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August 27th, 2018, 08:18 PM   #1
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Question Non-linear ordinary differential equation

Can anyone help me solve the following differential equation?

$\displaystyle \frac{dy}{dx} = \frac{(y-x)(1-y)}{(c-x)(1-2y+x)}$ 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.

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Last edited by skipjack; August 27th, 2018 at 08:21 PM.
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August 27th, 2018, 09:08 PM   #2
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$y = x$ is another obvious solution.
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August 27th, 2018, 09:38 PM   #3
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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.
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August 27th, 2018, 10:34 PM   #4
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Oops! Sorry about that... must have been getting sleepy!
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August 28th, 2018, 08:12 AM   #5
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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 $y = 1$.
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August 28th, 2018, 02:50 PM   #6
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Thanks skipjack. I am looking for a solution with any $c\leq1$ though.
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August 28th, 2018, 06:58 PM   #7
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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
\[y = \sum_{k = 0}^{\infty} c_k(x-x_0)^k\].

2. Assuming $y$ is also analytic, you have an expression for $y'$ as
\[y'(x) = \sum_{k=0}^\infty (k+1)c_{k+1}(x-x_0)^k \].

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 $\{c_0,\dotsc, c_{k-1}$. This uniquely determines the solution on any interval for which the solution is analytic.

6. A more careful analysis is required if $x_0 \approx c$.
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August 28th, 2018, 11:20 PM   #8
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Where was this ODE found?
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