Solve hua kisi se?
:P
xlog_e(\frac{x^y}{e^x})\frac {dy}{dx}=ylog_e(\frac {y^x}{e^y})
-
UP 0 DOWN 0 2 4
4 Answers
Not fully solved ; but tryin' to simplify a bit...
x\{y log(x) - x\}\frac{dy}{dx} = y \{xlog(y)-y\}
y^2 - xy \ log(\frac{y}{x})= x^2 \frac{dy}{dx}
Dividing by "xy" throughout-
\frac{y}{x} - log(\frac{y}{x})= \frac{x}{y} \frac{dy}{dx}
Subs. , y/x =t
t - log(t) = \frac{1}{t} \left( x\frac{dt}{dx} +t \right)
Which is reducing our Expression to -
\frac{dx}{x} = \frac{dt}{t^2 -t-t.log(t)}
Bhai, iske aage toh bas yehi soojh rahaa hai...
log(t) = u,
So, \frac{dx}{x} = \frac{du}{(e^u -u-1)}
Toh, now reduced to an integration Qn --
Find - \int \frac{du}{(e^u -u-1)}
From Soumik's hint
x(ylogx-x)dy/dx=y(xlogy-y)
(logx-x/y)dy/dx=logy-y/x
logx.dy/dx+y/x=logy+(x/y)dy/dx
(d/dx)(ylogx)=d/dx(xlogy)
Now you can solve it