{"id":811,"date":"2025-06-20T17:15:09","date_gmt":"2025-06-20T17:15:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=811"},"modified":"2025-07-22T15:16:12","modified_gmt":"2025-07-22T15:16:12","slug":"direction-fields-and-eulers-method-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-learn-it-3\/","title":{"raw":"Direction Fields and Euler's Method: Learn It 3","rendered":"Direction Fields and Euler’s Method: Learn It 3"},"content":{"raw":"

Equilibrium Solutions and Their Stability<\/h2>\r\n

Now consider the direction field for the differential equation [latex]y^{\\prime} = (x - 3)(y^2 - 4)[\/latex], shown below. This direction field has several interesting properties that reveal important information about the behavior of solutions.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Direction field for the differential equation y'=(x\u22123)(y2\u22124) showing two solutions. These solutions are very close together, but one is barely above the equilibrium solution y=\u22122 and the other is barely below the same equilibrium solution.[\/caption]\r\n

First, notice the horizontal dashes that appear all the way across the graph at [latex]y = -2[\/latex] and [latex]y = 2[\/latex]. This means that when [latex]y = -2[\/latex], we have [latex]y^{\\prime} = 0[\/latex]. Let's verify this by substituting into the differential equation:<\/p>\r\n

[latex](x - 3)(y^2 - 4) = (x - 3)((-2)^2 - 4) = (x - 3)(0) = 0 = y^{\\prime}[\/latex]<\/p>\r\n

Therefore [latex]y = -2[\/latex] is a solution to the differential equation. Similarly, [latex]y = 2[\/latex] is also a solution.<\/p>\r\n

These are the only constant-valued solutions to this differential equation. Here's why: If [latex]y = k[\/latex] is a constant solution, then [latex]y^{\\prime} = 0[\/latex]. Substituting into our equation gives:<\/p>\r\n

[latex]0 = (x - 3)(k^2 - 4)[\/latex]<\/p>\r\n

Since this must be true for all values of [latex]x[\/latex], we need [latex]k^2 - 4 = 0[\/latex], which gives us [latex]k = -2[\/latex] and [latex]k = 2[\/latex].<\/p>\r\n

These constant solutions are called equilibrium solutions<\/strong> to the differential equation.<\/p>\r\n\r\n

\r\n

equilibrium solutions<\/h3>\r\n

Consider the differential equation [latex]y^{\\prime} = f(x,y)[\/latex]. An equilibrium solution<\/strong> is any solution of the form [latex]y = c[\/latex], where [latex]c[\/latex] is a constant.<\/p>\r\n[latex]\\\\[\/latex]\r\n

To find equilibrium solutions: Set the right-hand side equal to zero and solve for constant values. An equilibrium solution [latex]y = k[\/latex] satisfies [latex]f(x,k) = 0[\/latex] for all values of [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/p>\r\n\r\n<\/section>\r\n

Stability of Equilibrium Solutions<\/h3>\r\n

An important characteristic of equilibrium solutions is whether other solutions approach them as asymptotes for large values of [latex]x[\/latex]. This leads us to classify equilibrium solutions by their stability<\/strong>.<\/p>\r\n\r\n

\r\n

stability classifications<\/h3>\r\n

Consider the differential equation [latex]y^{\\prime} = f(x,y)[\/latex] with equilibrium solution [latex]y = k[\/latex]:<\/p>\r\n\r\n

    \r\n \t
  • Asymptotically Stable<\/strong>: If there exists [latex]\\epsilon > 0[\/latex] such that solutions starting near [latex]k[\/latex] (within [latex]\\epsilon[\/latex]) approach [latex]k[\/latex] as [latex]x \\to \\infty[\/latex].<\/li>\r\n \t
  • Asymptotically Unstable<\/strong>: If there exists [latex]\\epsilon > 0[\/latex] such that solutions starting near [latex]k[\/latex] never approach [latex]k[\/latex] as [latex]x \\to \\infty[\/latex].<\/li>\r\n \t
  • Semi-stable<\/strong>: If the solution is neither stable nor unstable.<\/li>\r\n<\/ul>\r\n<\/section>Let's examine the behavior of solutions to [latex]y^{\\prime} = (x - 3)(y^2 - 4)[\/latex] with initial condition [latex]y(0) = 0.5[\/latex]. The figure below shows the direction field and corresponding solution.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Direction field for the initial-value problem y'=(x\u22123)(y2\u22124),y(0)=0.5.[\/caption]\r\n

    Notice that the solution values stay between [latex]y = -2[\/latex] and [latex]y = 2[\/latex] (our equilibrium solutions). As [latex]x[\/latex] approaches infinity, [latex]y[\/latex] approaches [latex]2[\/latex].<\/p>\r\n

    The behavior is similar if the initial value is higher than [latex]2[\/latex], say [latex]y(0) = 2.3[\/latex]. In this case, solutions decrease and approach [latex]y = 2[\/latex] as [latex]x[\/latex] approaches infinity. Therefore [latex]y = 2[\/latex] is asymptotically stable.<\/p>\r\n

    What happens when the initial value is below [latex]y = -2[\/latex]? The figure below illustrates this scenario with initial value [latex]y(0) = -3[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Direction field for the initial-value problem y'=(x\u22123)(y2\u22124),y(0)=\u22123.[\/caption]\r\n\r\nThe solution decreases rapidly toward negative infinity as [latex]x[\/latex] approaches infinity. If the initial value is slightly higher than [latex]-2[\/latex], the solution approaches [latex]2[\/latex] instead. Since solutions don't approach [latex]y = -2[\/latex] in either case, [latex]y = -2[\/latex] is asymptotically unstable.\r\n\r\n

    \r\n
    \r\n

    Create a direction field for the differential equation [latex]y^{\\prime} ={\\left(y - 3\\right)}^{2}\\left({y}^{2}+y - 2\\right)[\/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n

    \r\n

    The direction field is shown below.<\/p>\r\n\r\n

    <\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Direction field for the differential equation y'=(y\u22123)2(y2+y\u22122).[\/caption]<\/figure>\r\n

    The equilibrium solutions are [latex]y=-2,y=1[\/latex], and [latex]y=3[\/latex]. To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at [latex]y=-2[\/latex] the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to [latex]y=-2[\/latex] approach [latex]y=-2[\/latex], and the solution is stable. For the solution [latex]y=1[\/latex], all initial conditions above and below [latex]y=1[\/latex] are repelled (pushed away) from [latex]y=1[\/latex], so this solution is unstable. The solution [latex]y=3[\/latex] is semi-stable, because for initial conditions slightly greater than [latex]3[\/latex], the solution approaches infinity, and for initial conditions slightly less than [latex]3[\/latex], the solution approaches [latex]y=3[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n
    Analysis<\/strong><\/div>\r\n

    It is possible to find the equilibrium solutions to the differential equation by setting the right-hand side equal to zero and solving for [latex]y[\/latex]. This approach gives the same equilibrium solutions as those we saw in the direction field.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

    Equilibrium Solutions and Their Stability<\/h2>\n

    Now consider the direction field for the differential equation [latex]y^{\\prime} = (x - 3)(y^2 - 4)[\/latex], shown below. This direction field has several interesting properties that reveal important information about the behavior of solutions.<\/p>\n

    \"A
    Direction field for the differential equation y’=(x\u22123)(y2\u22124) showing two solutions. These solutions are very close together, but one is barely above the equilibrium solution y=\u22122 and the other is barely below the same equilibrium solution.<\/figcaption><\/figure>\n

    First, notice the horizontal dashes that appear all the way across the graph at [latex]y = -2[\/latex] and [latex]y = 2[\/latex]. This means that when [latex]y = -2[\/latex], we have [latex]y^{\\prime} = 0[\/latex]. Let’s verify this by substituting into the differential equation:<\/p>\n

    [latex](x - 3)(y^2 - 4) = (x - 3)((-2)^2 - 4) = (x - 3)(0) = 0 = y^{\\prime}[\/latex]<\/p>\n

    Therefore [latex]y = -2[\/latex] is a solution to the differential equation. Similarly, [latex]y = 2[\/latex] is also a solution.<\/p>\n

    These are the only constant-valued solutions to this differential equation. Here’s why: If [latex]y = k[\/latex] is a constant solution, then [latex]y^{\\prime} = 0[\/latex]. Substituting into our equation gives:<\/p>\n

    [latex]0 = (x - 3)(k^2 - 4)[\/latex]<\/p>\n

    Since this must be true for all values of [latex]x[\/latex], we need [latex]k^2 - 4 = 0[\/latex], which gives us [latex]k = -2[\/latex] and [latex]k = 2[\/latex].<\/p>\n

    These constant solutions are called equilibrium solutions<\/strong> to the differential equation.<\/p>\n

    \n

    equilibrium solutions<\/h3>\n

    Consider the differential equation [latex]y^{\\prime} = f(x,y)[\/latex]. An equilibrium solution<\/strong> is any solution of the form [latex]y = c[\/latex], where [latex]c[\/latex] is a constant.<\/p>\n

    [latex]\\\\[\/latex]<\/p>\n

    To find equilibrium solutions: Set the right-hand side equal to zero and solve for constant values. An equilibrium solution [latex]y = k[\/latex] satisfies [latex]f(x,k) = 0[\/latex] for all values of [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/p>\n<\/section>\n

    Stability of Equilibrium Solutions<\/h3>\n

    An important characteristic of equilibrium solutions is whether other solutions approach them as asymptotes for large values of [latex]x[\/latex]. This leads us to classify equilibrium solutions by their stability<\/strong>.<\/p>\n

    \n

    stability classifications<\/h3>\n

    Consider the differential equation [latex]y^{\\prime} = f(x,y)[\/latex] with equilibrium solution [latex]y = k[\/latex]:<\/p>\n

      \n
    • Asymptotically Stable<\/strong>: If there exists [latex]\\epsilon > 0[\/latex] such that solutions starting near [latex]k[\/latex] (within [latex]\\epsilon[\/latex]) approach [latex]k[\/latex] as [latex]x \\to \\infty[\/latex].<\/li>\n
    • Asymptotically Unstable<\/strong>: If there exists [latex]\\epsilon > 0[\/latex] such that solutions starting near [latex]k[\/latex] never approach [latex]k[\/latex] as [latex]x \\to \\infty[\/latex].<\/li>\n
    • Semi-stable<\/strong>: If the solution is neither stable nor unstable.<\/li>\n<\/ul>\n<\/section>\n

      Let’s examine the behavior of solutions to [latex]y^{\\prime} = (x - 3)(y^2 - 4)[\/latex] with initial condition [latex]y(0) = 0.5[\/latex]. The figure below shows the direction field and corresponding solution.<\/p>\n

      \"A
      Direction field for the initial-value problem y’=(x\u22123)(y2\u22124),y(0)=0.5.<\/figcaption><\/figure>\n

      Notice that the solution values stay between [latex]y = -2[\/latex] and [latex]y = 2[\/latex] (our equilibrium solutions). As [latex]x[\/latex] approaches infinity, [latex]y[\/latex] approaches [latex]2[\/latex].<\/p>\n

      The behavior is similar if the initial value is higher than [latex]2[\/latex], say [latex]y(0) = 2.3[\/latex]. In this case, solutions decrease and approach [latex]y = 2[\/latex] as [latex]x[\/latex] approaches infinity. Therefore [latex]y = 2[\/latex] is asymptotically stable.<\/p>\n

      What happens when the initial value is below [latex]y = -2[\/latex]? The figure below illustrates this scenario with initial value [latex]y(0) = -3[\/latex].<\/p>\n

      \"A
      Direction field for the initial-value problem y’=(x\u22123)(y2\u22124),y(0)=\u22123.<\/figcaption><\/figure>\n

      The solution decreases rapidly toward negative infinity as [latex]x[\/latex] approaches infinity. If the initial value is slightly higher than [latex]-2[\/latex], the solution approaches [latex]2[\/latex] instead. Since solutions don’t approach [latex]y = -2[\/latex] in either case, [latex]y = -2[\/latex] is asymptotically unstable.<\/p>\n

      \n
      \n

      Create a direction field for the differential equation [latex]y^{\\prime} ={\\left(y - 3\\right)}^{2}\\left({y}^{2}+y - 2\\right)[\/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.<\/p>\n<\/div>\n