correct option: B.
$y\frac{{dy}}{{dx}} + 4x = 0$
b
(b) A differential equation in which the dependent variable and its differential coefficient occur only in the first degree and are not multiplied together is called a linear differential equation.

Hence $y\frac{{dy}}{{dx}} + 4x = 0$  is non-linear differential equation.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If for the matrix $A=\left[\begin{array}{cc}\tan x & 1 \\ -1 & \tan x\end{array}\right], A+A^{\prime}=2 \sqrt{3}$, , then the value of $x \in\left[0, \frac{\pi}{2}\right]$ is :
If $A$ is a $m \times n$ matrix such that $A B$ and $B A$ are both defined, then $B$ is an
A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is:
Mark the correct alternative in the following question:
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P(X = r})}{\text{P(X = n} -\text{r})}$ is independent of n and r, then p equals:
The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is :
R is a relation on the set Z of integers and it is given by (x, y) ∈ R ⇔ | x - y | ≤ 1. Then, R is:
For two $3\times3$ matrices $A$ and $B$ , let $A+ B\, = 2B'$ and $3A + 2B\, = I_3$, where $B'$ is the transpose of $B$ and $I_3$ is $3\times3$ identity matrix. Then
Let $A=\{0,3,4,6,7,8,9,10\} \quad$ and $R$ be the relation defined on A such that $R =\{( x , y ) \in A \times A : x - y \quad$ is odd positive integer or $x-y=2\}$. The minimum number of elements that must be added to the relation $R$, so that it is a symmetric relation, is equal to $...........$.
Two fair dice, each with faces numbered $1,2,3,4,5$ and $6$ , are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number, then the value of $14 p$ is. . . . . 
Find the shortest distance between the given two lines :
$\frac{x+1}{1}=\frac{y+1}{-1}=\frac{z+1}{1}$ and $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{4}$.