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Find the general solution of the following differential equation: $$y(\cos(x)+\ln(y))+(x+ye^y)y'=0$$

I have not even been able to change this equation to a better form.

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    $\begingroup$ Thanks for your efforts to use MathJax! one helpful hint is that all trig functions, including $\cos$, look nicest when preceded immediately with a back slash: \cos (x). Also, the same holds for $\log, \ln$, and in this case we have \ln(y). $\endgroup$
    – amWhy
    Feb 23, 2018 at 21:22

1 Answer 1

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$$y(\cos(x)+\ln(y))+(x+ye^y)y'=0\\\cos(x)+\ln(y)+\frac{xy'}y+y'e^y=0\\\left(\sin x\right)'+\ln y+x(\ln y)'+\left(e^y\right)'=0\\\left(\sin x+e^y\right)'+(x\ln y)'=0\\\sin x+e^y+x\ln y=c$$where $c$ is a constant. The trick here is to try and rearrange things to give terms which look like the results of chain rule or product rule.

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