Question

Find the general solution of the differential equation: \( y \log_e y \frac{dx}{dy} + x = \frac{y^2}{2} \).

Hint:

Always double check whether a differential equation is linear in \(y\) or linear in \(x\). If you see single powers of \(x\) isolated alongside complicated expressions of \(y\), it's highly likely to be a linear equation in terms of \(\frac{dx}{dy}\) instead of the standard \(\frac{dy}{dx}\)!

Answer 1

Concept: Let us rearrange this equation into a first-order linear differential equation structure where \(x\) is the dependent variable and \(y\) is the independent parameter. The standard template form is: \[ \frac{dx}{dy} + P(y)x = Q(y) \] The steps to solve are:
• Find the Integrating Factor: \( IF = e^{\int P(y) \, dy} \)
• Multiply and write solution format: \( x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C \)

Step 1: Transform the equation into standard linear form.
The given equation is: \[ y \log_e y \frac{dx}{dy} + x = \frac{y^2}{2} \] Divide the entire equation by the coefficient of \(\frac{dx}{dy}\), which is \( y \log_e y \): \[ \frac{dx}{dy} + \frac{1}{y \log_e y}x = \frac{y^2}{2y \log_e y} = \frac{y}{2 \log_e y} \] Comparing this to the standard form \( \frac{dx}{dy} + P(y)x = Q(y) \), we identify: \[ P(y) = \frac{1}{y \log_e y}, \quad Q(y) = \frac{y}{2 \log_e y} \]

Step 2: Calculate the Integrating Factor (\(IF\)).
\[ IF = e^{\int P(y) \, dy} = e^{\int \frac{1}{y \log_e y} \, dy} \] To evaluate the integral \( \int \frac{1}{y \log_e y} dy \), use substitution \( u = \log_e y \), which gives \( du = \frac{1}{y} dy \): \[ \int \frac{1}{u} \, du = \log_e|u| = \log_e(\log_e y) \] Substituting this back into the power exponent: \[ IF = e^{\log_e(\log_e y)} = \log_e y \]

Step 3: Set up the general solution formula.
The general solution is given by: \[ x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C \] Substitute our values of \(IF\) and \(Q(y)\): \[ x \cdot (\log_e y) = \int \left(\frac{y}{2 \log_e y}\right) \cdot (\log_e y) \, dy + C \] Notice that the term \(\log_e y\) cancels out perfectly in the numerator and denominator of the integrand: \[ x \cdot (\log_e y) = \int \frac{y}{2} \, dy + C \]

Step 4: Perform the final integration.
\[ \int \frac{y}{2} \, dy = \frac{1}{2} \cdot \left(\frac{y^2}{2}\right) = \frac{y^2}{4} \] Thus, the final general solution equation is: \[ x \log_e y = \frac{y^2}{4} + C \]