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Mathematics
Differential Equations
Understanding Euler Method: Finding Initial Condition of y(0)=1
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[QUOTE="chwala, post: 6854494, member: 287397"] [B]TL;DR Summary:[/B] Kindly see the reference below; [URL]https://math.libretexts.org/Courses/Monroe_Community_College/MTH_225_Differential_Equations/3%3A_Numerical_Methods/3.1%3A_Euler's_Method[/URL] My interest is on Example 3.1.1 How did they determine the initial condition ##y(0)=1##or is it picked out of convenience? why not ##y(0)=0.95## or ##y(0)=0.99## The Euler method is straightforward to me; i.e ##y_{n+1}=y_n+ hf(t_0, y_0)## where the smaller the steps i.e ##h## size the better the approximation. My question is 'how does one go about in determining the initial condition ##y(0)=1## in this problem? am assuming that this has to be a point on the line. Assuming that i had this other problem; using it to elaborate my point, ##\dfrac{dy}{dt}=y## and given ##y(0)=1## then in this case if i am to find ##y(t)## (using separation of variables) then the Exact solution is, ##y(t)=e^{t + k}## and when ##x=0, y=1 ⇒k=0##. The exact value at ##t=4## is given by, ##y(4)=e^4≈54.598## on the other hand Euler method (Numerical method) with step size ##h=0.0125## will give us, ##y=53.26## ...In this case,the initial condition, ##y(0)=1## makes sense to me as ##(0,1)## is a point on the curve. I hope that i am clear on my argument. On the flip side, i attempted to find; ##f(x,y)## of the function ##\dfrac{dy}{dx}+2y=x^3e^{-2x}## and i ended up with; ##f(x,y)=ye^{2x}-\dfrac{x^4}{4} + y^2e^{2x}-x^3y-e^{2x}y## i am not sure whether its correct...still checking then i can post all my working. Cheers [/QUOTE]
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Mathematics
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Understanding Euler Method: Finding Initial Condition of y(0)=1
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