-1
$\begingroup$

Please help me solve this first-order non-linear differential equation.

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "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
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ 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. $\endgroup$ – mickep Sep 21 '15 at 6:39
  • $\begingroup$ How do you suggest that we help? $\endgroup$ – Did Sep 21 '15 at 8:09
4
$\begingroup$

Hint

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.

I am sure that you can take from here.

$\endgroup$

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