system of ordinary differential equations

In many problems one would have to find the functions y1(x), y2(x), , yn(x) satisfying the differential equationMathworldPlanetmath system of the form

{dy1dx=f1(x,y1,y2,,yn),dy2dx=f2(x,y1,y2,,yn),    dyndx=fn(x,y1,y2,,yn). (1)

Such a system is called normal systemPlanetmathPlanetmath.  The functions fi are supposed to be differentiableMathworldPlanetmathPlanetmath.  Usually there are also given some initial conditionsMathworldPlanetmath

y1(x0)=y01,y2(x0)=y02,,yn(x0)=y0n. (2)

The solving procedure may be as follows.

First we differentiate the first equation (1) with respect to the argumentMathworldPlanetmath x:


Here one substitutes the derivativesPlanetmathPlanetmath dyidx as they are given by the equations (1), getting the equation of the form


When one differentiates this equation and makes the substitutions as above, the result has the form


Then one can continue similarly and will finally come to the system

{dy1dx=f1(x,y1,y2,,yn),d2y1dx2=F2(x,y1,y2,,yn),    dny1dxn=Fn(x,y1,y2,,yn). (3)

The n-1 first equations (3) determine y2,y3,,yn as functions of x, y1, y1,,y1(n-1):

{y2=φ2(x,y1,y1,,y1(n-1)),y3=φ3(x,y1,y1,,y1(n-1)),    yn=φn(x,y1,y1,,y1(n-1)). (4)

These expressions of y2,y3,,yn are put into the last of the equations (3), and then one has an n’th order differential equation for solving y1:

dny1dxn=Φ(x,y1,y1,,y1(n-1)). (5)

Solving this gives the function

y1=ψ(x,C1,C2,,Cn). (6)

Differentiating this n-1 times yields the derivatives dy1dx,d2y1dx2,,dn-1y1dxn-1 as functions of x,C1,C2,,Cn.  These derivatives are put into the equations (4), giving the functions y2,y3,,yn:

{y2=ψ2(x,C1,C2,,Cn),    yn=ψn(x,C1,C2,,Cn). (7)

In the solution (6) and (7), one has still to consider the initial conditions (2); then the constants Ci of integration attain certain values.

Remark.  If the system (1) is linear, then also the equation (5) is linear.

Example.  Solve the functions y(x) and z(x) from the pair of differential equations

{dydx=x+y+z,dzdx= 2x-4y-3z (8)

subject to the initial conditions  y(0)=1  and  z(0)=0.

Differentiation of the first equation with respect to x gives

d2ydx2= 1+dydx+dzdx.

Setting to this the first derivatives from (8) turns it into

d2ydx2= 3x-3y-2z+1. (9)

Into this we put the expression

z=dydx-x-y (10)

got from the first equation (8), obtaing the second orderPlanetmathPlanetmath linear differential equation

d2ydx2+2dydx+y= 5x+1

with constant coefficients. The general solution of this last equation is

y=(C1+C2x)e-x+5x-9, (11)

and by (10) this yields

z=(C2-2C1-2C2x)e-x-6x-14. (12)

The initial conditions give from (11) and (12)

C1-9= 1,C2-2C1+14= 0,

whence  C1=10  and  C2=6  and thus the particular solution in question is



  • 1 N. Piskunov: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Teine köide. Viies trükk.  Kirjastus Valgus, Tallinn (1966).

Title system of ordinary differential equations
Canonical name SystemOfOrdinaryDifferentialEquations
Date of creation 2014-03-07 16:37:51
Last modified on 2014-03-07 16:37:51
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 8
Author pahio (2872)
Entry type Topic
Classification msc 34A05