Differential equations: y’= sin(cx) sin(yx), y(0)=1

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Discussion Overview

The discussion centers around the non-linear first order ordinary differential equation (ODE) given by y’= sin(cx) sin(yx) with the initial condition y(0)=1. Participants explore potential methods for solving or approximating this equation, including the use of software tools like Mathematica and MATLAB.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant identifies the equation as a non-linear first order ODE and expresses uncertainty about potentially missing details in its formulation.
  • Another participant provides a method involving Taylor series expansion, calculating derivatives at x=0 and suggesting that further derivatives could yield a power series approximation for y.
  • A third participant shares their experience with Mathematica, noting that symbolic solutions are not feasible and opting for a numerical solution using NDSolve, while also acknowledging the limitations of Mathematica in handling the problem.
  • A later reply expresses enthusiasm about the Taylor expansion approach and indicates a willingness to explore it further.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a definitive solution method, as multiple approaches are discussed, including Taylor series and numerical methods, with no agreement on which is preferable or more effective.

Contextual Notes

Some limitations include the complexity of the derivatives involved in the Taylor series expansion and the challenges faced by Mathematica in finding a symbolic solution. The discussion does not resolve these mathematical challenges.

Faema
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It’s a non-linear first order ode, the trig form of which is y’= sin(cx) sin(yx), y(0)=1. It’s possible that I’ve forgotten an important detail in the construction of the equation. If there’s someone who has some ideas about solving or approximating this, or who knows how to get mathematica or MATLAB to do something useful with it, I would be so grateful. Thank you in advance!
 
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Here's one way. Since y(0)= 1 and y'(x)= sin(cx)sin(xy), y'(0)= sin(0)sin(0)= 0.
Differentiating, y''(x)= ccos(cx)sin(xy)+ ysin(cx)cos(xy) and so y''(0)= 0. Differentiating again, y'''(x)= -c^2sin(cx)sin(xy)+ 2cycos(cx)cox(xy)- y^2sin(cx)sin(xy) and y'''(0)= 2cy(0)= 2c. So far, we have, for the Taylor's series about x=0, y(x)= 1+ 0x+ 0x^2+ 2c/3! x^3= 1+ (c/3)x^3. Although, obviously, the calculations are going to get more complicated, continuing in that way, you will have a power series approximatiom for y.
 
Faema said:
It’s a non-linear first order ode, the trig form of which is y’= sin(cx) sin(yx), y(0)=1. It’s possible that I’ve forgotten an important detail in the construction of the equation. If there’s someone who has some ideas about solving or approximating this, or who knows how to get mathematica or MATLAB to do something useful with it, I would be so grateful. Thank you in advance!

This is what I do: give it a few minutes of a little of this, little of that myself. Ok, no go. Then I shamelessly turn to Mathematica's DSolve which attempts to find a symbolic solution. Still no go. I'm feelin' good by now cus' Mathematica can't do it neither. Finally I stop trying to be a purist and just turn to Mathematica's NDSolve which computes a numeric solution of the IVP:

myc=1;
mysol=NDSolve[{y'[x]==Sin[myc x] Sin[y[x] x],y[0]=0},y,{x,0,1}];
Plot[y[x]/.mysol,{x,0,1}]

. . . yeah, I know. I'm not proud. Hall's got a good idea up there. If you're a Mathematica junkie, it wouldn't be hard to get maybe say 20-30 of those derivatives, maybe more, before Mathematica starts choking. Then form the series and compare it to the numerical results. Be good practice for when you really need to do something with the software.
 
Last edited:
This looks promising! Thankyou! Looking into a Taylor expansion..
 

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