Use the improved Euler’s method with tolerance to approximate the solution to dxdt=1+tsin(tx),x(0)=0,at t= 1. For a tolerance ofε=0.01 , use a stopping procedure based on the absolute error.

Short Answer

Expert verified

ϕ(1) = 1.25495

Step by step solution

01

Find the equation of approximation value

Here dxdt=1+tsin(tx),x(0)=0,

For ξ=0.01, x = 0, , c = 0, M = 10, h = 1

F=f(t,x)=1+tsin(tx)G=f(t+h,x+hF)=1+(t+h)sin((t+h)(x+hF))

02

Solve for x and t

Apply initial points t=0,x=0,h=1

F(0,0)=1G(0,0)=1.84147

tn+1=(tn+h)xn+1=xn+h2(F+G)t=1x=1.42074

Hence ϕ(1)=y(1,1)=1.42074

03

Evaluate the value of x and t

t=0,x=0,h=0.5

F(0,0)=1G(0,0)=1.1237

t=0+0.5=0.5x=0+0.25(1+1.1237)=0530925ϕ(0.5)=0.530925

04

Determine the value of x and t for the conditions.

t=0.5,x=0.530925,h=0.5

F(0.5,0.530925)=1.13118G(0.5,0.530925)=1.88962

t=0.5+0.5=1x=0.530925+0.25(1.13118+1.88962)=1.28613ϕ(1)=y(1,0.5)=1.28613

05

Determine the all-other values.

Apply the same procedure for all other values and the values are

ϕ(1)=y(1,0.25)=1.26026ϕ(1)=y(1,0.125)=1.25495

Since,

y(1,0.125)-y(1,0.25)=1.25495-1.26026=0.00531<0.01ϕ(t)=1.25495

Hence the solution is role="math" localid="1664306405053" ϕ(1) = 1.25495

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