Question

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

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

Hint:

When classifying a differential equation, always try to rearrange it into the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\) first, as it's a very common type. If terms can be separated into functions of only x and only y, it's linear.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to classify the given first-order differential equation into one of the standard types.

Step 2: Key Formula or Approach:
We need to check if the equation can be rearranged into the standard forms for each type:
-

Variable Separable: Can it be written as \(f(y)dy = g(x)dx\)?
-

Homogeneous: Can \(\frac{dy}{dx}\) be expressed as a function of \(\frac{y}{x}\)?
-

Linear: Can it be written as \(\frac{dy}{dx} + P(x)y = Q(x)\)?
-

Exact: Can it be written as \(M(x,y)dx + N(x,y)dy = 0\) where \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)?

Step 3: Detailed Explanation:
The given equation is \(\frac{dy}{dx} = -\frac{x+y}{1+x^2}\).
Let's split the fraction on the right-hand side:
\[ \frac{dy}{dx} = -\frac{x}{1+x^2} - \frac{y}{1+x^2} \]
Now, let's rearrange the terms to see if it matches the linear form by moving the term with y to the left side:
\[ \frac{dy}{dx} + \frac{1}{1+x^2}y = -\frac{x}{1+x^2} \]
This equation is exactly in the standard form of a first-order linear differential equation, \(\frac{dy}{dx} + P(x)y = Q(x)\), where:
- \(P(x) = \frac{1}{1+x^2}\)
- \(Q(x) = -\frac{x}{1+x^2}\)
It is not variable separable as we cannot group all \(x\) terms with \(dx\) and all \(y\) terms with \(dy\). It is also not homogeneous. Therefore, the correct classification is a first-order linear equation.

Step 4: Final Answer:
The differential equation is a First order Linear equation.