Exercise · ZW-EULER-S-METHOD-009 · ★★☆ Intermediate
With \(y'=y\), \(y(0)=1\) and \(h=0.2\), what is the approximation \(y_2\) to \(y(0.4)\)?
Choose one option.
Show answer
- D. \(1.44\)
Explanation
Since \(y_1=1.2\) and \(y_2=1.2+0.2\cdot 1.2=1.44\), the answer is \(1.44\). The value \(1.4\) appears when the initial slope is added twice, and \(1.2\) corresponds only to the first step.
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-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?