User Name Remember Me? Password

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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 $0  August 27th, 2018, 09:08 PM #2 Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259$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$from the solution. Last edited by skipjack; August 27th, 2018 at 10:35 PM. August 27th, 2018, 10:34 PM #4 Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259 Oops! Sorry about that... must have been getting sleepy! Thanks from topsquark August 28th, 2018, 08:12 AM #5 Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259 If$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: 669 Thanks: 440 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: 21,028 Thanks: 2259 Where was this ODE found? Tags differential, equation, nonlinear, ordinary Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mnizami Differential Equations 3 February 19th, 2017 04:45 PM zylo Calculus 5 November 30th, 2016 06:56 PM Abdus Salam Differential Equations 1 April 28th, 2015 06:39 PM Norm850 Differential Equations 2 January 31st, 2012 11:57 PM allison711 Differential Equations 3 February 8th, 2008 08:19 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      