Same steps for differential equations.
1ylnydy=dx∫1ylnydy=∫dx \begin{align*} \frac{1}{y\ln y}dy&=dx\\ \int\frac{1}{y\ln y}dy&=\int dx\\ \end{align*} ylny1dy∫ylny1dy=dx=∫dx
Let u=lnyu=\ln yu=lny, and notice ∫dx=x\displaystyle\int dx=x∫dx=x.
∫1udy=∫dxln∣u∣=x+Cln∣lny∣=x+Cy=keex \begin{align*} \int\frac{1}{u}dy&=\int dx\\ \ln|u|&=x+C\\ \ln|\ln y|&=x+C\\ y&=ke^{e^x} \end{align*} ∫u1dyln∣u∣ln∣lny∣y=∫dx=x+C=x+C=keex
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