Question

Find the general solution of the differential equation \( y \, dx - x \, dy + (x \log x) \, dx = 0 \).

Answer 1

Step 1: Write the differential equation clearly
The given differential equation is:
y dx - x dy + (x log x) dx = 0

Step 2: Rearrange terms
Group dx and dy terms:
(y + x log x) dx - x dy = 0

Step 3: Express in differential form
Rewrite as:
(y + x log x) dx = x dy

Step 4: Divide both sides by x dx
\(\frac{y + x \log x}{x} dx = dy\)
or
dy = \(\frac{y}{x} + \log x\) dx

Step 5: Recognize this as a linear differential equation
It can be written as:
\(\frac{dy}{dx} - \frac{y}{x} = \log x\)

Step 6: Find the integrating factor (IF)
IF = \(e^{-\int \frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}\)

Step 7: Multiply through by the integrating factor
\(\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{\log x}{x}\)
This simplifies to:
\(\frac{d}{dx} \left(\frac{y}{x}\right) = \frac{\log x}{x}\)

Step 8: Integrate both sides
\(\int \frac{d}{dx} \left(\frac{y}{x}\right) dx = \int \frac{\log x}{x} dx\)
\(\frac{y}{x} = \int \frac{\log x}{x} dx + C\)

Step 9: Evaluate the integral
Let \(I = \int \frac{\log x}{x} dx\)
Use substitution \(t = \log x\), so \(dt = \frac{1}{x} dx\)
Therefore, \(I = \int t dt = \frac{t^2}{2} + C_1 = \frac{(\log x)^2}{2} + C_1\)

Step 10: Write the general solution
\(\frac{y}{x} = \frac{(\log x)^2}{2} + C\)
Multiplying both sides by x:
\(y = x \left( \frac{(\log x)^2}{2} + C \right)\)

Final Answer:
\(y = \frac{x (\log x)^2}{2} + C x\)