Resolviendo una ecuación diferencial con una solución ingeniosa
Problema: Resolver $$ y'=\frac{(x+2y+1)y}{(3x+4y+1)x} $$ Solución: Reescribiendo la ecuación diferencial ordinaria, tenemos que $$(3x^2+4xy+x)dy-(xy+2y^2+y)dx=0$$ $$\implies(3x^{-1}y^2dy-x^{-2}y^3dx)+(4x^{-2}y^3dy-2x^{-3}y^4dx)+x^{-1}y^2\bigg(\frac{xdy-ydx}{x^2}\bigg)=0$$ $$\implies d\bigg(\frac{y^3}{x}\bigg)+d\bigg(\frac{y^4}{x^2}\bigg)+\frac{y^2}{x}d\bigg(\frac{y}{x}\bigg)=0$$ Haciendo, $\frac{y^3}{x}=u$ and $\frac{y^4}{x^2}=v$ gives $$du+dv+\sqrt{v}d\bigg(\frac{v}{u}\bigg)=0$$ $$\implies\frac{d(u+v)}{\sqrt{u+v}}+\frac{\sqrt{v}}{\sqrt{u+v}}d\bigg(\frac{v}{u}\bigg)=0$$ $$\implies\frac{d(u+v)}{\sqrt{u+v}}+\dfrac{\sqrt{\frac{v}{u}}}{\sqrt{1+\frac{v}{u}}}d\bigg(\frac{v}{u}\bigg)=0$$ Integrando, tenemos que $$2\sqrt{u+v}+\frac{\sqrt{v(u+v)}}{u}-\ln\bigg(\frac{\sqrt{v}+\sqrt{u+v}}{\sqrt{u}}\bigg)=C$$ $$\implies\boxed{\bigg(2+\frac{1}{y}\bigg)\sqrt{\frac{y^4}{x^2}+\frac{y^3}{x}}-\ln\bigg(\frac{y^2}{x}+\sqrt{\frac{y^4}{x^2}+\frac{y^3}{x}}\bigg)+\frac{1}{2}\ln\bigg(\frac{y^3}{x}\bigg