$$\frac{\mathrm dy}{\mathrm dx} = e^{yx}+\frac yx$$

I'm trying to solve the differential equation. Here is part of my attempt: enter image description here

I know that i have to integrate both sides of the equation but when I try to integrate my answer doesn´t match with the book´s answer.

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – AlexR
    Mar 4, 2015 at 20:56
  • $\begingroup$ Sorry where you read chain rule in fact it was used the product rule. $\endgroup$ Mar 4, 2015 at 20:59
  • 1
    $\begingroup$ Please check out the links I provided and convert the image into MathJax. You can basically use the familiar LaTeX commands, equations starting with $$ and use alignment like this: $$\begin{align*} a & = b \\ \Rightarrow a + c & = b + c \end{align*}$$ $$\begin{align*} a & = b \\ \Rightarrow a + c & = b + c \end{align*}$$ $\endgroup$
    – AlexR
    Mar 4, 2015 at 21:02
  • $\begingroup$ What's the book answer? $\endgroup$
    – science
    Mar 4, 2015 at 21:22
  • $\begingroup$ y=-x Lg*lg*(C/x) $\endgroup$ Mar 4, 2015 at 21:26

1 Answer 1


Given the fact that the book answer you cited is $y(x)=-x~\ln\ln\dfrac Cx~,~$ it follows that you've

misspelled the original equation, meaning that the question is actually supposed to read $y'(x)=$

$=\exp\bigg(\dfrac yx\bigg)+\dfrac yx~,~$ which begs for a substitution of the form $u(x)=\dfrac{y(x)}x~,~$ yielding $(x\cdot u)~'$

$=e^u+u.~$ But $(x\cdot u)~'=x\cdot u'+u\iff x~\dfrac{du}{dx}=e^u\iff\dfrac{dx}x=\dfrac{du}{e^u}~.$ I believe you can

take it from here.

  • $\begingroup$ As I said before I didn´t misspelled the original funcion. It is as in the book. $\endgroup$ Mar 7, 2015 at 21:52
  • $\begingroup$ @ViniciusL.Beserra: If so, then the book itself contains a typographic error. You can see this for yourself by noticing that the solution presented in the book does not fulfill the equation. $\endgroup$
    – Lucian
    Mar 8, 2015 at 1:40
  • $\begingroup$ Thanks Lucian, but in fact I would like someone unmark my question as -1. $\endgroup$ Mar 9, 2015 at 14:20

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