2
$\begingroup$

Solve the following differential equation: $$ (e^x+2\ln y)ydx+xdy=0 $$

It is clear that the equation is not exact. So, I tried to express $y'$: $$ \begin{aligned} e^xy+2y\ln y+xy'=0\iff \begin{cases} \left[ \begin{aligned} &y\equiv0\\ &y\equiv e^{-{1/2}} \end{aligned} \right. \ \ \text{if}\ \ x=0\\ y'=-\frac{e^x}{x}\cdot y-\frac{1}{x}\cdot2y\ln y\ \ \ \text{otherwise} \end{cases} \end{aligned} $$ The problem is that the differential equation seems to be non-linear, and I don't know the ways of solving those. Maybe there's an easier way of solving the initial differential equation?

$\endgroup$
1
  • $\begingroup$ This equation is exact if you mutliply by integrating factor $\mu (x,y)=\dfrac x y$ $\endgroup$ Feb 21, 2020 at 18:30

1 Answer 1

2
$\begingroup$

Maybe there's an easier way of solving the initial differential equation? Yes. This equation becomes exact if you mutliply by integrating factor $$\mu (x,y)=\dfrac x y$$


$$(e^x+2\ln y)ydx+xdy=0$$ Divide by $y$ note that $ d(\ln y )=\dfrac {dy}{y}$ $$(e^x+2\ln y)dx+xd \ln y=0$$ Substitute $u= \ln y$ $$(e^x+2u)dx+xdu=0$$ Multiply by $x$: $$xe^xdx+2uxdx+x^2du=0$$ $$xe^xdx+d(ux^2)=0$$ Integrate. $$xe^x-e^x+x^2 \ln y =C$$

$\endgroup$
4
  • 1
    $\begingroup$ In particular, multiplication by $g:=x$ is mandated by requiring$$0=[ge^x+2gu]_u-(xg)_x=g_u(e^x+2u)+g-xg_x,$$so it helps to take $g_u=0$ and $g_x=g/x$, i.e. $g\propto x$. $\endgroup$
    – J.G.
    Feb 21, 2020 at 18:19
  • $\begingroup$ Yes that's true @J.G. $\endgroup$ Feb 21, 2020 at 18:20
  • 1
    $\begingroup$ Easily comprehensible and neat solution. Thank you! $\endgroup$
    – Bonrey
    Feb 21, 2020 at 19:08
  • $\begingroup$ @Bonrey Thank you... you're welcomed. $\endgroup$ Feb 21, 2020 at 19:14

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.