3 years ago
I have it down to
ln(y)+(x/y)y'=y'ln(x)+y/x
The problem is factoring out y', which leads to either
y'=(ln(y)-(y/x))/(ln(x)-(x/y))
or to
y'=((y/x)-ln(y))/((x/y)-ln(x))
Am I missing something?
2 Answers
Best Answer
briannaciamaichelo Staff answered 3 years ago
briannaciamaichelo Staff answered 3 years ago
x^y=y^x -> ylog x=xlog y -> log xdy+(y)/(x)dx=log ydx+(x)/(y)dy ->
(log x-(x)/(y))dy=(log y-(y)/(x))dx -> (dy)/(dx)=(log y-(y)/(x))/(log x-(x)/(y))
Best Answer
Ben Staff answered 3 years ago
Ben Staff answered 3 years ago
Write
e^(yln x)=e^(xln y)
Then
((y)/(x)+y'ln x)e^(yln x)=(ln y+(xy')/(y))e^(xln y)
rearranging, we get
(x^yln x-(x)/(y)y^x)y'=y^xln y-(y)/(x)x^y
Use y^x=x^y and get
y'=(ln y-(y/x))/(ln x-(x/y))