Approximating Solutions to Differential Equations with Euler's Method

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SUMMARY

The discussion focuses on using Euler's Method to approximate solutions to the differential equation y' + y = 3 + x with the initial condition y(0) = 1. The participants clarify the iterative formula, yn+1 = yn + f(x0, y0)(x - x0), and demonstrate the calculation for y(0.1) resulting in an approximate value of 1.2. The confusion regarding the variables t and x is addressed, confirming they represent the same variable in this context. The discussion emphasizes the importance of correctly applying the method to obtain accurate approximations.

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  • Understanding of first-order differential equations
  • Familiarity with Euler's Method for numerical approximation
  • Basic algebra for iterative calculations
  • Knowledge of initial value problems
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Homework Statement


y'+y=3+x y(0)=1
(a) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2,
0.3, and 0.4 using the Euler method with h = 0.1.

Homework Equations


yn+1 = yn + f(x0, y0)(x-x0). Adjusting 0 for the next number as we go up

The Attempt at a Solution


I'm not looking to have the whole thing solved I'm just looking to solve the very first one and if I figure it out I can go from there.
So I solved for y which gets me
y= -e^-x +x +2. I checked this solution on wolfram alpha.
y1=y0 + (-e^-x +x +2)*h= 1 +(-1+0+2)*.1=1.1. According the textbook its 1.2. It feels so close.
 
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I think you solved the Diff Eq. and the question asked you to apply the method.
Start with y' = 3+ x - y. I am confused by your mixing of t and x. I will assume that they are supposed to be the same variable.
y(0) = 1, x0 = 0.
Then y(.1) ~ y(0) + (3+x0 -y0)(.1) = 1 + (3-1)*.1 = 1.2.
Next, use y(.1) and x(.1) in the formula to iterate to the approximation for y(.2). And on and on.
 
RUber said:
I think you solved the Diff Eq. and the question asked you to apply the method.
Start with y' = 3+ x - y. I am confused by your mixing of t and x. I will assume that they are supposed to be the same variable.
y(0) = 1, x0 = 0.
Then y(.1) ~ y(0) + (3+x0 -y0)(.1) = 1 + (3-1)*.1 = 1.2.
Next, use y(.1) and x(.1) in the formula to iterate to the approximation for y(.2). And on and on.

Yup, I knew whatever error I made it had to be as obvious as possible. Thank you so much.
 

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