Question #278484

.1.1 Use Euler’s method with step size h = 0.1 to approximate the solution to the initial valueproblem: y'=x y, y(1)=4, atthepointsx=1.1,1.2and1.3,correcttofivedecimal

places.

2.1.2 If the analytical solution to the initial value problem in (2.1.1) is

(4) y = 1 (x2 + 7), determine

4

the %Error in the numerical method where x =1.3

2.2 A chicken cools down from 100 °C to 60 °C within 10 


1
Expert's answer
2021-12-14T15:58:05-0500

2.1.1

y=xy,y(1)=4y'=x\sqrt y,y(1)=4


Euler method

y1=y0+hf(x0,y0)=4+0.1f(1,4)=4.2y_1=y_0+hf(x_0,y_0)=4+0.1f(1,4)=4.2

 y2=y1+hf(x1,y1)=4.2+0.1f(1.1,4.2)=4.42543y_2=y_1+hf(x_1,y_1)=4.2+0.1f(1.1,4.2)=4.42543

y3=y2+hf(x2,y2)=4.42543+0.1f(1.2,4.42543)=4.67787y_3=y_2+hf(x_2,y_2)=4.42543+0.1f(1.2,4.42543)=4.67787

y(1.3)=4.67787y(1.3)=4.67787


2.1.2

y=(x2+7)/4\sqrt y=(x^2+7)/4

y(1.3)=(1.32+7)2/16=4.71976y(1.3)=(1.3^2+7)^2/16=4.71976


error = 4.719764.677874.71976=0.0089=0.89%\frac{4.71976-4.67787}{4.71976}=0.0089=0.89\%


2.2

A chicken cools down from 100 °C to 60 °C within 10 minutes of being placed in a stream of air

at 20°C20\degree C


by Newton's Law of Cooling:

rate of cooling:

dTdt=k(TTenv)\frac{dT}{dt}=-k(T-T_{env})

where k is constant,

Tenv is temperature of the environment


then:

dTdt=k(T20)\frac{dT}{dt}=-k(T-20)


T=cekt20T=ce^{-kt}-20

T(0)=c20=100T(0)=c-20=100

c=120c=120

T(10)=120e10k20=60T(10)=120e^{-10k}-20=60

k=ln(80/120)/10=0.04k=-ln(80/120)/10=0.04


T(t)=120e0.04t20T(t)=120e^{-0.04t}-20


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