Question

Consider the ordinary differential equation: \[ \frac{1}{2} \frac{dy}{dx} + \frac{y}{x} = 1. { If } y = \frac{2}{3} { at } x = 1, { then the value of } y { at } x = 3 { is } \_\_\_\_\_\_ { (rounded off to the nearest integer).} \]

Hint:

To solve linear first-order differential equations, use the integrating factor method. Multiply through by the integrating factor, integrate, and apply initial conditions to find the constant of integration.

Answer 1

We are given the differential equation: \[ \frac{1}{2} \frac{dy}{dx} + \frac{y}{x} = 1 \] Step 1: Multiply the entire equation by 2 to simplify: \[ \frac{dy}{dx} + \frac{2y}{x} = 2 \] Step 2: Recognize the equation as a linear first-order ODE. This is in the form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] Where \( P(x) = \frac{2}{x} \) and \( Q(x) = 2 \). Step 3: Find the integrating factor: The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = x^2 \] Step 4: Multiply through by the integrating factor: \[ x^2 \frac{dy}{dx} + 2xy = 2x^2 \] Now, the left-hand side is the derivative of \( x^2 y \): \[ \frac{d}{dx} (x^2 y) = 2x^2 \] Step 5: Integrate both sides: \[ x^2 y = \int 2x^2 \, dx = \frac{2x^3}{3} + C \] Thus: \[ y = \frac{2x}{3} + \frac{C}{x^2} \] Step 6: Use the initial condition to find \( C \): We are given that \( y = \frac{2}{3} \) when \( x = 1 \): \[ \frac{2}{3} = \frac{2(1)}{3} + \frac{C}{1^2} \] \[ \frac{2}{3} = \frac{2}{3} + C \] So, \( C = 0 \). Step 7: Find the value of \( y \) at \( x = 3 \): Substitute \( C = 0 \) into the equation for \( y \): \[ y = \frac{2x}{3} \] Now, evaluate at \( x = 3 \): \[ y = \frac{2(3)}{3} = 2 \] Thus, the value of \( y \) at \( x = 3 \) is: \[ \boxed{2} \]