anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is this the ODE?\[\left(\frac{1}{x+y}+2y(x+1)\right)y'-\frac{y}{x(x+y)}=-y^2\]
anonymous
  • anonymous
yes!
anonymous
  • 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 \[\begin{cases}N=\dfrac{\partial\Psi}{\partial y}\\[1ex]M=\dfrac{\partial\Psi}{\partial x}\end{cases}\]

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anonymous
  • anonymous
The ODE is actually exact, so you can proceed with the method shown here: https://en.wikipedia.org/wiki/Exact_differential_equation
anonymous
  • anonymous
so the method of a total differential?
anonymous
  • anonymous
Right
anonymous
  • 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
anonymous
  • 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.
anonymous
  • 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 !
anonymous
  • anonymous
You're welcome!
anonymous
  • anonymous
i can't solve it :(
anonymous
  • 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 $$\frac{\partial \Phi}{\partial y}=\frac1{x+y}+2y(x+1)\implies \Phi(x,y)=\log(x+y)+(x+1)y^2+f(x)\\\frac{\partial\Phi}{\partial x}=y^2-\frac1x+\frac1{x+y}=\implies \Phi(x,y)=xy^2-\log x+\log(x+y)+g(y)$$
anonymous
  • anonymous
now we can look at those together to determine that our solution is of the form $$\Phi(x,y)=C\\(x+1)y^2+\log(x+y)-\log x=C$$
anonymous
  • 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 \((x+1)y^2+\log(x+y)-\log x=0\)
anonymous
  • anonymous
@OOOPS ?
anonymous
  • anonymous
I'm asking what you mean by this: ". Waiting for you for a long time @oldrin.bataku"
anonymous
  • anonymous
How to get this? Please, explain me \[\Phi(x,y)=C\\(x+1)y^2+\log(x+y)-\log x=C\]
anonymous
  • anonymous
@OOOPS well we have that $$\left[\frac1{x+y}+2y(x+1)\right]dy+\left[y^2-\frac{y}{x(x+y)}\right]dy=0\\d\Phi=0\\\ \implies \Phi(x,y)=C$$
anonymous
  • 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)
anonymous
  • 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
IrishBoy123
  • 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
anonymous
  • anonymous
okay thank a lot all! :)
IrishBoy123
  • IrishBoy123
personally, i think this is a really dodgy/interesting DE it totally blows up around (0,0)
anonymous
  • anonymous
yes there is a reason why our mathematics assistent (doctor in physics) chose that one on the exam :p

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