Find all solutions of \[ xy'+y=x \] for \(x>0\).
Solution
1. The differential equation is equivalent to\[ (xy)'=x. \]
By the product rule, \[ (xy)'=xy'+y. \] Thus the given equation \[ xy'+y=x \] is exactly the displayed derivative identity.
2. Every solution on \((0,\infty)\) has the form\[ \boxed{ y(x)=\frac{x}{2}+\frac{C}{x} } \]for some constant \(C\in\RR\).
Integrating the identity in step 1 on the connected interval \((0,\infty)\) gives \[ xy = \frac{x^2}{2}+C \] for some constant \(C\). Since \(x>0\), division by \(x\) gives \[ y(x)=\frac{x}{2}+\frac{C}{x}. \]
3. Every function displayed in step 2 is a solution.
For \[ y(x)=\frac{x}{2}+\frac{C}{x}, \] one has \[ y'(x)=\frac12-\frac{C}{x^2}. \] Therefore \[ xy'(x)+y(x) = x\left(\frac12-\frac{C}{x^2}\right) + \frac{x}{2}+\frac{C}{x} = x. \]
4. Q.E.D.
Steps 2 and 3 give exactly all solutions on \(x>0\).