Given the differential equation:
dy/dx = y(1+x), y(0)=1,
Use Euler's method with step size .1 to approximate y(.3).
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
Is this the answer or am I going about this the wrong way? If so, please steer me in the right direction. Thanks.
You are on the right track! Euler's method is a numerical method used to approximate the solution of a differential equation.
In Euler's method, we start with an initial condition (in this case, y(0) = 1) and use the slope at that point to approximate the value of the function at the next point. We then repeat this process iteratively to approximate the value at subsequent points.
The formula for Euler's method is:
y(i+1) = y(i) + h * F(x(i), y(i))
where i is the index of the current point, h is the step size, F is the function representing the slope, and x(i) and y(i) are the coordinates of the current point.
Applying this to your differential equation, we have:
F(x, y) = y(1 + x)
Using the given data, let's proceed step by step:
Step 1:
We have x0 = 0, y0 = 1, and h = 0.1.
Using the formula, we can calculate:
y1 = y0 + h * F(x0, y0)
= 1 + 0.1 * (1 + 0 * 1)
= 1 + 0.1 * 1
= 1 + 0.1
= 1.1
Step 2:
Now we have x1 = 0.1, y1 = 1.1, and h = 0.1.
Using the formula, we can calculate:
y2 = y1 + h * F(x1, y1)
= 1.1 + 0.1 * (1.1 * (1 + 0.1))
= 1.1 + 0.1 * (1.1 * 1.1)
= 1.1 + 0.1 * 1.21
= 1.1 + 0.121
= 1.221
Step 3:
Now we have x2 = 0.2, y2 = 1.221, and h = 0.1.
Using the formula, we can calculate:
y3 = y2 + h * F(x2, y2)
= 1.221 + 0.1 * (1.221 * (1 + 0.2))
= 1.221 + 0.1 * (1.221 * 1.2)
= 1.221 + 0.1 * 1.4652
= 1.221 + 0.14652
= 1.36752
So, the approximate value of y(0.3) using Euler's method with a step size of 0.1 is 1.36752.
Keep in mind that Euler's method introduces some error, and using a smaller step size (h) will result in a more accurate approximation.