Problem Statement
Find the general solution of the differential equation: $(x^2 + 1)\frac{dy}{dx} + 2xy = \sqrt{x^2 + 4}$
Verified Solution & Marking Scheme
Rewrite in standard first-order linear form
Divide by $(x^2 + 1)$:
$\frac{dy}{dx} + \left(\frac{2x}{x^2 + 1}\right) y = \frac{\sqrt{x^2 + 4}}{x^2 + 1}$
This is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2x}{x^2 + 1}$ and $Q(x) = \frac{\sqrt{x^2 + 4}}{x^2 + 1}$.
Find the Integrating Factor (I.F.)
$\text{I.F.} = e^{\int P(x) \, dx} = e^{\int \frac{2x}{x^2 + 1} \, dx} = e^{\ln(x^2 + 1)} = x^2 + 1$
Multiply and integrate
The general solution is given by:
$y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$
$y(x^2 + 1) = \int \left( \frac{\sqrt{x^2 + 4}}{x^2 + 1} \right) (x^2 + 1) \, dx + C = \int \sqrt{x^2 + 4} \, dx + C$
Apply standard integral formula
Using $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln|x + \sqrt{x^2 + a^2}|$ with $a = 2$:
$y(x^2 + 1) = \frac{x}{2}\sqrt{x^2 + 4} + 2\ln|x + \sqrt{x^2 + 4}| + C$
$y = \frac{1}{x^2 + 1} \left[ \frac{x}{2}\sqrt{x^2 + 4} + 2\ln|x + \sqrt{x^2 + 4}| + C \right]$