Question

Consider the ordinary differential equation:

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:

\[ \boxed{2} \]