Solve the differential equation #(x y'-y)y=(x+y)^2 e^(-y/x)# ?

Question

1361 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T00:51:50

#log_e x + C = e^(y/x)/(1+y/x)^2#

Making #y(x)= x lambda(x)# and substituting into the differential equation we have

#x^2(xlambda(x)lambda'(x)-(1+lambda(x))^2e^(-lambda(x)))=0#

then we have

#(dlambda)/(dx)=((1+lambda)^2)/(xlambda)e^(-lambda)# which is separable then

#dx/x=(lambda dlambda)/(1+lambda)^2 e^(lambda)#

integrating we have

#log_e x + C= e^(lambda)/(1+lambda)# or

#log_e x + C = e^(y/x)/(1+y/x)^2#