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(a) Program your computer algebra system, using Euler's method with step 0.01, to calculate $ y(2), $ where $ y $ is the solution of the initial-value problem

$ y' = x^3 - y^3 y(0) = 1 $

(b) Check your work by using the CAS to draw the solution curve.

a) $f(2)=1.90195$

b) The blue curve in the graph below is the solution curve for the initial condition $y(0)=1$

Differential Equations

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in this problem were asked to solve a differential equation using oilers method. And that differential equation is dy dx equals X to the third minus Y to the third with Y zero equals 1. We have our step size His .01 And we're looking for Y of two. So I wrote a code in Mathematica and this is the output that I got. You can see I have the initial condition the mhm step size in the function. And when I plugged it all in I got About 1.9. So that's what I would say for the first part is that Wife too Is approximately 1.9. And then for the next part it asked to plot the solution curve. So I did that again in Mathematica. And this is the I'll put that I got so at around to get here and when I asked for the specific value, I got again 1.899 which is really close to the value, Wife two is approximately 1.9.