Question

The differential equation \(\displaystyle \frac{dy}{dx}=-\left(\frac{x+y}{1+x^2}\right)\) is

• A. of Variable separable form
• B. First order Linear equation
• C. Homogeneous
• D. Exact differential Equation

Hint:

A differential equation of the form \(\frac{dy}{dx}+P(x)y=Q(x)\) is called a first order linear differential equation.

Answer 1

The Correct Option is B

We are given: \[ \frac{dy}{dx}=-\left(\frac{x+y}{1+x^2}\right). \] First distribute the negative sign: \[ \frac{dy}{dx}=-\frac{x+y}{1+x^2}. \] Separate the numerator: \[ \frac{dy}{dx}=-\frac{x}{1+x^2}-\frac{y}{1+x^2}. \] Now bring the term containing \(y\) to the left side: \[ \frac{dy}{dx}+\frac{y}{1+x^2}=-\frac{x}{1+x^2}. \] This is of the form: \[ \frac{dy}{dx}+P(x)y=Q(x). \] Here, \[ P(x)=\frac{1}{1+x^2} \] and \[ Q(x)=-\frac{x}{1+x^2}. \] Since it is in the standard form of a first order linear differential equation, the given differential equation is a first order linear equation. Therefore, the correct answer is: \[ \text{First order Linear equation}. \]