Question

PORFAVOR AYUDENME CON ESTE EJERCICIO ES LA NUMERO 64 SE LOS PIDOO

Answer 1

$\dfrac{dy}{dx}=\dfrac{e^xy}{e^x+2y}=\dfrac{y}{1+2ye^{-x}}\\\\\\\text{Hacemos cambio de variable}\\ \\z=ye^{-x}\to y=e^xz\\\\dy=e^xz~dx+ e^x~dz\\\\\\\text{De la ecuaci\'on tenemos:}\\\\\dfrac{y}{1+2ye^{-x}}dx=e^xz~dx+ e^x~dz\\\\\\\dfrac{e^{x}z}{1+2z}dx=e^xz~dx+ e^x~dz\\\\\\\dfrac{z}{1+2z}dx=z~dx+ dz\\\\\\-\dfrac{2z^2}{1+2z}dx=dz\\\\\\dx=-\dfrac{1+2z}{2z^2}dz\\\\\\x=\dfrac{1}{z}-2\ln z+C\\\\\\\boxed{x=\dfrac{e^x}{y}-2\ln\dfrac{y}{e^x}-1}$