Linear ODE with integrating factor x and y(1) = 1

Question

For $x>0$ , solve $y'+\dfrac{y}{x}=\dfrac{1}{x^2}$ subject to $y(1)=1$ .

Short answer

$y=\dfrac{1+\ln x}{x}$ . Multiplying by the integrating factor $x$ turns the left side into $(xy)'$ , so $(xy)'=1/x$ and $xy=\ln x+C$ ; $y(1)=1$ gives $C=1$ .

Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 3.5 Ordinary Differential Equations, 'First-order linear ODE', p.47; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Separable Equations: Technique and Initial Conditions
The Fundamental Theorem of Calculus
Indefinite Integrals, the Net Change Theorem, and the Substitution Rule