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Determine the recursive formulas for the Taylor methodof order 4 for the initial value problem$$ y^{\prime}=x^{2}+y, \quad y(0)=0 $$

$y_{0}=0$ & $x_{0}=0$

Calculus 2 / BC

Chapter 3

Mathematical Models and Numerical Methods Involving First-Order Equations

Section 7

Higher-Order Numerical Methods:Taylor and Runge-Kutta

Differential Equations

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A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Determine the recursive fo…

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The solution to the initia…

okay. We want to determine the recursive formless for the tailor method of order for in this problem. So P equals four for the initial value problem. Why prime equals X squared plus y With initial condition, why have zero equal to zero? So first we can start off by writing. Why not equal to zero and ex, not equal to zero. And then we can write the first form love for X end as exempt us one equals x end plus h. And since we're going all the way out to order four, we have that. Why n plus one is equal to why end plus each times f where we're assuming that efforts evaluated at the point x n comma y and and similarly, before we can expand out the rest of the terms, including the partial derivatives and all of these derivatives and functions are assumed to be evaluated at the point X and come a white. And and I'm just not writing them here because we want to save space and be concise here because it gets a little lengthy. So then plus H cubed over six times f x x, where this term in this set of brackets is given as the derivative of the previous bracketed term with respect to X. So we're going from this function f of X comma y and then we get the multi very a chain rule to get the derivative with respect to X as X plus y times f And then we used this chain will again on this back editor to get f x x Times start plus f y so plus f x y times f plus f why times f x and then plus each the fourth. We're 24 times f x x x plus f x x y times F plus f X y times f x plus F X Y times FX Again because we're taking the view of the second term here to get uh export times f x plus F y times f x x, and that's coming from the product rule in the multivariate case. So we close this and then we can continue by simplifying this right hand side after plugging in the derivatives. That's ah needed, and we will get why and plus one is equal to why end plus each times X and squared, plus white and plus H squared over two times to X and plus X and squared pus y n plus two h cubed over three times X n plus age, the fourth over 12. And this is the final form for Why n Plus One as well as the equation for X n plus one. Given the initial conditions, why not equal to zero X not equal to zero?

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