# system of ordinary differential equations

In many problems one would have to find the functions $y_{1}(x)$, $y_{2}(x)$, $\ldots$, $y_{n}(x)$ satisfying the differential equation  system of the form

 $\displaystyle\begin{cases}\frac{dy_{1}}{dx}\;=\;f_{1}(x,y_{1},y_{2},\ldots,y_{% n}),\\ \frac{dy_{2}}{dx}\;=\;f_{2}(x,y_{1},y_{2},\ldots,y_{n}),\\ \ldots\qquad\ldots\qquad\ldots\\ \frac{dy_{n}}{dx}\;=\;f_{n}(x,y_{1},y_{2},\ldots,y_{n}).\end{cases}$ (1)
 $\displaystyle y_{1}(x_{0})=y_{01},\quad y_{2}(x_{0})=y_{02},\;\;\ldots,\;y_{n}% (x_{0})=y_{0n}.$ (2)

The solving procedure may be as follows.

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

 $\frac{d^{2}y_{1}}{dx^{2}}\;=\;\frac{\partial f_{1}}{\partial x}+\frac{\partial f% _{1}}{\partial y_{1}}\frac{dy_{1}}{dx}+\ldots+\frac{\partial f_{1}}{\partial y% _{n}}\frac{dy_{n}}{dx}$

Here one substitutes the derivatives  $\frac{dy_{i}}{dx}$ as they are given by the equations (1), getting the equation of the form

 $\frac{d^{2}y_{1}}{dx^{2}}\;=\;F_{2}(x,y_{1},\ldots,y_{n}).$

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

 $\frac{d^{3}y_{1}}{dx^{3}}\;=\;F_{3}(x,y_{1},\ldots,y_{n}).$

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

 $\displaystyle\begin{cases}\frac{dy_{1}}{dx}\;=\;f_{1}(x,y_{1},y_{2},\ldots,y_{% n}),\\ \frac{d^{2}y_{1}}{dx^{2}}\;=\;F_{2}(x,y_{1},y_{2},\ldots,y_{n}),\\ \ldots\qquad\ldots\qquad\ldots\\ \frac{d^{n}y_{1}}{dx^{n}}\;=\;F_{n}(x,y_{1},y_{2},\ldots,y_{n}).\end{cases}$ (3)

The $n\!-\!1$ first equations (3) determine $y_{2},y_{3},\ldots,y_{n}$ as functions of $x$, $y_{1}$, $y_{1}^{\prime},\ldots,y_{1}^{(n-1)}$:

 $\displaystyle\begin{cases}y_{2}\;=\;\varphi_{2}(x,y_{1},y_{1}^{\prime},\ldots,% y_{1}^{(n-1)}),\\ y_{3}\;=\;\varphi_{3}(x,y_{1},y_{1}^{\prime},\ldots,y_{1}^{(n-1)}),\\ \ldots\qquad\ldots\qquad\ldots\\ y_{n}\;=\;\varphi_{n}(x,y_{1},y_{1}^{\prime},\ldots,y_{1}^{(n-1)}).\end{cases}$ (4)

These expressions of $y_{2},y_{3},\ldots,y_{n}$ are put into the last of the equations (3), and then one has an $n$’th order differential equation for solving $y_{1}$:

 $\displaystyle\frac{d^{n}y_{1}}{dx^{n}}\;=\;\Phi(x,y_{1},y_{1}^{\prime},\ldots,% y_{1}^{(n-1)}).$ (5)

Solving this gives the function

 $y_{1}\;=\;\psi(x,C_{1},C_{2},\ldots,C_{n})}$ (6)

Differentiating this $n\!-\!1$ times yields the derivatives $\frac{dy_{1}}{dx},\frac{d^{2}y_{1}}{dx^{2}},\ldots,\frac{d^{n-1}y_{1}}{dx^{n-1}}$ as functions of $x,C_{1},C_{2},\ldots,C_{n}$.  These derivatives are put into the equations (4), giving the functions $y_{2},y_{3},\ldots,y_{n}$:

 $\displaystyle\begin{cases}y_{2}\;=\;\psi_{2}(x,C_{1},C_{2},\ldots,C_{n}),\\ \ldots\qquad\ldots\qquad\ldots\\ y_{n}\;=\;\psi_{n}(x,C_{1},C_{2},\ldots,C_{n}).\end{cases}$ (7)

In the solution (6) and (7), one has still to consider the initial conditions (2); then the constants $C_{i}$ 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

 $\displaystyle\begin{cases}\frac{dy}{dx}\;=\;\,x\!+\!y\!+\!z,\\ \frac{dz}{dx}\;=\;2x\!-\!4y\!-\!3z\end{cases}$ (8)

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

Differentiation of the first equation with respect to $x$ gives

 $\frac{d^{2}y}{dx^{2}}\;=\;1+\frac{dy}{dx}+\frac{dz}{dx}.$

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

 $\displaystyle\frac{d^{2}y}{dx^{2}}\;=\;3x-3y-2z+1.$ (9)

Into this we put the expression

 $\displaystyle z\;=\;\frac{dy}{dx}\!-\!x\!-\!y$ (10)

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

 $\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+y\;=\;5x\!+\!1$

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

 $\displaystyle y=(C_{1}+C_{2}x)e^{-x}+5x-9,$ (11)

and by (10) this yields

 $\displaystyle z\;=\;(C_{2}-2C_{1}-2C_{2}x)e^{-x}-6x-14.$ (12)

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

 $C_{1}-9\;=\;1,\qquad C_{2}-2C_{1}+14\;=\;0,$

whence  $C_{1}=10$  and  $C_{2}=6$  and thus the particular solution in question is

 $\displaystyle\begin{cases}y\;=\;(6x+10)e^{-x}+5x-9,\\ z\;=\;-(12x+14)e^{-x}-6x+14.\end{cases}$

## References

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

Title system of ordinary differential equations SystemOfOrdinaryDifferentialEquations 2014-03-07 16:37:51 2014-03-07 16:37:51 pahio (2872) pahio (2872) 8 pahio (2872) Topic msc 34A05