Question

Solve the differential equation \( y e^y dx = (y^3 + 2x e^y) dy \), when \( y(0) = 1 \).

Hint:

• If a differential equation is not linear in \( y \), check if it is linear in \( x \) by treating \( \frac{dx}{dy} \) as the primary derivative.
• Properties of logarithms in exponents are crucial for simplifying Integrating Factors.

Answer 1

Concept:

• Linear differential equation of form \( \frac{dx}{dy} + P(y)x = Q(y) \).
• Integrating Factor \( IF = e^{\int P(y) dy} \).

Step 1: Rewrite the equation in linear form in \( x \)
Divide by \( y e^y dy \):
\( \frac{dx}{dy} = \frac{y^3 + 2x e^y}{y e^y} \)
\( \frac{dx}{dy} = \frac{y^2}{e^y} + \frac{2x}{y} \)
\( \frac{dx}{dy} - \frac{2}{y} x = y^2 e^{-y} \)

Step 2: Find the Integrating Factor
Here \( P(y) = -\frac{2}{y} \).
\( IF = e^{\int -\frac{2}{y} dy} = e^{-2 \ln y} = e^{\ln y^{-2}} = \frac{1}{y^2} \)

Step 3: Find the general solution
\( x \cdot IF = \int Q(y) \cdot IF dy \)
\( x \cdot \frac{1}{y^2} = \int y^2 e^{-y} \cdot \frac{1}{y^2} dy \)
\( \frac{x}{y^2} = \int e^{-y} dy = -e^{-y} + C \)

Step 4: Apply initial conditions
Given \( y(0) = 1 \), so at \( x = 0, y = 1 \):
\( \frac{0}{1} = -e^{-1} + C \implies C = \frac{1}{e} \)
Final solution: \( \frac{x}{y^2} = \frac{1}{e} - e^{-y} \implies x = y^2 \left( \frac{1}{e} - e^{-y} \right) \)