Question

An integrating factor of the differential equation \( (x^2+1)\frac{dy}{dx} + xy = x^3 \) is

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

Hint:

For a linear DE \(y' + P(x)y = Q(x)\), the integrating factor is \(I.F. = e^{\int P(x)dx}\).
Remember common integrals, e.g., \(\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C\).
\(e^{\ln A} = A\).

Answer 1

The Correct Option is C

The given first-order linear differential equation is \( (x^2+1)\frac{dy}{dx} + xy = x^3 \). First, write it in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \). Divide by \((x^2+1)\): \[ \frac{dy}{dx} + \frac{x}{x^2+1}y = \frac{x^3}{x^2+1} \] Here, \( P(x) = \frac{x}{x^2+1} \). The integrating factor (I.F.) is \( e^{\int P(x)dx} \). Calculate \( \int P(x)dx = \int \frac{x}{x^2+1} dx \). Let \(u = x^2+1\), then \(du = 2x dx\), so \(x dx = \frac{1}{2}du\). \[ \int \frac{x}{x^2+1} dx = \int \frac{1}{u} \cdot \frac{1}{2}du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(x^2+1) \] (Since \(x^2+1>0\), absolute value is not needed). This can be written as \( \ln((x^2+1)^{1/2}) = \ln\sqrt{x^2+1} \). So, the Integrating Factor is: \[ I.F. = e^{\ln\sqrt{x^2+1}} = \sqrt{x^2+1} \] This matches option (c). Option (d) \( e^{\log(1+x^2)} \), if \(\log\) is natural log, is \(1+x^2\). \[ \boxed{\sqrt{1+x^2}} \]