2
$\begingroup$

The question is to solve the differential equation given below $$y' = \frac{xy-y^2\ln{y}}{x^2\ln{x}- xy} $$ First I tried simple substitution but that didn't work , then I tried to cross multiply and arrange things in a manner so that I could (maybe ) solve the question , but that didn't help either.

$\endgroup$
  • $\begingroup$ Try seeing if it is exact, and if not see if there is an integrating factor to make it exact. $\endgroup$ – Dave Dec 5 '16 at 15:11
  • $\begingroup$ I dont think its exact and it no way a linear form. $\endgroup$ – Tejus Dec 5 '16 at 15:40
2
$\begingroup$

An addition to the implicit solution of Ng Chung Tak :

If we change $\enspace\displaystyle e^{kxy}=x^x y^y\enspace$ to $\enspace\displaystyle w=z^{1/z}\enspace$ with $\enspace\displaystyle z:=e^{kx}/y\enspace$ and $\enspace\displaystyle w:=x^{xe^{-kx}}\enspace$ then we know that there are solutions only possible for $\enspace\displaystyle 0<w\leq e^{1/e}$ .

If we change $\enspace\displaystyle e^{kxy}=x^x y^y\enspace$ to $\enspace\displaystyle w=ze^z\enspace$ with $\enspace\displaystyle w:=-xe^{-kx}\ln x\enspace$ and $\enspace\displaystyle z:=-kx+\ln y\enspace$ then we can solve $\enspace\displaystyle w=ze^z\enspace$ for $\enspace z\enspace$ with Lamberts W-function .

$\endgroup$
8
$\begingroup$

\begin{align*} \frac{dy}{dx} &= \frac{y(x-y\ln y)}{x(x\ln x-y)} \\ &= \frac{x^2y^2 \left( \dfrac{1}{xy}-\dfrac{\ln y}{x^2} \right)} {x^2y^2 \left( \dfrac{\ln x}{y^2}-\dfrac{1}{xy} \right)} \\ 0 &= \left( \frac{1}{xy}-\frac{\ln y}{x^2} \right)dx+ \left( \frac{1}{xy}-\frac{\ln x}{y^2} \right)dy \\ \frac{\partial}{\partial y} \left( \frac{1}{xy}-\frac{\ln y}{x^2} \right) &= -\frac{1}{xy^2}-\frac{1}{x^2y} \tag{$\partial_y M$} \\ \frac{\partial}{\partial x} \left( \frac{1}{xy}-\frac{\ln x}{y^2} \right) &= -\frac{1}{x^2y}-\frac{1}{xy^2} \tag{$\partial_x N$} \\ 0 &= \left( \frac{\ln y}{x}+\frac{\ln x}{y} \right)' \\ k &= \frac{\ln y}{x}+\frac{\ln x}{y} \\ kxy &= x\ln x+y\ln y \\ e^{kxy} &= x^xy^y \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.