Use Euler's method with n = 5 on the interval 0 ≤ t ≤ $\frac { 1 } { 2 }$ to approximate the solution f(t) to $y ^ { \prime } = 3 y , y ( 0 ) = 1$ Is the following the correct answer? $\begin{array} { l }
\mathrm { t } _ { 0 } = 0 ; \quad \mathrm { y } _ { 0 } = 1 \
\mathrm { t } _ { 1 } = 0.1 ; \quad \mathrm { y } _ { 1 } = 1.3 \
\mathrm { t } _ { 2 } = 0.2 ; \quad \mathrm { y } _ { 2 } = 1.69 \
\mathrm { t } 3 = 0.3 ; \quad \mathrm { y } _ { 3 } = 2.197 \
\mathrm { t } _ { 4 } = 0.4 ; \quad \mathrm { y } _ { 4 } = 2.8561 \
5 = 0.5 ; \quad \text { y5 } = 3.71293 \
\end{array}$

Question

Use Euler's method with n = 5 on the interval 0 ≤ t ≤ $\frac { 1 } { 2 }$ to approximate the solution f(t) to $y ^ { \prime } = 3 y , y ( 0 ) = 1$ Is the following the correct answer? $\begin{array} { l } 
\mathrm { t } _ { 0 } = 0 ; \quad \mathrm { y } _ { 0 } = 1 \
\mathrm { t } _ { 1 } = 0.1 ; \quad \mathrm { y } _ { 1 } = 1.3 \
\mathrm { t } _ { 2 } = 0.2 ; \quad \mathrm { y } _ { 2 } = 1.69 \
\mathrm { t } 3 = 0.3 ; \quad \mathrm { y } _ { 3 } = 2.197 \
\mathrm { t } _ { 4 } = 0.4 ; \quad \mathrm { y } _ { 4 } = 2.8561 \
5 = 0.5 ; \quad \text { y5 } = 3.71293 \
\end{array}$

Quizplus · Accepted Answer

Euler's method is a numerical method for approximating the solution to a differential equation. It is a first-order method, which means that it uses the value of the solution at the previous point in time to approximate the value at the current point in time. The formula for Euler's method is:$$y_{n+1} = y_n + h f(t_n, y_n)$$where $h$ is the step size, $t_n$ is the current point in time, and $y_n$ is the value of the solution at $t_n$.In this problem, we are given the differential equation $y ^ { \prime } = 3 y$ and the initial condition $y ( 0 ) = 1$. We are also given the step size $h = 0.1$. Using Euler's method, we can approximate the solution to this differential equation as follows:$$y_1 = y_0 + h f(t_0, y_0) = 1 + 0.1 (3)(1) = 1.3$$$$y_2 = y_1 + h f(t_1, y_1) = 1.3 + 0.1 (3)(1.3) = 1.69$$$$y_3 = y_2 + h f(t_2, y_2) = 1.69 + 0.1 (3)(1.69) = 2.197$$$$y_4 = y_3 + h f(t_3, y_3) = 2.197 + 0.1 (3)(2.197) = 2.8561$$$$y_5 = y_4 + h f(t_4, y_4) = 2.8561 + 0.1 (3)(2.8561) = 3.71293$$Therefore, the correct answer is:$$\begin{array} { l } \mathrm { t } _ { 0 } = 0 ; \quad \mathrm { y } _ { 0 } = 1 \\mathrm { t } _ { 1 } = 0.1 ; \quad \mathrm { y } _ { 1 } = 1.3 \\mathrm { t } _ { 2 } = 0.2 ; \quad \mathrm { y } _ { 2 } = 1.69 \\mathrm { t } 3 = 0.3 ; \quad \mathrm { y } _ { 3 } = 2.197 \\mathrm { t } _ { 4 } = 0.4 ; \quad \mathrm { y } _ { 4 } = 2.8561 \5 = 0.5 ; \quad \text { y5 } = 3.71293 \\end{array}$$

Use Euler's Method with N = 5 on the Interval 12\frac { 1 } { 2 }21​

Use Euler's Method with N = 5 on the Interval $\frac { 1 } { 2 }$