Question

The equation of the curve passing through \( (1, 0) \) and which has slope \( \left( 1 + \frac{y}{x} \right) \) at \( (x, y) \), is:

• A. \( y = x e^x \)
• B. \( y = x + \log x \)
• C. \( y = x - \log x \)
• D. \( y = x + 2 \log x \)
• E. \( y = x \log x \)

Hint:

When solving first-order linear differential equations, always look for an integrating factor to simplify the equation. In this case, the equation was reduced using the standard form and the integrating factor \( \frac{1}{x} \).

Answer 1

Answer: (E)

Step 1: Start with the given slope \( \frac{dy}{dx} = 1 + \frac{y}{x} \). 
Step 2: Rearrange the equation to separate variables: \[ \frac{dy}{dx} = 1 + \frac{y}{x} \quad \Rightarrow \quad \frac{dy}{dx} - \frac{y}{x} = 1. \] Step 3: This is a first-order linear differential equation. The standard form is: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \( P(x) = -\frac{1}{x} \) and \( Q(x) = 1 \). 
Step 4: To solve this, find the integrating factor \( I(x) \): \[ I(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}. \] Step 5: Multiply the differential equation by the integrating factor: \[ \frac{1}{x} \left( \frac{dy}{dx} - \frac{y}{x} \right) = \frac{1}{x} \cdot 1 \quad \Rightarrow \quad \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{1}{x}. \] Step 6: Now integrate both sides with respect to \( x \): \[ \int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int \frac{1}{x} dx \quad \Rightarrow \quad \frac{y}{x} = \log x + C. \] Step 7: Solve for \( y \): \[ y = x \log x + Cx. \] Step 8: Now, use the initial condition \( y(1) = 0 \) to find \( C \): \[ 0 = 1 \cdot \log 1 + C \cdot 1 \quad \Rightarrow \quad C = 0. \] Therefore, the solution is: \[ y = x \log x. \]