yln⁡x−xy′⇔yln⁡x=xdydx⇔dyy=ln⁡xxdx∫dyy=∫ln⁡xxdxln|y|=ln2x2+Celn|y|=eln2x2+C|y|=eln2x2×eC∧eC=c1y=celn2x2

∫ln⁡xxdx Let ln⁡x=u thus, we have dxx=du ∫ln⁡xxdx=∫udu =u22+C =ln2x2+C Therefore, y=celn2x2 is the solution

yln⁡x−xy′⇔yln⁡x=xdydx⇔dyy=ln⁡xxdx∫dyy=∫ln⁡xxdxln|y|=ln2x2+Celn|y|=eln2x2+C|y|=eln2x2×eC∧eC=c1y=celn2x2

∫ln⁡xxdx Let ln⁡x=u thus, we have dxx=du ∫ln⁡xxdx=∫udu =u22+C =ln2x2+C Therefore, y=celn2x2 is the solution

Key to "yln⁡x−xy′=0?"

Juan Hewlett

Answered question

2021-12-13

y \ln x - x y^{'} = 0

Raymond Foley

Beginner2021-12-14Added 39 answers

y \ln x - x y^{'} \Leftrightarrow y \ln x = x \frac{d y}{d x} \Leftrightarrow \frac{d y}{y} = \frac{\ln x}{x} d x

\int \frac{d y}{y} = \int \frac{\ln x}{x} d x

\ln | y | = \frac{\ln^{2} x}{2} + C

e^{\ln | y |} = e^{\frac{\ln^{2} x}{2} + C}

| y | = e^{\frac{\ln^{2} x}{2}} \times e^{C} \land e^{C} = c_{1}

y = c e^{\frac{\ln^{2} x}{2}}

Daniel Cormack

Beginner2021-12-15Added 34 answers

\int \frac{\ln x}{x} d x

Let

\ln x = u

thus, we have

\frac{d x}{x} = d u

\int \frac{\ln x}{x} d x = \int u d u

= \frac{u^{2}}{2} + C

= \frac{\ln^{2} x}{2} + C

Therefore,

y = c e^{\frac{\ln^{2} x}{2}}

is the solution

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