Question

If the differential equation $$ x \, dy + (y + y^2 x) \, dx = 0 $$ with condition \( y = 1 \) at \( x = 1 \), then the solution is: 

• A. \( y = \frac{x}{1 + \log x} \)
• B. \( y = \frac{1 + \log x}{x} \)
• C. \( y = x(1 + \log x) \)
• D. \( y = \frac{1}{x(1 + \log x)} \)

Hint:

For first-order linear differential equations, use the integrating factor \( e^{\int P(x)dx} \).

Answer 1

The Correct Option is D

Step 1: Rewrite the Differential Equation Given: \[ x dy + (y + y^2 x) dx = 0 \] Rewriting in standard form: \[ \frac{dy}{dx} = -\frac{y + y^2 x}{x} \] Factor out \( y \): \[ \frac{dy}{dx} = - y \left(\frac{1 + yx}{x} \right) \] This is a separable differential equation. 

Step 2: Separate the Variables Rearrange: \[ \frac{dy}{y(1 + xy)} = -\frac{dx}{x} \] Use partial fraction decomposition for the left-hand side. Let: \[ \frac{1}{y(1+xy)} = \frac{A}{y} + \frac{B}{1+xy} \] Multiplying both sides by \( y(1+xy) \), we get: \[ 1 = A(1+xy) + By \] Substituting \( y = 0 \), we get \( A = 1 \). For \( y = -\frac{1}{x} \), we get \( B = -\frac{1}{x} \). Thus, rewriting: \[ \frac{1}{y(1+xy)} = \frac{1}{y} - \frac{x}{1+xy} \] 

Step 3: Integrate Both Sides \[ \int \left( \frac{1}{y} - \frac{x}{1+xy} \right) dy = -\int \frac{dx}{x} \] Integrating separately: \[ \ln |y| - \ln |1 + xy| = -\ln |x| + C \] 

Step 4: Simplify the Expression \[ \ln \left| \frac{y}{1+xy} \right| = -\ln |x| + C \] Taking exponentials: \[ \frac{y}{1+xy} = \frac{C}{x} \] Rearrange: \[ y = \frac{C}{x - Cx y} \] Solving for \( y \): \[ y(1 + Cx) = \frac{C}{x} \] \[ y = \frac{C}{x(1 + Cx)} \] 

Step 5: Apply Initial Condition \( y(1) = 1 \) Substituting \( x = 1 \), \( y = 1 \): \[ 1 = \frac{C}{1(1 + C)} \] \[ C + 1 = C \] Solving for \( C \), we find \( C = 1 \). Thus, the final solution is: \[ y = \frac{1}{x(1 + \log x)} \]