Question

The integrating factor of the differential equation $(1 + x^2) \frac{dy}{dx} + 2xy = \cos x$ is:

• A. $e^{x^2}$
• B. $1 + x^2$
• C. $\ln(1 + x^2)$
• D. $\frac{1}{1 + x^2}$

Hint:

If the numerator is the exact derivative of the denominator, the integral is simply the natural log of the denominator, yielding a direct simplification with the base $e$.

Answer 1

The Correct Option is B

Step 1: Aim to find the integrating factor.
For a linear equation the integrating factor is $e^{\int P\,dx}$, but first the equation must be in the neat form $\frac{dy}{dx} + Py = Q$. So we tidy it up.

Step 2: Divide through by the leading term.
The equation is $(1 + x^2)\frac{dy}{dx} + 2xy = \cos x$. Divide every term by $(1 + x^2)$.
\[ \frac{dy}{dx} + \frac{2x}{1 + x^2}\,y = \frac{\cos x}{1 + x^2} \]

Step 3: Identify $P$.
Comparing with the standard form, $P = \frac{2x}{1 + x^2}$.

Step 4: Integrate $P$.
Notice the top $2x$ is exactly the derivative of the bottom $1 + x^2$. So the integral is a logarithm.
\[ \int \frac{2x}{1 + x^2}\,dx = \ln(1 + x^2) \]

Step 5: Build the integrating factor.
Raise $e$ to this integral. The $e$ and $\ln$ cancel.
\[ \text{I.F.} = e^{\ln(1 + x^2)} = 1 + x^2 \]

Step 6: State the answer.
So the integrating factor is $1 + x^2$.
\[ \boxed{1 + x^2} \]