Solve the differential equation xy' + y = ln y'
Solve the differential equation $xy' + y = \ln y'$.
Explanation of Method
Setting \(p=y'\) turns the left side into \(\\frac{d}{dx}(xy)\). Treating \(p\) as the independent variable (using \(dy=p\\,dx\)) converts the equation into a first-order linear ODE for \(x(p)\), which is solved by an integrating factor and then substituted back to give the parametric solution.
Introduce \(p=y'\)
The equation becomes \(xp+y=\\ln p\), i.e. \(y=\\ln p - xp\).
Differentiate with respect to \(p\)
Using \(dy=p\\,dx\) so that \(\\frac{dy}{dp}=p\\frac{dx}{dp}\), differentiating \(y=\\ln p-xp\) with respect to \(p\) gives\n\\\[ p\\frac{dx}{dp} = \\frac1p - x - p\\frac{dx}{dp} \\;\\Longrightarrow\\; 2p\\frac{dx}{dp}+px = \\frac1p. \\\]
Solve the linear equation for \(x(p)\)
In standard form, \(\\frac{dx}{dp}+\\frac{x}{2p}=\\frac{1}{2p^2}\). The integrating factor is \(\\mu=\\sqrt p\), giving \(\\frac{d}{dp}(x\\sqrt p)=\\frac{1}{2p^{3/2}}\). Integrating,\n\\\[ x\\sqrt p = C - \\frac{1}{\\sqrt p} \\;\\Longrightarrow\\; x = \\frac{C\\sqrt p - 1}{p}. \\\]
Recover \(y\)
Substituting into \(y=\\ln p - xp\) gives \(y=\\ln p - C\\sqrt p + 1\).
Independent verification
Parametrizing \(x(p)=\\frac{C\\sqrt p-1}{p}\), \(y(p)=\\ln p - C\\sqrt p+1\) and computing \(dy/dx=(dy/dp)/(dx/dp)\) symbolically confirms it equals \(p\) identically in \(p\) and \(C\), and that \(xp+y-\\ln p \\equiv 0\). Both checks reduce to exactly zero, confirming the parametric solution.
Result
Final answer:
Verified — the source's answer is correct. The general solution in parametric form (parameter \(p=y'>0\)) is \[ \\boxed{x = \\frac{C\\sqrt p - 1}{p}, \\qquad y = \\ln p - C\\sqrt p + 1} \] equivalently the implicit relation \(y+xp=\\ln p\) with \(p=y'\).