Use Euler's method with step size0.4 to estimatey(2) , where y(x) is the solution of the initial-value problem y'=4x+y^2, y(0)=1.

. y(2)=

Just plug in the values given. Below is a table showing iteration number, (x,y) and next y value. Looks like y(2) = 405.0307

0: (0.0000,1.0000) -> 1.4000
1: (0.4000,1.4000) -> 2.8240
2: (0.8000,2.8240) -> 7.2940
3: (1.2000,7.2940) -> 30.4949
4: (1.6000,30.4949) -> 405.0307
5: (2.0000,405.0307) -> 66028.1757

Looks like y grows pretty fast. As is to be expected, since y' grows like y^2.

Use Euler's method with step size0.4 to estimate y(2) , where y(x) is the solution of the initial-value problem y'=5x+y^2, y(0)=1.

Y(1)=

To use Euler's method, we need to take steps from the initial value and approximate the value of the solution at the desired endpoint. Here's how we can do it step by step:

1. Determine the number of steps needed. In this case, we want to estimate y(2) with a step size of 0.4. So the number of steps is calculated as (2 - 0) / 0.4 = 5.

2. Set up a table to keep track of the steps and the corresponding approximations. We start with the initial value y(0) = 1 and increment x by the step size, which is 0.4.

| x_i | y_i |
|---------|---------|
| 0.0 | 1.0 |
| 0.4 | |
| 0.8 | |
| 1.2 | |
| 1.6 | |
| 2.0 | |

3. Now, let's calculate the approximations using the formula for Euler's method:
y_i+1 = y_i + h * f(x_i, y_i)
where h is the step size, f(x_i, y_i) is the derivative evaluated at (x_i, y_i).

For our problem, the derivative is given as f(x, y) = 4x + y^2.

With the given data, we have:
x_1 = 0.0 + 0.4 = 0.4
y_1 = y_0 + 0.4 * f(x_0, y_0) = 1.0 + 0.4 * (4 * 0.0 + 1.0^2) = 1.4

Continue this calculation for the remaining steps:

x_2 = 0.4 + 0.4 = 0.8
y_2 = y_1 + 0.4 * f(x_1, y_1)

x_3 = 0.8 + 0.4 = 1.2
y_3 = y_2 + 0.4 * f(x_2, y_2)

x_4 = 1.2 + 0.4 = 1.6
y_4 = y_3 + 0.4 * f(x_3, y_3)

x_5 = 1.6 + 0.4 = 2.0
y_5 = y_4 + 0.4 * f(x_4, y_4)

4. Plug the values obtained into the table.

| x_i | y_i |
|---------|---------|
| 0.0 | 1.0 |
| 0.4 | 1.4 |
| 0.8 | ... |
| 1.2 | ... |
| 1.6 | ... |
| 2.0 | ... |

5. Continue steps 3 and 4 until you reach the desired endpoint. In this case, we want to estimate y(2), so we fill in the table until x = 2.0.

| x_i | y_i |
|---------|---------|
| 0.0 | 1.0 |
| 0.4 | 1.4 |
| 0.8 | 1.96 |
| 1.2 | 2.616 |
| 1.6 | 3.429 |
| 2.0 | ... |

6. The estimated value of y(2) is the value of y_i at x = 2.0. From the table, we find y(2) ≈ 3.429.

Therefore, y(2) ≈ 3.429.