{"id":812,"date":"2025-06-20T17:15:13","date_gmt":"2025-06-20T17:15:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=812"},"modified":"2025-09-10T15:00:19","modified_gmt":"2025-09-10T15:00:19","slug":"direction-fields-and-eulers-method-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-learn-it-4\/","title":{"raw":"Direction Fields and Euler's Method: Learn It 4","rendered":"Direction Fields and Euler’s Method: Learn It 4"},"content":{"raw":"

Euler\u2019s Method<\/h2>\r\n

Sometimes you need to find approximate solutions to differential equations that are difficult or impossible to solve analytically. Euler's Method<\/strong> provides a systematic approach for finding these numerical approximations using the concept of linear approximation.<\/p>\r\n

Let's start with a simple example to see how this works.<\/p>\r\n\r\n

\r\n

Consider the initial-value problem:<\/p>\r\n

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

We can solve this exactly by integrating: [latex]y = x^2 - 3x + C[\/latex].<\/p>\r\n

Using the initial condition gives us [latex]y = x^2 - 3x + 3[\/latex].<\/p>\r\n\r\n<\/section>The solution for this initial-value problem appears as the parabola in the figure below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]\"A Figure 10. Euler\u2019s Method for the initial-value problem [latex]{y}^{\\prime }=2x - 3,y\\left(0\\right)=3[\/latex].[\/caption]The red graph consists of line segments that approximate the solution to the initial-value problem. Notice how the process works:\r\n

    \r\n \t
  • Start at the initial value [latex](0, 3)[\/latex]<\/li>\r\n \t
  • Use the differential equation to find the slope at each point<\/li>\r\n \t
  • Move forward by a fixed step size (here [latex]0.5[\/latex])<\/li>\r\n \t
  • Draw a line segment with that slope<\/li>\r\n<\/ul>\r\nThis approach is the basis of Euler\u2019s Method.\r\n\r\nBefore we state Euler\u2019s Method as a theorem, let\u2019s consider another initial-value problem.\r\n\r\n
    \r\n

    Let's work through a more complex example that we can't solve analytically:<\/p>\r\n

    [latex]y^{\\prime} = x^2 - y^2, \\quad y(-1) = 2[\/latex]<\/p>\r\n

    At the point [latex](-1, 2)[\/latex], the slope of the solution is [latex]y^{\\prime} = (-1)^2 - (2)^2 = -3[\/latex]. We define [latex]x_0 = -1[\/latex] and [latex]y_0 = 2[\/latex]. Since the slope at this point equals [latex]-3[\/latex], we can use linear approximation to approximate [latex]y[\/latex] near [latex](-1, 2)[\/latex]:<\/p>\r\n

    [latex]L(x) = y_0 + f'(x_0)(x - x_0)[\/latex]<\/p>\r\n

    Here [latex]x_0 = -1[\/latex], [latex]y_0 = 2[\/latex], and [latex]f'(x_0) = -3[\/latex], so the linear approximation becomes:<\/p>\r\n

    [latex]L(x) = 2 + (-3)(x - (-1)) = 2 - 3x - 3 = -3x - 1[\/latex]<\/p>\r\n

    Now we choose a step size [latex]h = 0.1[\/latex]. Incrementing [latex]x_0[\/latex] by [latex]h[\/latex] gives our next [latex]x[\/latex] value:<\/p>\r\n

    [latex]x_1 = x_0 + h = -1 + 0.1 = -0.9[\/latex]<\/p>\r\n

    We substitute [latex]x_1 = -0.9[\/latex] into the linear approximation to calculate [latex]y_1[\/latex]:<\/p>\r\n

    [latex]y_1 = L(x_1) = -3(-0.9) - 1 = 1.7[\/latex]<\/p>\r\n

    Therefore the approximate [latex]y[\/latex] value for the solution when [latex]x = -0.9[\/latex] is [latex]y = 1.7[\/latex].<\/p>\r\n

    We repeat the process using [latex]x_1 = -0.9[\/latex] and [latex]y_1 = 1.7[\/latex] to calculate [latex]x_2[\/latex] and [latex]y_2[\/latex].<\/p>\r\n

    The new slope is [latex]y^{\\prime} = (-0.9)^2 - (1.7)^2 = -2.08[\/latex].<\/p>\r\n

    First, [latex]x_2 = x_1 + h = -0.9 + 0.1 = -0.8[\/latex]. Using linear approximation:<\/p>\r\n

    [latex]L(x) = y_1 + f'(x_1)(x - x_1) = 1.7 + (-2.08)(x - (-0.9)) = 1.7 - 2.08x - 1.872 = -2.08x - 0.172[\/latex]<\/p>\r\n

    Finally, we substitute [latex]x_2 = -0.8[\/latex] into the linear approximation:<\/p>\r\n

    [latex]y_2 = L(x_2) = -2.08(-0.8) - 0.172 = 1.492[\/latex]<\/p>\r\n

    Therefore the approximate value of the solution is [latex]y = 1.492[\/latex] when [latex]x = -0.8[\/latex].<\/p>\r\n

    This is the idea behind Euler's Method. Repeating these steps gives us the following approximations:<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]n[\/latex]<\/th>\r\n[latex]x_n[\/latex]<\/th>\r\n[latex]y_n[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex]-1[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex]-0.9[\/latex]<\/td>\r\n[latex]1.7[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex]-0.8[\/latex]<\/td>\r\n[latex]1.492[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]3[\/latex]<\/td>\r\n[latex]-0.7[\/latex]<\/td>\r\n[latex]1.3334[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]4[\/latex]<\/td>\r\n[latex]-0.6[\/latex]<\/td>\r\n[latex]1.2046[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]5[\/latex]<\/td>\r\n[latex]-0.5[\/latex]<\/td>\r\n[latex]1.0955[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]6[\/latex]<\/td>\r\n[latex]-0.4[\/latex]<\/td>\r\n[latex]1.0004[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]7[\/latex]<\/td>\r\n[latex]-0.3[\/latex]<\/td>\r\n[latex]0.9164[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]8[\/latex]<\/td>\r\n[latex]-0.2[\/latex]<\/td>\r\n[latex]0.8414[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]9[\/latex]<\/td>\r\n[latex]-0.1[\/latex]<\/td>\r\n[latex]0.7746[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]10[\/latex]<\/td>\r\n[latex]0[\/latex]<\/td>\r\n[latex]0.7156[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section>
    \r\n

    Euler\u2019s method<\/h3>\r\n

    Consider the initial-value problem<\/p>\r\n

    [latex]y^{\\prime} =f\\left(x,y\\right),y\\left({x}_{0}\\right)={y}_{0}[\/latex].<\/p>\r\n

    To approximate a solution to this problem using Euler\u2019s method, define<\/p>\r\n\r\n

    [latex]\\begin{array}{c}{x}_{n}={x}_{0}+nh\\hfill \\\\ {y}_{n}={y}_{n - 1}+hf\\left({x}_{n - 1},{y}_{n - 1}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n

    Here [latex]h>0[\/latex] represents the step size and [latex]n[\/latex] is an integer, starting with [latex]1[\/latex]. The number of steps taken is counted by the variable [latex]n[\/latex].<\/p>\r\n\r\n<\/section>

    Choosing Step Size<\/strong>[latex]\\\\[\/latex]Typically use [latex]h = 0.1[\/latex] or [latex]h = 0.05[\/latex]. Smaller step sizes give more accurate results but require more computation. The tradeoff between accuracy and efficiency is illustrated below.[caption id=\"\" align=\"aligncenter\" width=\"858\"]\"Two Figure 11. Euler\u2019s method for the initial-value problem [latex]{y}^{\\prime }=2x - 3,y\\left(0\\right)=3[\/latex] with (a) a step size of [latex]h=0.5[\/latex]; and (b) a step size of [latex]h=0.25[\/latex].[\/caption]<\/section>
    \r\n
    \r\n

    Consider the initial-value problem<\/p>\r\n\r\n

    [latex]{y}^{\\prime }={x}^{3}+{y}^{2},y\\left(1\\right)=-2[\/latex].<\/div>\r\n \r\n

    Using a step size of [latex]0.1[\/latex], generate a table with approximate values for the solution to the initial-value problem for values of [latex]x[\/latex] between [latex]1[\/latex] and [latex]2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n

    \r\n

    Start by identifying the value of [latex]h[\/latex], then figure out what [latex]f\\left(x,y\\right)[\/latex] is. Then use the formula for Euler\u2019s Method to calculate [latex]{y}_{1},{y}_{2}[\/latex], and so on.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n

    \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]n[\/latex]<\/th>\r\n[latex]{x}_{n}[\/latex]<\/th>\r\n[latex]{y}_{n}={y}_{n - 1}+hf\\left({x}_{n - 1},{y}_{n - 1}\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex]1[\/latex]<\/td>\r\n[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex]1.1[\/latex]<\/td>\r\n[latex]{y}_{1}={y}_{0}+hf\\left({x}_{0},{y}_{0}\\right)=-1.5[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex]1.2[\/latex]<\/td>\r\n[latex]{y}_{2}={y}_{1}+hf\\left({x}_{1},{y}_{1}\\right)=-1.1419[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]3[\/latex]<\/td>\r\n[latex]1.3[\/latex]<\/td>\r\n[latex]{y}_{3}={y}_{2}+hf\\left({x}_{2},{y}_{2}\\right)=-0.8387[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]4[\/latex]<\/td>\r\n[latex]1.4[\/latex]<\/td>\r\n[latex]{y}_{4}={y}_{3}+hf\\left({x}_{3},{y}_{3}\\right)=-0.5487[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]5[\/latex]<\/td>\r\n[latex]1.5[\/latex]<\/td>\r\n[latex]{y}_{5}={y}_{4}+hf\\left({x}_{4},{y}_{4}\\right)=-0.2442[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]6[\/latex]<\/td>\r\n[latex]1.6[\/latex]<\/td>\r\n[latex]{y}_{6}={y}_{5}+hf\\left({x}_{5},{y}_{5}\\right)=0.0993[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]7[\/latex]<\/td>\r\n[latex]1.7[\/latex]<\/td>\r\n[latex]{y}_{7}={y}_{6}+hf\\left({x}_{6},{y}_{6}\\right)=0.5099[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]8[\/latex]<\/td>\r\n[latex]1.8[\/latex]<\/td>\r\n[latex]{y}_{8}={y}_{7}+hf\\left({x}_{7},{y}_{7}\\right)=1.0272[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]9[\/latex]<\/td>\r\n[latex]1.9[\/latex]<\/td>\r\n[latex]{y}_{9}={y}_{8}+hf\\left({x}_{8},{y}_{8}\\right)=1.7159[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]10[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n[latex]{y}_{10}={y}_{9}+hf\\left({x}_{9},{y}_{9}\\right)=2.6962[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    [ohm_question hide_question_numbers=1]311303[\/ohm_question]<\/section>","rendered":"

    Euler\u2019s Method<\/h2>\n

    Sometimes you need to find approximate solutions to differential equations that are difficult or impossible to solve analytically. Euler’s Method<\/strong> provides a systematic approach for finding these numerical approximations using the concept of linear approximation.<\/p>\n

    Let’s start with a simple example to see how this works.<\/p>\n

    \n

    Consider the initial-value problem:<\/p>\n

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

    We can solve this exactly by integrating: [latex]y = x^2 - 3x + C[\/latex].<\/p>\n

    Using the initial condition gives us [latex]y = x^2 - 3x + 3[\/latex].<\/p>\n<\/section>\n

    The solution for this initial-value problem appears as the parabola in the figure below.<\/p>\n

    \"A
    Figure 10. Euler\u2019s Method for the initial-value problem [latex]{y}^{\\prime }=2x - 3,y\\left(0\\right)=3[\/latex].<\/figcaption><\/figure>\n

    The red graph consists of line segments that approximate the solution to the initial-value problem. Notice how the process works:<\/p>\n

      \n
    • Start at the initial value [latex](0, 3)[\/latex]<\/li>\n
    • Use the differential equation to find the slope at each point<\/li>\n
    • Move forward by a fixed step size (here [latex]0.5[\/latex])<\/li>\n
    • Draw a line segment with that slope<\/li>\n<\/ul>\n

      This approach is the basis of Euler\u2019s Method.<\/p>\n

      Before we state Euler\u2019s Method as a theorem, let\u2019s consider another initial-value problem.<\/p>\n

      \n

      Let’s work through a more complex example that we can’t solve analytically:<\/p>\n

      [latex]y^{\\prime} = x^2 - y^2, \\quad y(-1) = 2[\/latex]<\/p>\n

      At the point [latex](-1, 2)[\/latex], the slope of the solution is [latex]y^{\\prime} = (-1)^2 - (2)^2 = -3[\/latex]. We define [latex]x_0 = -1[\/latex] and [latex]y_0 = 2[\/latex]. Since the slope at this point equals [latex]-3[\/latex], we can use linear approximation to approximate [latex]y[\/latex] near [latex](-1, 2)[\/latex]:<\/p>\n

      [latex]L(x) = y_0 + f'(x_0)(x - x_0)[\/latex]<\/p>\n

      Here [latex]x_0 = -1[\/latex], [latex]y_0 = 2[\/latex], and [latex]f'(x_0) = -3[\/latex], so the linear approximation becomes:<\/p>\n

      [latex]L(x) = 2 + (-3)(x - (-1)) = 2 - 3x - 3 = -3x - 1[\/latex]<\/p>\n

      Now we choose a step size [latex]h = 0.1[\/latex]. Incrementing [latex]x_0[\/latex] by [latex]h[\/latex] gives our next [latex]x[\/latex] value:<\/p>\n

      [latex]x_1 = x_0 + h = -1 + 0.1 = -0.9[\/latex]<\/p>\n

      We substitute [latex]x_1 = -0.9[\/latex] into the linear approximation to calculate [latex]y_1[\/latex]:<\/p>\n

      [latex]y_1 = L(x_1) = -3(-0.9) - 1 = 1.7[\/latex]<\/p>\n

      Therefore the approximate [latex]y[\/latex] value for the solution when [latex]x = -0.9[\/latex] is [latex]y = 1.7[\/latex].<\/p>\n

      We repeat the process using [latex]x_1 = -0.9[\/latex] and [latex]y_1 = 1.7[\/latex] to calculate [latex]x_2[\/latex] and [latex]y_2[\/latex].<\/p>\n

      The new slope is [latex]y^{\\prime} = (-0.9)^2 - (1.7)^2 = -2.08[\/latex].<\/p>\n

      First, [latex]x_2 = x_1 + h = -0.9 + 0.1 = -0.8[\/latex]. Using linear approximation:<\/p>\n

      [latex]L(x) = y_1 + f'(x_1)(x - x_1) = 1.7 + (-2.08)(x - (-0.9)) = 1.7 - 2.08x - 1.872 = -2.08x - 0.172[\/latex]<\/p>\n

      Finally, we substitute [latex]x_2 = -0.8[\/latex] into the linear approximation:<\/p>\n

      [latex]y_2 = L(x_2) = -2.08(-0.8) - 0.172 = 1.492[\/latex]<\/p>\n

      Therefore the approximate value of the solution is [latex]y = 1.492[\/latex] when [latex]x = -0.8[\/latex].<\/p>\n

      This is the idea behind Euler’s Method. Repeating these steps gives us the following approximations:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      [latex]n[\/latex]<\/th>\n[latex]x_n[\/latex]<\/th>\n[latex]y_n[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
      [latex]0[\/latex]<\/td>\n[latex]-1[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n<\/tr>\n
      [latex]1[\/latex]<\/td>\n[latex]-0.9[\/latex]<\/td>\n[latex]1.7[\/latex]<\/td>\n<\/tr>\n
      [latex]2[\/latex]<\/td>\n[latex]-0.8[\/latex]<\/td>\n[latex]1.492[\/latex]<\/td>\n<\/tr>\n
      [latex]3[\/latex]<\/td>\n[latex]-0.7[\/latex]<\/td>\n[latex]1.3334[\/latex]<\/td>\n<\/tr>\n
      [latex]4[\/latex]<\/td>\n[latex]-0.6[\/latex]<\/td>\n[latex]1.2046[\/latex]<\/td>\n<\/tr>\n
      [latex]5[\/latex]<\/td>\n[latex]-0.5[\/latex]<\/td>\n[latex]1.0955[\/latex]<\/td>\n<\/tr>\n
      [latex]6[\/latex]<\/td>\n[latex]-0.4[\/latex]<\/td>\n[latex]1.0004[\/latex]<\/td>\n<\/tr>\n
      [latex]7[\/latex]<\/td>\n[latex]-0.3[\/latex]<\/td>\n[latex]0.9164[\/latex]<\/td>\n<\/tr>\n
      [latex]8[\/latex]<\/td>\n[latex]-0.2[\/latex]<\/td>\n[latex]0.8414[\/latex]<\/td>\n<\/tr>\n
      [latex]9[\/latex]<\/td>\n[latex]-0.1[\/latex]<\/td>\n[latex]0.7746[\/latex]<\/td>\n<\/tr>\n
      [latex]10[\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]0.7156[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n
      \n

      Euler\u2019s method<\/h3>\n

      Consider the initial-value problem<\/p>\n

      [latex]y^{\\prime} =f\\left(x,y\\right),y\\left({x}_{0}\\right)={y}_{0}[\/latex].<\/p>\n

      To approximate a solution to this problem using Euler\u2019s method, define<\/p>\n

      [latex]\\begin{array}{c}{x}_{n}={x}_{0}+nh\\hfill \\\\ {y}_{n}={y}_{n - 1}+hf\\left({x}_{n - 1},{y}_{n - 1}\\right).\\hfill \\end{array}[\/latex]<\/div>\n

      Here [latex]h>0[\/latex] represents the step size and [latex]n[\/latex] is an integer, starting with [latex]1[\/latex]. The number of steps taken is counted by the variable [latex]n[\/latex].<\/p>\n<\/section>\n

      Choosing Step Size<\/strong>[latex]\\\\[\/latex]Typically use [latex]h = 0.1[\/latex] or [latex]h = 0.05[\/latex]. Smaller step sizes give more accurate results but require more computation. The tradeoff between accuracy and efficiency is illustrated below.<\/p>\n
      \"Two
      Figure 11. Euler\u2019s method for the initial-value problem [latex]{y}^{\\prime }=2x - 3,y\\left(0\\right)=3[\/latex] with (a) a step size of [latex]h=0.5[\/latex]; and (b) a step size of [latex]h=0.25[\/latex].<\/figcaption><\/figure>\n<\/section>\n
      \n
      \n

      Consider the initial-value problem<\/p>\n

      [latex]{y}^{\\prime }={x}^{3}+{y}^{2},y\\left(1\\right)=-2[\/latex].<\/div>\n

       <\/p>\n

      Using a step size of [latex]0.1[\/latex], generate a table with approximate values for the solution to the initial-value problem for values of [latex]x[\/latex] between [latex]1[\/latex] and [latex]2[\/latex].<\/p>\n<\/div>\n