Question

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

Answer 1

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

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

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

Step 4: Isolate dy
Divide both sides by x:
\(\frac{y + x \log x}{x} dx = dy\)
This can be written as:
dy = \(\frac{y}{x} + \log x\) dx

Step 5: Identify as a linear differential equation
The equation is in the form:
\(\frac{dy}{dx} - \frac{y}{x} = \log x\)

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

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

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

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

Step 10: State the general solution
Substitute the evaluated integral back:
\(\frac{y}{x} = \frac{(\log x)^2}{2} + C\)
Multiply by x to solve for y:
\(y = x \left( \frac{(\log x)^2}{2} + C \right)\)

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