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Determine ODE again from Slope Field
dgl-08-02
multiple
84
randRangeExclude(-8,8,[0,1]) randRangeExclude(-4,4,[-1,0,1]) -C*A

The slope field of the ODE y' = Ay + {\color{teal}b} is given below.

Determine \color{teal}b.

style({ stroke: "black", strokeWidth: 1.5 }); graphInit({ range: [[-8, 8], [-10,10]], scale: [22, 22], axisArrows: "->", tickStep: 1, }); label([0,C], "\\color{red}C", "left"); // Vektorfeld for(var i = -8; i <= 8; i+=1) { for(var j = -9; j <= 9; j+=1) { var dy = A/7.5* j+ B/7.5; line([i - 0.25, j - dy*0.25], [i + 0.25, j + dy*0.25], { arrows: "", strokeWidth: 1, stroke: "gray" }); } }

x \color{teal}b = B

At a given point (x_0,y_0), the slope field shows a tiny piece of the tangent of a solution of the ODE, on which (x_0,y_0) lies.

The slope of the tangent is determined by the value of the right-hand side of the ODE at the point (x_0,y_0).

At the point \color{red}y_0 = y_\infty = C the tangent slopes are zero.

We solve {\color{red}y'_\infty} = 0= negParens(A) \cdot {\color{red}negParens(C)} + \color{teal}b to find \color{teal}b.

Thus, we get \color{teal}b = B.