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Chapter 1: Q5.3-12E (page 1)

In Problems 10鈥�13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

t2y''+y=t+2;y(1)=1,y'(1)=-1on[1,2]

Short Answer

Expert verified

The solution is:

y(1.25)=0.75y(1.5)=0.625y(1.75)=0.6y(2)=0.6548

Step by step solution

01

Transform the equation

Here h=0.25 on [0,2].

The equations can be written as:

x1(t)=y(t)x2(t)=y'(t)=x'1

The transformation of the equation is:

localid="1664089567642" x'1(t)=x2(t)x'2(t)=-x1+t+2t2

The initial conditions are:

x1(1)=y1(1)=1=x1,0x2(1)=y'(1)=-1=x2,0

02

Apply Euler’s method

Now,

xn+1=xn+hf(tn,xn)

tn+1=tn+ht1=1+0.25=1.25x1(1.25)=x1,1=0.75x2(1.25)=x2,1=-0.5

And

tn+1=tn+ht2=1.25+0.25=1.5x1(1.5)=x1,2=0.625x2(1.5)=x2,2=-0.1

t3=1.5+0.25=1.75x1(1.75)=x1,3=0.6x2(1.75)=x2,3=-0.219

t4=1.75+0.25=2x1(2)=x1,4=0.6548x2(2)=x2,4=0.4765

Thus, this is the required result

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