Problem Statement
Consider the initial value problem $(x^2 + 1)\frac{dy}{dx} + 2xy = 4x^3$, with $y(0) = 2$.
**(a)** [5 marks] By recognizing an exact derivative or finding an integrating factor, solve the differential equation to find $y$ in terms of $x$.
**(b)** [3 marks] State the domain of the solution and evaluate $\lim_{x \to \infty} \frac{y(x)}{x^2}$.
Verified Solution & Marking Scheme
Part (a) [M1]
Recognizing LHS is product rule: $\frac{d}{dx}[(x^2 + 1)y] = (x^2 + 1)\frac{dy}{dx} + 2xy$
Part (a) [A1]
$\frac{d}{dx}[(x^2 + 1)y] = 4x^3$
Part (a) [M1]
Integrate both sides: $(x^2 + 1)y = \int 4x^3\,dx = x^4 + C$
Part (a) [M1]
Substitute initial condition $y(0) = 2$: $(0 + 1)(2) = 0 + C \implies C = 2$
Part (a) [A1]
$y(x) = \frac{x^4 + 2}{x^2 + 1}$
Part (b) [A1]
Domain: Since $x^2 + 1 \neq 0$ for all real $x$, domain is $x \in \mathbb{R}$.
Part (b) [M1]
$\lim_{x \to \infty} \frac{y}{x^2} = \lim_{x \to \infty} \frac{x^4 + 2}{x^4 + x^2}$
Part (b) [A1]
$= \lim_{x \to \infty} \frac{1 + 2/x^4}{1 + 1/x^2} = 1$