Exercise · ZW-EULER-S-METHOD-008 · ★★☆ Intermediate
For \(y'=t^2\), \(y(1)=0\), with \(h=0.5\), what is \(y_1\) at \(t=1.5\)?
Choose one option.
Show answer
- A. \(0.5\)
Explanation
We have \(y_1=0+0.5\cdot 1^2=0.5\), because the slope is evaluated at \(t_0=1\). The value \(0.625\) uses \(t_1=1.5\) in the slope, while \(0.25\) omits the factor \(h\) and \(1.125\) mixes both errors.
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-009 With y'=y, y(0)=1 and h=0.2, what is the approximation y_2 to y(0.4)?