Use Taylor series method (4th degree) to obtain the solution of y' = - xy correct to four decimal places for x=0.5 with y(0)=1 with h=0.1. Use 4 decimal places.

To address this problem, we'll apply the fourth-degree Taylor series method to solve the given differential equation with an intitial condition and a step size. We aim to find the value of y at x = 0.5 with a precision of four decimal places.

Given:

[tex]y' = -xy; \ y(0) = 1; \ h = 0.1[/tex]

First, we need to derive the necessary derivatives of y based on the given differential equation. The Taylor series expansion to the fourth degree for a function y(x) is given by:

[tex]y(x+h)=y(x)+hy'(x)+\dfrac{h^2}{2!}y''(x)+\dfrac{h^3}{3!}y'''(x)+\dfrac{h^4}{4!}y''''(x)[/tex]

Calculate our higher-order derivatives:

y' = -xy
y'' = -y - xy'
y''' = -2y' - xy''
y'''' = -3y'' - xy'''

We will evaluate these derivatives at x = 0 (where y = 1) and then use the Taylor series to find y₀(0.1), y₁(0.2), y₂(0.3), y₃(0.4), and finally y₄(0.5). Let's calculate these values:

y₀' = -x₀y₀
y₀'' = -y₀ - x₀y₀'
y₀''' = -2y₀' - x₀y₀''
y₀'''' = -3y₀'' - x₀y₀'''

Using the initial condition y(0) = 1:

y₀'(0) = -(0)(1) = 0
y₀''(0) = -(1) - (0)(0)' = -1
y₀'''(0) = - 2(0) - (0)(-1) = 0
y₀''''(0) = -3(-1) - (0)(0) = 3

Now the Taylor series:

[tex]y_1=y_0(x)+hy_0'(x)+\dfrac{h^2}{2!}y_0''(x)+\dfrac{h^3}{3!}y_0'''(x)+\dfrac{h^4}{4!}y_0''''(x)[/tex]

[tex]\Longrightarrow y_1=1+(0.1)(0)+\dfrac{(0.1)^2}{2!}(-1)+\dfrac{(0.1)^3}{3!}(0)+\dfrac{(0.1)^4}{4!}(3)\\\\\\\\\therefore y_1 \approx 0.9950[/tex]

Repeat the above process until we reach a step of 4:

y₁' = -x₁y₁
y₁'' = -y₁ - x₁y₁'
y₁''' = -2y₁' - x₁y₁''
y₁'''' = -3y₁'' - x₁y₁'''

Using the condition, y₁(0.1) = 0.9950:

y₁'(0) = -(0.1)(0.9950) = -0.0995
y₁''(0) = -(0.9950) - (0.1)(-0.0995) = -0.9851
y₁'''(0) = -2(-0.0995) - (0.1)(-0.9851) = 0.2975
y₁''''(0) = -3(-0.9851) - (0.1)(0.2975) =

Taylor series:

[tex]y_2=y_1(x)+hy_1'(x)+\dfrac{h^2}{2!}y_1''(x)+\dfrac{h^3}{3!}y_1'''(x)+\dfrac{h^4}{4!}y_1''''(x)[/tex]

[tex]\Longrightarrow y_2=0.9950+(0.1)(-0.0995)+\dfrac{(0.1)^2}{2!}(-0.9851)+\dfrac{(0.1)^3}{3!}(0.2975)+\dfrac{(0.1)^4}{4!}(2.926)\\\\\\\\[/tex]

[tex]\therefore y_2 \approx 0.9802[/tex]

Using the condition, y₂(0.2) = 0.9802:

y₂' = -x₂y₂ = -0.1960
y₂'' = -y₂ - x₂y₂' = -0.9410
y₂''' = -2y₂' - x₂y₂'' = 0.5803
y₂'''' = -3y₂'' - x₂y₂''' = 2.707

Taylor series:

∴ y₃ ≈ 0.9560

Using the condition, y₃(0.3) = 0.9560:

y₃' = -x₃y₃ = -0.2868
y₃'' = -y₃ - x₃y₃' = -0.8700
y₃''' = -2y₃' - x₃y₃'' = 0.8346
y₃'''' = -3y₃'' - x₃y₃''' = 2.360

Taylor series:

∴ y₄ ≈ 0.9231

Using the condition, y₄(0.4) = 0.9231:

y₄' = -x₄y₄ = -0.3692
y₄'' = -y₄ - x₄y₄' = -0.7754
y₄''' = -2y₄' - x₄y₄'' = 1.0487
y₄'''' = -3y₄'' - x₄y₄''' = 1.907

Taylor series:

∴ y₅(0.5) ≈ 0.8825

Thus, the solution for the differential equation is approximately 0.8825.