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 August 27th, 2018, 08:18 PM #1 Newbie   Joined: Aug 2018 From: Australia Posts: 3 Thanks: 0 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 $0  August 27th, 2018, 09:08 PM #2 Global Moderator Joined: Dec 2006 Posts: 19,700 Thanks: 1804$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$from the solution. Last edited by skipjack; August 27th, 2018 at 10:35 PM.  August 27th, 2018, 10:34 PM #4 Global Moderator Joined: Dec 2006 Posts: 19,700 Thanks: 1804 Oops! Sorry about that... must have been getting sleepy! Thanks from topsquark  August 28th, 2018, 08:12 AM #5 Global Moderator Joined: Dec 2006 Posts: 19,700 Thanks: 1804 If$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$y = 1$. Thanks from topsquark  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: 469 Thanks: 261 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\$. Thanks from topsquark and Goomback
 August 28th, 2018, 11:20 PM #8 Global Moderator   Joined: Dec 2006 Posts: 19,700 Thanks: 1804 Where was this ODE found?

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