Respuesta :

LammettHash
Rewrite the ODE as

[tex]y\log y\,\mathrm dx+(x-\log y)\,\mathrm dy=0\iff y\log y\dfrac{\mathrm dx}{\mathrm dy}+x=\log y[/tex]
[tex]\dfrac{\mathrm dx}{\mathrm dy}=\dfrac1{y\log y}x=\dfrac1y[/tex]

so that it is now linear in [tex]x[/tex]. An integrating factor would

[tex]\mu(y)=\exp\left(\displaystyle\int\frac{\mathrm dy}{y\log y}\right)=e^{\log(\log y)}=\log y[/tex]

Multiply both sides by [tex]\mu(y)[/tex] to get

[tex]\log y\dfrac{\mathrm dx}{\mathrm dy}=\dfrac1yx=\dfrac{\log y}y[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dy}[x\log y]=\dfrac{\log y}y[/tex]
[tex]x\log y=\displaystyle\int\frac{\log y}y\,\mathrm dy[/tex]
[tex]x\log y=\dfrac12\log^2y+C[/tex]
[tex]x=\dfrac12\log y+\dfrac C{\log y}[/tex]