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August 27th, 2018, 08:18 PM  #1 
Newbie Joined: Aug 2018 From: Australia Posts: 3 Thanks: 0  Nonlinear ordinary differential equation
Can anyone help me solve the following differential equation? $\displaystyle \frac{dy}{dx} = \frac{(yx)(1y)}{(cx)(12y+x)}$ with $x,y \in [0, 1]$ and c is a constant with $0<c\leq1$. I know the obvious solution $y=1$ but I was wondering whether there are any other solutions. Also, I am looking for a generic solution that depends on any $c$ within the specified interval. Thanks. Last edited by skipjack; August 27th, 2018 at 08:21 PM. 
August 27th, 2018, 09:08 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
$y = x$ is another obvious solution.

August 27th, 2018, 09:38 PM  #3 
Newbie Joined: Aug 2018 From: Australia Posts: 3 Thanks: 0 
No it's not. y=x gives $\frac{dy}{dx}=0$ from the equation and $y'=1$ from the solution.
Last edited by skipjack; August 27th, 2018 at 10:35 PM. 
August 27th, 2018, 10:34 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
Oops! Sorry about that... must have been getting sleepy!

August 28th, 2018, 08:12 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
If $c$ = 1, the equation has the "solution" $3(x  1)^2(y  1)^4  2(x  1)^3(y  1)^3 = \text{B}$, where $\text{B}$ is a constant. However, satisfying $x,y \in$[$0,1$] (and allowing $x$ to have any value in that interval) requires that $\text{B} = 0$, which causes that solution to reduce to $y = (2x + 1)/3$ or $y = 1$. 
August 28th, 2018, 02:50 PM  #6 
Newbie Joined: Aug 2018 From: Australia Posts: 3 Thanks: 0 
Thanks skipjack. I am looking for a solution with any $c\leq1$ though.

August 28th, 2018, 06:58 PM  #7 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 
This can be solved (mostly) straightforward by power series expansion. It might seem a bit trickier if you haven't handled rational derivatives before but its actually no harder than the usual case. A sketch for the general method follows. 1. Given a DE of the form $y'(x) = \frac{f(x,y)}{g(x,y)}$ where $f,g$ are explicitly known and analytic (i.e. have a known power series expansion. Begin by taking an ansatz of the form \[y = \sum_{k = 0}^{\infty} c_k(xx_0)^k\]. 2. Assuming $y$ is also analytic, you have an expression for $y'$ as \[y'(x) = \sum_{k=0}^\infty (k+1)c_{k+1}(xx_0)^k \]. 3. Plugging $y,y'$ into the DE and multiplying by $g$, you have the equality on the level of power series \[y'(x)g(x,y) = f(x,y) \] where we note that every term shown here can be expressed as a power series in $x$ since we know $f,g$ explicitly. 4. Plug in the initial data constraint, $y = y(x_0) = y_0$, to obtain $c_0 = y_0$. 5. Once the initial data is specified, for each $k \geq 1$, $c_k$ can be isolated in the equation from (3) to solve for it recursively in terms of $\{c_0,\dotsc, c_{k1}$. This uniquely determines the solution on any interval for which the solution is analytic. 6. A more careful analysis is required if $x_0 \approx c$. 
August 28th, 2018, 11:20 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
Where was this ODE found?


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differential, equation, nonlinear, ordinary 
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