system of ordinary differential equations

In many problems one would have to find the functions$y_1(x)$, $y_2(x)$, $\mathrm{\dots }$, $y_n(x)$ satisfying the differential equationsystem of the form

$\{\begin{array}{cc}\frac{d{y}_1}{dx}=f_1(x,y_1,y_2,\mathrm{\dots },y_n),\hfill & \\ \frac{d{y}_2}{dx}=f_2(x,y_1,y_2,\mathrm{\dots },y_n),\hfill & \\ \mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hfill & \\ \frac{d{y}_n}{dx}=f_n(x,y_1,y_2,\mathrm{\dots },y_n).\hfill & \end{array}$(1)

Such a system is called normal system.  The functions $f_i$ are supposedto be differentiable.  Usually there are alsogiven some initial conditions

$y_1(x_0)=y_{01},y_2(x_0)=y_{02},\mathrm{\dots },y_n(x_0)=y_{0n}.$(2)

The solving procedure may be as follows.

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

$$\frac{d^2{y}_1}{d{x}^2}=\frac{\partial f_1}{\partial x}+\frac{\partial f_1}{\partial y_1}\frac{d{y}_1}{dx}+\mathrm{\dots }+\frac{\partial f_1}{\partial y_n}\frac{d{y}_n}{dx}$$

Here one substitutes the derivatives $\frac{d{y}_i}{dx}$ as theyare given by the equations (1), getting the equation of theform

$$\frac{d^2{y}_1}{d{x}^2}=F_2(x,y_1,\mathrm{\dots },y_n).$$

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

$$\frac{d^3{y}_1}{d{x}^3}=F_3(x,y_1,\mathrm{\dots },y_n).$$

Then one can continue similarly and will finally come to thesystem

$\{\begin{array}{cc}\frac{d{y}_1}{dx}=f_1(x,y_1,y_2,\mathrm{\dots },y_n),\hfill & \\ \frac{d^2{y}_1}{d{x}^2}=F_2(x,y_1,y_2,\mathrm{\dots },y_n),\hfill & \\ \mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hfill & \\ \frac{d^n{y}_1}{d{x}^n}=F_n(x,y_1,y_2,\mathrm{\dots },y_n).\hfill & \end{array}$(3)

The $n-1$ first equations (3) determine $y_2,y_3,\mathrm{\dots },y_n$as functions of $x$, $y_1$, $y_1^{\prime },\mathrm{\dots },y_1^{(n-1)}$:

$\{\begin{array}{cc}{y}_2={\phi }_2(x,y_1,y_1^{\prime },\mathrm{\dots },y_1^{(n-1)}),\hfill & \\ y_3={\phi }_3(x,y_1,y_1^{\prime },\mathrm{\dots },y_1^{(n-1)}),\hfill & \\ \mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hfill & \\ y_n={\phi }_n(x,y_1,y_1^{\prime },\mathrm{\dots },y_1^{(n-1)}).\hfill & \end{array}$(4)

These expressions of $y_2,y_3,\mathrm{\dots },y_n$ are put into the lastof the equations (3), and then one has an $n$’th orderdifferential equation for solving $y_1$:

${\displaystyle \frac{d^n{y}_1}{d{x}^n}}=\mathrm{\Phi }(x,y_1,y_1^{\prime },\mathrm{\dots },y_1^{(n-1)}).$(5)

Solving this gives the function

$y_1=\psi (x,C_1,C_2,\mathrm{\dots },C_n).$(6)

Differentiating this $n-1$ times yields the derivatives$\frac{d{y}_1}{dx},\frac{d^2{y}_1}{d{x}^2},\mathrm{\dots },\frac{d^{n-1}{y}_1}{d{x}^{n-1}}$ as functions of$x,C_1,C_2,\mathrm{\dots },C_n$.  These derivatives are put into theequations (4), giving the functions $y_2,y_3,\mathrm{\dots },y_n$:

$\{\begin{array}{cc}{y}_2={\psi }_2(x,C_1,C_2,\mathrm{\dots },C_n),\hfill & \\ \mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hspace{1em}\hspace{1em}\mathrm{\dots }\hfill & \\ y_n={\psi }_n(x,C_1,C_2,\mathrm{\dots },C_n).\hfill & \end{array}$(7)

In the solution (6) and (7), one has still to consider the initialconditions (2); then the constants $C_i$ of integration attaincertain values.

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

Example.  Solve the functions $y(x)$ and $z(x)$ from the pairof differential equations

$\{\begin{array}{cc}\frac{dy}{dx}=x+y+z,\hfill & \\ \frac{dz}{dx}=\mathrm{\hspace{0.33em}2}x-4y-3z\hfill & \end{array}$(8)

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

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

$$\frac{d^{2}y}{d{x}^2}=\mathrm{\hspace{0.33em}1}+\frac{dy}{dx}+\frac{dz}{dx}.$$

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

${\displaystyle \frac{d^{2}y}{d{x}^2}}=\mathrm{\hspace{0.33em}3}x-3y-2z+1.$(9)

Into this we put the expression

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

gotfrom the first equation (8), obtaing the second order lineardifferential equation

$$\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+y=\mathrm{\hspace{0.33em}5}x+1$$

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

$y=(C_1+C_{2}x)e^{-x}+5x-9,$(11)

and by (10) this yields

$z=(C_2-2C_1-2C_{2}x)e^{-x}-6x-14.$(12)

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

$$C_1-9=\mathrm{\hspace{0.33em}1},C_2-2C_1+14=\mathrm{\hspace{0.33em}0},$$

whence  $C_1=10$  and  $C_2=6$  and thus theparticular solution in question is

$\{\begin{array}{cc}y=(6x+10)e^{-x}+5x-9,\hfill & \\ z=-(12x+14)e^{-x}-6x+14.\hfill & \end{array}$