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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is this the ODE?\[\left(\frac{1}{x+y}+2y(x+1)\right)y'\frac{y}{x(x+y)}=y^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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 \[\begin{cases}N=\dfrac{\partial\Psi}{\partial y}\\[1ex]M=\dfrac{\partial\Psi}{\partial x}\end{cases}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The ODE is actually exact, so you can proceed with the method shown here: https://en.wikipedia.org/wiki/Exact_differential_equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the method of a total differential?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Checking to see if an ODE is exact is usually the goto 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok , 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 !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\left[\frac1{x+y}+2y(x+1)\right]dy+\left[y^2\frac{y}{x(x+y)}\right]dx=0$$ with \(y(1)=0\) now consider $$\frac{y}{x(x+y)}=\frac{x+yx}{x(x+y)}=\frac1x\frac1{x+y}$$ so it follows $$\frac{\partial \Phi}{\partial y}=\frac1{x+y}+2y(x+1)\implies \Phi(x,y)=\log(x+y)+(x+1)y^2+f(x)\\\frac{\partial\Phi}{\partial x}=y^2\frac1x+\frac1{x+y}=\implies \Phi(x,y)=xy^2\log x+\log(x+y)+g(y)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we can look at those together to determine that our solution is of the form $$\Phi(x,y)=C\\(x+1)y^2+\log(x+y)\log x=C$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0imposing \(y(1)=0\) picks out our solution: $$(1+1)\cdot0+\log(1+0)\log(1)=C\\0+00=C\\0=C$$ so our solution is \((x+1)y^2+\log(x+y)\log x=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm asking what you mean by this: ". Waiting for you for a long time @oldrin.bataku"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How to get this? Please, explain me \[\Phi(x,y)=C\\(x+1)y^2+\log(x+y)\log x=C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@OOOPS well we have that $$\left[\frac1{x+y}+2y(x+1)\right]dy+\left[y^2\frac{y}{x(x+y)}\right]dy=0\\d\Phi=0\\\ \implies \Phi(x,y)=C$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2yx \frac{ 1 }{ x+y} +g'(y)= \frac{ 1 }{ x+y }+2yx+2y\] \[g'(y)= \frac{ 2 }{ x+y } +2y\] g(y)= 2 log(x+y) +y² so function(x;y)= xy² logx+3log(x+y) +y²=C inserting y(1)=0 0log1+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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thank a lot all! :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0personally, i think this is a really dodgy/interesting DE it totally blows up around (0,0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes there is a reason why our mathematics assistent (doctor in physics) chose that one on the exam :p
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