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 differential equation \[ \frac{dy}{dx}+Py=Q, \] the integrating factor is \[ e^{\int P\,dx}. \] Always first divide the equation by the coefficient of \(\frac{dy}{dx}\).

Answer 1

The Correct Option is C

Step 1: Convert the differential equation into standard linear form.
Given equation is \[ (x^2+1)\frac{dy}{dx}+xy=x^3 \] Dividing throughout by \[ x^2+1, \] we get \[ \frac{dy}{dx}+\frac{x}{x^2+1}y=\frac{x^3}{x^2+1} \] This is of the form \[ \frac{dy}{dx}+Py=Q \] where \[ P=\frac{x}{x^2+1} \]

Step 2: Use the formula for integrating factor.
For a linear differential equation, \[ \frac{dy}{dx}+Py=Q, \] the integrating factor is \[ I.F.=e^{\int P\,dx} \] So, \[ I.F.=e^{\int \frac{x}{x^2+1}\,dx} \]

Step 3: Evaluate the integral.
Let \[ u=x^2+1 \] Then, \[ du=2x\,dx \] So, \[ \int \frac{x}{x^2+1}\,dx = \frac{1}{2}\log(x^2+1) \] Therefore, \[ I.F.=e^{\frac{1}{2}\log(x^2+1)} \] \[ I.F.=(x^2+1)^{\frac{1}{2}} \] \[ I.F.=\sqrt{1+x^2} \]

Step 4: Final conclusion.
Hence, the integrating factor is \[ \boxed{\sqrt{1+x^2}} \]