Math: Euler's Method for Differential Equations
Approximating solutions to differential equations step by step
Math: Euler's Method for Differential Equations
Approximating solutions to differential equations step by step
Math - Grade 9-12
- 1
Use Euler's Method with step size h = 0.5 to approximate y(1) for the differential equation dy/dx = x + y, given y(0) = 1.
You need two steps because you are going from x = 0 to x = 1 with h = 0.5.
Starting at (0, 1), the slope is 0 + 1 = 1, so y(0.5) is approximately 1 + 0.5(1) = 1.5. At (0.5, 1.5), the slope is 0.5 + 1.5 = 2, so y(1) is approximately 1.5 + 0.5(2) = 2.5. - 2
Use Euler's Method with step size h = 0.25 to approximate y(0.5) for dy/dx = 2x - y, given y(0) = 1.
At (0, 1), the slope is 2(0) - 1 = -1, so y(0.25) is approximately 1 + 0.25(-1) = 0.75. At (0.25, 0.75), the slope is 2(0.25) - 0.75 = -0.25, so y(0.5) is approximately 0.75 + 0.25(-0.25) = 0.6875. - 3
A table for dy/dx = y - x starts with x = 0 and y = 2. Complete the next two Euler steps using h = 1, and find the approximate value of y(2).
Each step increases x by 1 and updates y using the slope at the start of the interval.
At (0, 2), the slope is 2 - 0 = 2, so y(1) is approximately 2 + 1(2) = 4. At (1, 4), the slope is 4 - 1 = 3, so y(2) is approximately 4 + 1(3) = 7. - 4
Use Euler's Method with h = 0.1 to approximate y(0.2) for dy/dx = x squared + y, given y(0) = 1.
At (0, 1), the slope is 0 squared + 1 = 1, so y(0.1) is approximately 1 + 0.1(1) = 1.1. At (0.1, 1.1), the slope is 0.1 squared + 1.1 = 1.11, so y(0.2) is approximately 1.1 + 0.1(1.11) = 1.211. - 5
For dy/dx = 3 - y with y(0) = 0, use Euler's Method with step size h = 0.5 to approximate y(1.5).
This problem takes three Euler steps because 1.5 divided by 0.5 equals 3.
At (0, 0), the slope is 3, so y(0.5) is approximately 1.5. At (0.5, 1.5), the slope is 1.5, so y(1) is approximately 2.25. At (1, 2.25), the slope is 0.75, so y(1.5) is approximately 2.625. - 6
A student uses Euler's Method for dy/dx = x - 2y with y(0) = 1 and h = 0.5. The student writes y(0.5) = 1 + 0.5(0 - 2) = 0. Is the student's first step correct? Explain.
The student's first step is correct. At the starting point (0, 1), the slope is 0 - 2(1) = -2, so y(0.5) is approximately 1 + 0.5(-2) = 0. - 7
Use the slope field for dy/dx = x + y and the starting point (0, 1). With h = 0.5, draw the first two Euler line segments and state the approximate value of y(1).
Use the slope at the left endpoint of each interval, not the slope at the end of the segment.
The first slope at (0, 1) is 1, so the first segment goes to approximately (0.5, 1.5). The second slope at (0.5, 1.5) is 2, so the next segment goes to approximately (1, 2.5). Therefore, y(1) is approximately 2.5. - 8
Use Euler's Method with h = 0.2 to approximate y(0.4) for dy/dx = y cos(x), given y(0) = 2. Round your final answer to three decimal places.
At (0, 2), the slope is 2 cos(0) = 2, so y(0.2) is approximately 2 + 0.2(2) = 2.4. At (0.2, 2.4), the slope is 2.4 cos(0.2), which is approximately 2.352. Therefore, y(0.4) is approximately 2.4 + 0.2(2.352) = 2.870. - 9
The exact solution of a differential equation is y = e^x, with y(0) = 1. Euler's Method is used on dy/dx = y with h = 0.5 to estimate y(1). Find the Euler estimate and compare it to the exact value e, using e approximately 2.718.
For dy/dx = y, the slope at each step is the current y-value.
At (0, 1), the slope is 1, so y(0.5) is approximately 1.5. At (0.5, 1.5), the slope is 1.5, so y(1) is approximately 2.25. The exact value is about 2.718, so the Euler estimate is 0.468 too low. - 10
Explain why using a smaller step size often gives a better Euler's Method approximation. Use the idea of tangent line segments in your answer.
Think about whether a short straight segment or a long straight segment follows a curve more closely.
A smaller step size often gives a better approximation because each tangent line segment is used for a shorter interval. This usually reduces the amount of error caused by following a straight line while the true solution curve is bending.