2. Suppose we are given a differential equation and initial condition: Then we can approximate the solution to the differential equation by its linearization (which is “close enough” in a short interval about x ). 0 (solution curve) The basis of Euler’s method is to “string together” many linearizations to approximate a curve.

3. Now, let’s specify a new value for the independent variable: If dx is small, then we have a new linearization: From the point, which lies exactly on the solution curve, we have obtained the point, which lies very close to the point on the solution curve.

4. Second Step: We use the point and the slope of the solution curve through. Setting, we use the linearization of the solution curve through to calculate This gives us the next approximation to values along the solution curve. Continue the pattern to find the third approximation: Let’s see this process graphically…

5. Three steps in the Euler approximation to the solution of the initial value problem, True solution curve Euler approximation Error

6. Using Euler’s Method Find the first three approximations using Euler’s method for the initial value problem starting at with We have:

7. Using Euler’s Method Find the first three approximations using Euler’s method for the initial value problem starting at with First Approximation

8. Using Euler’s Method Find the first three approximations using Euler’s method for the initial value problem starting at with Second Approximation

9. Using Euler’s Method Find the first three approximations using Euler’s method for the initial value problem starting at with Third Approximation

10. Using Euler’s Method Use Euler’s method to solve the given initial value problem on the interval starting at with. Compare the approximations to the values of the exact solution. Let me show you a new calculator program!!!

11. xy (Euler)y (exact)Error 0–2–20 0.1–1.8000–1.80480.0048 0.2–1.6100–1.61870.0087 0.3–1.4290–1.44080.0118 0.4–1.2561–1.27030.0142 0.5–1.0905–1.10650.0160 0.6–0.9314–0.94880.0174 0.7–0.7783–0.79660.0183 0.8–0.6305–0.64930.0189 0.9–0.4874–0.50660.0191 1.0–0.3487–0.36790.0192

12. Other Practice Problems Use differentiation and substitution to show that the given function is the exact solution of the given initial value problem. Initial Condition: The same!

13. Other Practice Problems Use analytic methods to find the exact solution of the given initial value problem.

14. Other Practice Problems Use analytic methods to find the exact solution of the given initial value problem. Initial Condition: Solution: or

15. Other Practice Problems Use analytic methods to find the exact solution of the given initial value problem. Initial Condition: Solution: