Question

Integrating factor for the differential equation \((x^2 + y^2 + x) \, dx + xy \, dy = 0\) is:

• A. \(y\)
• B. \(x\)
• C. \(e^x\)
• D. \(e^{xy}\)

Hint:

Exam Tip:
To find an integrating factor:

• If \(\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = f(x)\), then I.F. = \(e^{\int f(x) dx}\).
• If \(\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = g(y)\), then I.F. = \(e^{\int g(y) dy}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the integrating factor for the given differential equation. The equation is: \[ (x^2 + y^2 + x) \, dx + xy \, dy = 0 \] Here, \(M = x^2 + y^2 + x\), \(N = xy\).
Check if it is exact: \(\frac{\partial M}{\partial y} = 2y\), \(\frac{\partial N}{\partial x} = y\).
Not exact. We need an integrating factor.

Step 2: Key Formula or Approach:
We can try to find an integrating factor that is a function of \(x\) alone, \(y\) alone, or a combination.
Compute: \[ \frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = \frac{1}{xy} (2y - y) = \frac{1}{xy} \cdot y = \frac{1}{x} \] This is a function of \(x\) only.
So, the integrating factor is \(e^{\int \frac{1}{x} dx} = e^{\ln x} = x\).

Step 3: Detailed Explanation:
Since \(\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)\) is a function of \(x\) only, the integrating factor is \(e^{\int f(x) dx} = e^{\ln x} = x\).
Check: Multiply the equation by \(x\): \[ x(x^2 + y^2 + x) \, dx + x^2 y \, dy = 0 \] Now, \(M = x^3 + xy^2 + x^2\), \(N = x^2 y\).
\(\frac{\partial M}{\partial y} = 2xy\), \(\frac{\partial N}{\partial x} = 2xy\).
Now it is exact.
So, the integrating factor is \(x\).

Step 4: Final Answer:
Therefore, option (B) is correct.