相平面法的画图包括解析法和等倾线法。我使用等倾线法，因为相平面法只使用最高二阶系统，故首先假设一个通用的二阶微分方程：X’’+aX’+bX^2+cX=0因为X’’=（dx’/dx）*（dx/dt）=X’*（dx’/dx）。
代入微分方程得：
X’*（dx’/dx）+ aX’+bX^2+cX=0.
因为斜率k=（dx’/dx）。设X’=y，则上式可化为：
k=-(b*x^2+c*x+a*y)/y
当输入任意一个x1，y1时，则可相应的确定（x1，y1）处的斜率k1，在x1加derta（程序中设定为0.001）确定x2，x2与斜率k所成直线的交点即为y2，则可相应确定k2（如图一）。以此类推，即可以斜率所成的小线段逼近相平面图。
假设输入微分方程为：X’’+X=0(The drawing of phase plane method includes analytic method and equal inclination line method. I use the isocline method, because the phase plane method uses only the highest two order system, so the first assumption of a general two order differential equation: X '' +aX '' +bX^2+cX=0 X 'because = (DX' /dx) * (dx/dt) * =X '(DX' /dx).
By substituting differential equations:
X '* (DX' /dx) + aX '+bX^2+cX=0.
Because the slope is k= (DX '/dx). If X '=y is set, then the upper form can be reduced to..:
K=- (b*x^2+c*x+a*y) /y
When the input to any x1, Y1, can be determined (x1, Y1) at the slope of K1, in X1 and derta (program set to 0.001) to determine the X2, X2 and K to the line of intersection of the slope is Y2, it can determine the corresponding K2 (Figure 1). And so on, that is, the small line segment can be approximated by the slope of the phase plane graph.
Suppose the input differential equation is: X '' +X=0)