# First order non-linear differential equation [closed]

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y+xy^2}{x-5y+y^2 \mathrm{cos}(y)}$$

## closed as off-topic by mickep, Did, user1551, user147263, drhabSep 21 '15 at 11:57

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mickep, Did, user1551, drhab
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• Hi and welcome to Math.SE. It is here mandatory to give your own thoughts and ideas on problems. What have you tried here? Have you solved any similar problems? What methods are you familiar with? If you answer those questions, it will be more likely that you get better help. In this particular example, I guess one would be happy to obtain an equation $f(x,y)=0$ that describes the solutions. – mickep Sep 21 '15 at 6:39
• How do you suggest that we help? – Did Sep 21 '15 at 8:09

Written as $$\frac{dy}{dx} = \frac{y+xy^2}{x-5y+y^2 \cos(y)}$$ the differential equation seems extremely complex because of the simultaneous presence of $y$ and $\cos(y)$.
So, let us reverse it and write it as $$\frac{dx}{dy} = \frac{x-5y+y^2 \cos(y)}{y+xy^2}$$ Now, because of the denominator, let us make a change of variable $x=z-\frac 1 y$ (which makes $y+xy^2=y^2z$), $\frac{dx}{dy}= \frac{dz}{dy}+\frac{1}{y^2}$. So, in terms of $z$, the differential equation write (after minor simplifications) $$\frac{\frac{1}{y^3}+\frac{5}{y}-\cos (y)}{z}+\frac{dz}{dy}=0$$ which seems to be quite simple.