## anonymous one year ago anyone knows how to solve this differential equation? [(1/(x+y))+2y(x+1)]y'-[y/(x(x+y))]=-y² with y(1)=0 thanks

1. anonymous

Is this the ODE?$\left(\frac{1}{x+y}+2y(x+1)\right)y'-\frac{y}{x(x+y)}=-y^2$

2. anonymous

yes!

3. anonymous

You can do some rearranging to write this ODE in the form $N(x,y)\,y'+M(x,y)=0$ where $$\Psi(x,y)=C$$ is the solution you're looking for that satisfies $\begin{cases}N=\dfrac{\partial\Psi}{\partial y}\\[1ex]M=\dfrac{\partial\Psi}{\partial x}\end{cases}$

4. anonymous

The ODE is actually exact, so you can proceed with the method shown here: https://en.wikipedia.org/wiki/Exact_differential_equation

5. anonymous

so the method of a total differential?

6. anonymous

Right

7. anonymous

hmm going to try it tomorrow , now too tired. i was trying bernouilli . but how do you recognize that to know the solution? because i tried bernouilli then i tried to put it in standard form , something like y'+a(x)y=0 but it seemed so hard

8. anonymous

Checking to see if an ODE is exact is usually the go-to method for me if I don't see any nice substitutions right away. The partial derivatives are easy to compute and thus easy to see they're equal. I doubt the Bernoulli/linear approach will work because there's no rearranging you can make to get the appropriate form: $y'=\dfrac{\dfrac{y}{x}-y^2(x+y)}{1+2y(x+1)(x+y)}$ I haven't spent too much time looking at this, so there might be a clever substitution you can make to avoid the exact/total route.

9. anonymous

ok , thanks man. that helps a lot in solving those difficult ones. I'm going to try it tomorrow and see if i can find it ! I keep you updated. Again thanks !

10. anonymous

You're welcome!

11. anonymous

i can't solve it :(

12. anonymous

$$\left[\frac1{x+y}+2y(x+1)\right]dy+\left[y^2-\frac{y}{x(x+y)}\right]dx=0$$ with $$y(1)=0$$ now consider $$\frac{y}{x(x+y)}=\frac{x+y-x}{x(x+y)}=\frac1x-\frac1{x+y}$$ so it follows $$\frac{\partial \Phi}{\partial y}=\frac1{x+y}+2y(x+1)\implies \Phi(x,y)=\log(x+y)+(x+1)y^2+f(x)\\\frac{\partial\Phi}{\partial x}=y^2-\frac1x+\frac1{x+y}=\implies \Phi(x,y)=xy^2-\log x+\log(x+y)+g(y)$$

13. anonymous

now we can look at those together to determine that our solution is of the form $$\Phi(x,y)=C\$$x+1)y^2+\log(x+y)-\log x=C 14. anonymous imposing \(y(1)=0$$ picks out our solution:$$(1+1)\cdot0+\log(1+0)-\log(1)=C\\0+0-0=C\\0=C so our solution is $$(x+1)y^2+\log(x+y)-\log x=0$$

15. anonymous

@OOOPS ?

16. anonymous

I'm asking what you mean by this: ". Waiting for you for a long time @oldrin.bataku"

17. anonymous

How to get this? Please, explain me $\Phi(x,y)=C\$$x+1)y^2+\log(x+y)-\log x=C$ 18. anonymous @OOOPS well we have that \left[\frac1{x+y}+2y(x+1)\right]dy+\left[y^2-\frac{y}{x(x+y)}\right]dy=0\\d\Phi=0\\\ \implies \Phi(x,y)=C 19. anonymous and we found \(\Phi(x,y)=(x+1)y^2+\log(x+y)-\log x$$ because its the 'smallest' solution that satisfies both of those partial derivatives (there are others that differ by a constant term)

20. anonymous

$2yx- \frac{ 1 }{ x+y} +g'(y)= \frac{ 1 }{ x+y }+2yx+2y$ $g'(y)= \frac{ 2 }{ x+y } +2y$ g(y)= 2 log(x+y) +y² so function(x;y)= xy² -logx+3log(x+y) +y²=C inserting y(1)=0 0-log1+3log1=C C= 0 (because log1=0) So my question now is , i did derivate to y at the end. Is this also correct? thank you and the little trick in the beginning is really smart! need to remember that

21. IrishBoy123

you have the wrong sign in your first line, it should be $$+\frac{1}{x+y}$$ which gives $$g'(y) = 2y, \ g(y) = y^2$$ this ties in with the solution posted above

22. anonymous

okay thank a lot all! :)

23. IrishBoy123

personally, i think this is a really dodgy/interesting DE it totally blows up around (0,0)

24. anonymous

yes there is a reason why our mathematics assistent (doctor in physics) chose that one on the exam :p