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In Problems 10–13, 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.

y''=t2-y2;y(0)=0,y'(0)=1on[0,1]

Short Answer

Expert verified

The required solution is;

y(0.25)=0.25y(0.5)=0.5y(0.75)=0.75y(1)=1

Step by step solution

01

Transform the equation

Here h=0.25 on [0,1].

The equations can be written as;

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

The transformation of the equation is;

x'1(t)=x2(t)x'2(t)=t2-x12

The initial conditions are;

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

02

Apply Euler’s method

Now,

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

tn+1=tn+ht1=0+0.25=0.25x1(0.25)=x1,1=0.25x2(0.25)=x2,1=1

And

tn+1=tn+ht2=0.25+0.25=0.5x1(0.5)=x1,2=0.5x2(0.5)=x2,2=1

t3=0.5+0.25=0.75x1(0.75)=x1,3=0.75x2(0.75)=x2,3=1

t4=0.75+0.25=1x1(1)=x1,4=1x2(1)=x2,4=1

This is the required result.

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