Solve the problem $y ^ { \prime } = x ^ { 2 } y ^ { 2 } , y ( 0 ) = 1$ numerically for $y ( 0.2 )$ using $h = 0.1$
A) 1.0
B) 1.001
C) 1.01
D) 1.02
E) 1.002

Question

Quizplus · Accepted Answer

Using Euler's method for numerical integration, we start with the initial condition $y(0) = 1$. The formula for Euler's method is $y_{n+1} = y_n + h f(x_n, y_n)$, where $f(x, y) = x^2y^2$, and $h = 0.1$.1. For the first step, at $x_0 = 0$, $y_0 = 1$, $f(0, 1) = 0^2 \cdot 1^2 = 0$. So, $y_1 = y_0 + 0.1 \cdot 0 = 1$.2. For the second step, at $x_1 = 0.1$, $y_1 = 1$, $f(0.1, 1) = 0.1^2 \cdot 1^2 = 0.01$. So, $y_2 = y_1 + 0.1 \cdot 0.01 = 1 + 0.001 = 1.001$.Therefore, $y(0.2) \approx 1.001$.

Solve the Problem y′=x2y2,y(0)=1y ^ { \prime } = x ^ { 2 } y ^ { 2 } , y ( 0 ) = 1y′=x2y2,y(0)=1

Solve the Problem $y ^ { \prime } = x ^ { 2 } y ^ { 2 } , y ( 0 ) = 1$