Question

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

Answer 1

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

Answer 2

yes!

Answer 3

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
\[\begin{cases}N=\dfrac{\partial\Psi}{\partial y}\\[1ex]M=\dfrac{\partial\Psi}{\partial x}\end{cases}\]

Answer 4

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

Answer 5

so the method of a total differential?

Answer 6

Right

Answer 7

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

Answer 8

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.

Answer 9

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 !

Answer 10

You're welcome!

Answer 11

i can't solve it :(

Answer 12

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

Answer 13

now we can look at those together to determine that our solution is of the form $$\Phi(x,y)=C\\(x+1)y^2+\log(x+y)-\log x=C$$

Answer 14

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 \((x+1)y^2+\log(x+y)-\log x=0\)

Answer 15

@OOOPS ?

Answer 16

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

Answer 17

How to get this? Please, explain me \[\Phi(x,y)=C\\(x+1)y^2+\log(x+y)-\log x=C\]

Answer 18

@OOOPS well we have that $$\left[\frac1{x+y}+2y(x+1)\right]dy+\left[y^2-\frac{y}{x(x+y)}\right]dy=0\\d\Phi=0\\\ \implies \Phi(x,y)=C$$

Answer 19

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)

Answer 20

\[2yx- \frac{ 1 }{ x+y} +g'(y)= \frac{ 1 }{ x+y }+2yx+2y\]

\[g'(y)= \frac{ 2 }{ x+y } +2y\]

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

Answer 21

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

Answer 22

okay thank a lot all! :)

Answer 23

personally, i think this is a really dodgy/interesting DE

it totally blows up around (0,0)

Answer 24

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