The solution of the differential equation dy/dx + y/x = x^2, subject to the condition y(1) = 1, is:
Correct: A
The differential equation is of the form dy/dx + Py = Q, where P = 1/x and Q = x^2. The integrating factor (IF) is e^(∫P dx) = e^(∫(1/x) dx) = e^(ln x) = x. Multiplying the equation by the IF, we get x(dy/dx) + y = x^3 => d/dx (xy) = x^3. Integrating both sides, we get xy = ∫x^3 dx = x^4/4 + C. Since y(1) = 1, we have (1)(1) = (1)^4/4 + C => 1 = 1/4 + C => C = 3/4. Thus, xy = x^4/4 + 3/4 => y = (x^4 + 3)/(4x).