Question

Solve the differential equation: \[ \frac{dy}{dx} + \frac{y}{x} = x^2, \quad y = 1 \text{ when } x = 1 \]

Hint:

Linear differential equations follow \( \frac{dy}{dx} + P(x)y = Q(x) \) with integrating factor \( e^{\int P(x) dx} \).

Answer 1

Step 1: Identify as linear differential equation. Standard form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] where \( P(x) = \frac{1}{x} \) and \( Q(x) = x^2 \). 

Step 2: Compute integrating factor (IF). \[ \mu(x) = e^{\int \frac{1}{x} dx} = e^{\ln x} = x \] Multiply throughout by \( x \): \[ x \frac{dy}{dx} + y = x^3 \] 

Step 3: Solve by integration. \[ \frac{d}{dx} (xy) = x^3 \] \[ xy = \int x^3 dx = \frac{x^4}{4} + C \] 

Step 4: Apply initial condition \( y(1) = 1 \). \[ 1(1) = \frac{1}{4} + C \] \[ C = \frac{3}{4} \] Thus, \[ y = \frac{x^4}{4x} + \frac{3}{4x} \] \[ y = \frac{x^3}{4} + \frac{3}{4x} \]