Answer :
Certainly! To estimate [tex]\( y(1) \)[/tex] using Euler's method for the initial value problem [tex]\( y' = -5x + y^2 \)[/tex] with [tex]\( y(0) = 1 \)[/tex] and a step size [tex]\( h = 0.2 \)[/tex], follow these steps:
### Step-by-Step Solution
1. Initialize the variables:
- Start with [tex]\( x_0 = 0 \)[/tex]
- [tex]\( y_0 = 1 \)[/tex]
- Step size [tex]\( h = 0.2 \)[/tex]
2. Euler's method formula:
- The general formula for the next value in Euler’s method is:
[tex]\[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \][/tex]
where [tex]\( f(x, y) = -5x + y^2 \)[/tex]
3. Perform iterations until [tex]\( x \)[/tex] reaches 1:
- First step (from [tex]\( x_0 = 0 \)[/tex] to [tex]\( x_1 = 0.2 \)[/tex]):
[tex]\[ f(x_0, y_0) = f(0, 1) = -5(0) + 1^2 = 1 \][/tex]
[tex]\[ y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.2 \cdot 1 = 1.2 \][/tex]
- Second step (from [tex]\( x_1 = 0.2 \)[/tex] to [tex]\( x_2 = 0.4 \)[/tex]):
[tex]\[ f(x_1, y_1) = f(0.2, 1.2) = -5(0.2) + 1.2^2 = -1 + 1.44 = 0.44 \][/tex]
[tex]\[ y_2 = y_1 + h \cdot f(x_1, y_1) = 1.2 + 0.2 \cdot 0.44 = 1.2 + 0.088 = 1.288 \][/tex]
- Third step (from [tex]\( x_2 = 0.4 \)[/tex] to [tex]\( x_3 = 0.6 \)[/tex]):
[tex]\[ f(x_2, y_2) = f(0.4, 1.288) = -5(0.4) + 1.288^2 = -2 + 1.658944 = -0.341056 \][/tex]
[tex]\[ y_3 = y_2 + h \cdot f(x_2, y_2) = 1.288 + 0.2 \cdot -0.341056 = 1.288 - 0.0682112 = 1.2197888 \][/tex]
- Fourth step (from [tex]\( x_3 = 0.6 \)[/tex] to [tex]\( x_4 = 0.8 \)[/tex]):
[tex]\[ f(x_3, y_3) = f(0.6, 1.2197888) = -5(0.6) + 1.2197888^2 = -3 + 1.4878534994 = -1.5121465006 \][/tex]
[tex]\[ y_4 = y_3 + h \cdot f(x_3, y_3) = 1.2197888 + 0.2 \cdot -1.5121465006 = 1.2197888 - 0.3024293 = 0.9173592 \][/tex]
- Fifth step (from [tex]\( x_4 = 0.8 \)[/tex] to [tex]\( x_5 = 1.0 \)[/tex]):
[tex]\[ f(x_4, y_4) = f(0.8, 0.9173592) = -5(0.8) + 0.9173592^2 = -4 + 0.8415456482 = -3.1584543518 \][/tex]
[tex]\[ y_5 = y_4 + h \cdot f(x_4, y_4) = 0.9173592 + 0.2 \cdot -3.1584543518 = 0.9173592 - 0.6316908704 = 0.2856777248 \][/tex]
Thus, after applying Euler’s method with a step size of 0.2, the approximate value of [tex]\( y(1) \)[/tex] is:
[tex]\[ y(1) \approx 0.2856777247248985 \][/tex]
This concludes the estimation of [tex]\( y(1) \)[/tex] using Euler's method for the given initial value problem.
### Step-by-Step Solution
1. Initialize the variables:
- Start with [tex]\( x_0 = 0 \)[/tex]
- [tex]\( y_0 = 1 \)[/tex]
- Step size [tex]\( h = 0.2 \)[/tex]
2. Euler's method formula:
- The general formula for the next value in Euler’s method is:
[tex]\[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \][/tex]
where [tex]\( f(x, y) = -5x + y^2 \)[/tex]
3. Perform iterations until [tex]\( x \)[/tex] reaches 1:
- First step (from [tex]\( x_0 = 0 \)[/tex] to [tex]\( x_1 = 0.2 \)[/tex]):
[tex]\[ f(x_0, y_0) = f(0, 1) = -5(0) + 1^2 = 1 \][/tex]
[tex]\[ y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.2 \cdot 1 = 1.2 \][/tex]
- Second step (from [tex]\( x_1 = 0.2 \)[/tex] to [tex]\( x_2 = 0.4 \)[/tex]):
[tex]\[ f(x_1, y_1) = f(0.2, 1.2) = -5(0.2) + 1.2^2 = -1 + 1.44 = 0.44 \][/tex]
[tex]\[ y_2 = y_1 + h \cdot f(x_1, y_1) = 1.2 + 0.2 \cdot 0.44 = 1.2 + 0.088 = 1.288 \][/tex]
- Third step (from [tex]\( x_2 = 0.4 \)[/tex] to [tex]\( x_3 = 0.6 \)[/tex]):
[tex]\[ f(x_2, y_2) = f(0.4, 1.288) = -5(0.4) + 1.288^2 = -2 + 1.658944 = -0.341056 \][/tex]
[tex]\[ y_3 = y_2 + h \cdot f(x_2, y_2) = 1.288 + 0.2 \cdot -0.341056 = 1.288 - 0.0682112 = 1.2197888 \][/tex]
- Fourth step (from [tex]\( x_3 = 0.6 \)[/tex] to [tex]\( x_4 = 0.8 \)[/tex]):
[tex]\[ f(x_3, y_3) = f(0.6, 1.2197888) = -5(0.6) + 1.2197888^2 = -3 + 1.4878534994 = -1.5121465006 \][/tex]
[tex]\[ y_4 = y_3 + h \cdot f(x_3, y_3) = 1.2197888 + 0.2 \cdot -1.5121465006 = 1.2197888 - 0.3024293 = 0.9173592 \][/tex]
- Fifth step (from [tex]\( x_4 = 0.8 \)[/tex] to [tex]\( x_5 = 1.0 \)[/tex]):
[tex]\[ f(x_4, y_4) = f(0.8, 0.9173592) = -5(0.8) + 0.9173592^2 = -4 + 0.8415456482 = -3.1584543518 \][/tex]
[tex]\[ y_5 = y_4 + h \cdot f(x_4, y_4) = 0.9173592 + 0.2 \cdot -3.1584543518 = 0.9173592 - 0.6316908704 = 0.2856777248 \][/tex]
Thus, after applying Euler’s method with a step size of 0.2, the approximate value of [tex]\( y(1) \)[/tex] is:
[tex]\[ y(1) \approx 0.2856777247248985 \][/tex]
This concludes the estimation of [tex]\( y(1) \)[/tex] using Euler's method for the given initial value problem.
