Question

Qual a única resposta correta como solução da ED : dydx=yx+1 ?

Answer 1

$dy=(xy+1)~dx\\ \\ (xy+1)~dx-dy=0\\ \\ (xy+1)u(x,y)~dx-u(x,y)dy=0\\ \\ P(x,y)=(xy+1)u(x,y)~~,~~Q(x,y)=-u(x,y)\\ \\ \to P_y=xu+(xy+1)u_y~~,~~Q_x=-u_x\\ \\ \texttt{conviene: }u=u(x)\\ \\ \to P_y=xu~~,~~Q_x=-u_x\to xu=-u_x=\dfrac{du}{dx}\\ \\ x~dx=-\dfrac{du}{u}\\ \\ \dfrac{x^2}{2}=-\ln u\to u=e^{-x^2/2}$


$\texttt{Ecuaci\'on equivalente }(xy+1)e^{-x^2/2}~dx-e^{-x^2/2}dy=0\\ \\ \\ f_x=(xy+1)e^{-x^2/2}~~,~~f_y=-e^{-x^2/2}\\ \\ f_y=-e^{-x^2/2}\to f(x,y)=-e^{-x^2/2}y+\phi(x)\to f_x=xye^{-x^2/2}+\phi_x\\ \\ \\ (xy+1)e^{-x^2/2}=xye^{-x^2/2}+\phi_x\\ \\ \\ \phi_x=e^{-x^2/2}\\ \\ \displaystyle \phi(x)=\int_{0}^{x}e^{-t^2/2}~dt\\ \\ \\ \texttt{soluci\'on}:\\ \\ e^{-x^2/2}y+\int_{0}^{x}e^{-t^2/2}~dt=C\\ \\ \\ \boxed{\boxed{y=C\cdot e^{x^2/2}-e^{x^2/2}\int_{0}^{x}e^{-t^2/2}~dt}}$