Given the differential equation:

dy/dx = y(1+x), y(0)=1,
Use Euler's method with step size .1 to approximate y(.3).

... please check this for me! no one has responded to this question yet.. thanks.

y' = y(1+x), y'(0) = 1(1+0)=1
->the solution has slope 1 at the point
(0,1); x0=0, y0=1, h=0.1, F(x,y)=y(1+x)

y1=y0+h*F(x0,y0)
y1=1 + 0.1(1(1+0))
y1=1.01

y2=1.01 + 0.1(1.01(1+0.1))
y2=1.02111

y3=1.02111 + 0.1(1.02111(1+0.2))
y3=1.1436432

To check whether the calculations are correct, we can double-check by performing the calculations ourselves. Here's how we can use Euler's method to approximate y(.3):

Step 1: Given the differential equation dy/dx = y(1+x), and the initial condition y(0) = 1, we first compute the slope at the initial point (0,1), which is y'(0) = 1(1+0) = 1.

Step 2: We set up the Euler's method by choosing a step size h = 0.1. We start with the initial point (x0, y0) = (0, 1), and we compute the next point using the formula:
y1 = y0 + h * F(x0, y0),
where F(x, y) = y(1+x).

Step 3: Plugging in the values, we have:
y1 = 1 + 0.1*(1(1+0)) = 1 + 0.1*(1) = 1 + 0.1 = 1.1.

Step 4: We continue this process until we reach the desired point. Doing the calculations iteratively, we find:
y2 = 1.1 + 0.1*(1.1(1+0.1)) = 1.1 + 0.1*(1.1*1.1) = 1.1 + 0.1*1.21 ≈ 1.221.

y3 = 1.221 + 0.1*(1.221(1+0.2)) = 1.221 + 0.1*(1.221*1.2) = 1.221 + 0.1*1.4652 ≈ 1.368.

Thus, the approximation of y(0.3) using Euler's method with a step size of 0.1 is approximately 1.368.

Comparing this with your calculations:
y1 = 1.01 (your calculation)
y2 = 1.02111 (your calculation)
y3 = 1.1436432 (your calculation)

It seems that there is an error in your calculations. The correct values for y1, y2, and y3, as shown above, are different from what you obtained.

Therefore, using Euler's method, the approximation for y(0.3) should be around 1.368.