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Euler's method is based on the fact that the tangent line gives a good local approximation for the function. But why restrict ourselves to linear approximants when higher degree polynomial approximants are available? For example, we can use the Taylor polynomial of degree about = No, which is defined by P.(x) = y(x) + y (xo)(x – Xa) + 21 (x- This polynomial is the nth partial sum of the Taylor series representation (te) (x – xa + + ax To determine the Taylor series for the solution $(x) to the initial value problem dy/dx = f(x,y), y(x) = %. we need only determine the values of the derivatives of 6 (assuming they exist) at that is $(x). &'(x).....The initial condition gives the first value o xo) = y. Using the equation y = f(x,y), we find $'(x) = f(xo. Yol. To determined" (to). we differentiate the equation y = f(x,y) implicitly with respect to x to obtain af of dy af af ay dx and thereby we can compute "(x). (a) Compute the Taylor polynomials of degree 4 for the solutions to the given initial value problems. Use these Taylor polynomials to approximate the solution atx=1. y = = = 2y: y0) = 1. (D = y(2 – y): 10) = 4 . (b) Compare the use of Euler's method with that of the Taylor series to approximate the solution $(x) to the initial value problem dy dx y = cos x - sinx, y) = 2. Do this by completing Table 1.3 on page 31. Give the approximations for o(1) and (3) to the nearest thousandth. Verify that $(x) = COS + and use this formula together with a calculator or tables to find the exact values of $(x) to the nearest thousandth. Finally, decide which of the first four methods in Table 1.3 will yield the closest approximation to (10) and give the reasons for your choice. Remember that the com- putation of trigonometric functions must be done in the radian mode.) (c) Compute the Taylor polynomial of degree 6 for the solution to the Airy equation Approximation of 6 (1) Approximation of (3) Method Euler's method using steps of size 0.1 Euler's method using steps of size 0.01 Taylor polynomial of degree 2 Taylor polynomial of degree 5 Exact value of $(x) to nearest thousandth with the initial conditions y(0) - 1.7(0) = 0. Do you see how, in general, the Taylor series method for an ith-order differential equation will employ each of the initial conditions mentioned in Definition 3, Section 1.2?

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(a) Compute the Taylor polynomials of degree 4 for the solutions to the given initial value problems. Use these Taylor polynomials to approximate the solution at x = 1. recall that the Taylor polynomials of degree 4 about origin as follows: : (x)= y(0) +y'(0)x+ *(0)x2 + (0x +- 2! 4! 1 1 1 ()x* 3! given that = x-2y,y(0)=1. rewrite the above equation as, y'(x) = x-2y(x) =y"(x) =1-2y'(x), y(x)=-2y"(x), 2.((x)=-2y"(x). at x=0 y(0) = 1, V'(0)=0-2y(0)=-2, y" (0)=1-2y' (O) = 5, Jump (O)=-2y"(0)=-10, di(4(O)=-2 yım (0) = 20 substitute these values in the below equation. P(x)= y(0) +y'(0)x+ x2 2! 3! 1 = 1-2x+ 2! 3! 4! 5 5 5 1-2x+ 2 Use these Taylor polynomials to approximate the solution at x = 1. 5 (1)=1-2(1)+ 1 1 + (0)** 4!" 1 2 3 4 --x +-X 1)+ (13-319=E(1 - 1 $0.67 آنا |

(9) given that = y(2-y), y(0) = 4. rewrite the above equation as, x'(x) = 2y (x)-[y(x)] =y"(x) = 2y'(x)-2y(x)y'(x), so (x) = 2y"(x)– 2y(x)»"(x)- 2[v'(x)]”, "(x)= 2y." ( x ) -2y(x) y(x) - 65" (x)}"(x). at x = 0 y(0) = 4, (0) = 2y(0)-[y(0))*=-8, y" (0) = 2y'(0)-2y(0)y' (0) = 48, ja o (O) = 2y" (0)-2y(0)y" (0)-2["(0)] 2.(*)(O) = 2y (0)-2y(0), (0-6y"(0);"(0) = 4800. substitute these values in the below equation. 1 P(x) = y(0) + y (0)x+; 2! 3! 4! = 4-8x+=(48)x +(-416)x + (4800)x* 2! 3! 4! 208 4 - 8x+24x2 -x+ 200x 3 Use these Taylor polynomials to approximate the solution at x = 1. P. (1)= 4–8(1)+24(1)2- 208 (1+200(1)* 452 105.67 3 1 1 1