Exercise · ZW-EULER-S-METHOD-007 · ★★☆ Intermediate
With \(y'=-2y\), \(y(0)=1\) and \(h=0.1\), what is \(y_2\)?
Choose one option.
Show answer
- B. \(0.64\)
Explanation
First \(y_1=1+0.1(-2)=0.8\), then \(y_2=0.8+0.1(-1.6)=0.64\). The value \(0.72\) comes from computing the second slope with \(y_0\), and \(0.6\) from subtracting \(0.2\) only once.
Review the concept: Euler's method
Euler's method approximates the solution of an initial value problem \(y'=f(t,y)\), \(y(t_0)=y_0\), advancing in steps of size \(h\) using the slope at the current point.
\[y_{n+1}=y_n+h\,f(t_n,y_n)\]
More practice problems: euler's method
- ZW-EULER-S-METHOD-001 For the problem y'=2t, y(0)=0, what is the Euler formula for y_n+1?
- ZW-EULER-S-METHOD-002 What is the global convergence order of Euler's method?
- ZW-EULER-S-METHOD-003 For y'=y, y(0)=1, what is the value of y_1 with h=0.5?
- ZW-EULER-S-METHOD-004 What is the value of f(t,y) in the problem y'=t+y?
- ZW-EULER-S-METHOD-005 How many steps does Euler need to go from t=0 to t=1 with h=0.25?
- ZW-EULER-S-METHOD-006 Apply one Euler step to y'=t-y, y(0)=2, with h=0.1. What is y_1?
- ZW-EULER-S-METHOD-008 For y'=t^2, y(1)=0, with h=0.5, what is y_1 at t=1.5?
- ZW-EULER-S-METHOD-009 With y'=y, y(0)=1 and h=0.2, what is the approximation y_2 to y(0.4)?