Exercise · ZW-EULER-S-METHOD-006 · ★★☆ Intermediate
Apply one Euler step to \(y'=t-y\), \(y(0)=2\), with \(h=0.1\). What is \(y_1\)?
Choose one option.
Show answer
- C. \(1.8\)
Explanation
Since \(f(0,2)=0-2=-2\), we get \(y_1=2+0.1\cdot(-2)=1.8\). Using a positive sign gives \(2.2\), and evaluating \(f\) at \(t_1\) gives a different value.
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-007 With y'=-2y, y(0)=1 and h=0.1, what is y_2?
- 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)?