Math 2280-2 Problem 4.3.10 Invoke some packages that are needed with(DEtools): Define the slope function f := (t,x) -> Matrix([[1., 2.],[1., 0.]]).x + Vector([0.,exp(-t)]);

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xtrue:= t-> (1/9)*Vector([2*exp(2*t) - 2*exp(-t) - 6*t*exp(-t),exp(2*t) - exp(-t) + 6*t*exp(-t)]);

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Initial condition x0:=Vector([0,0]);

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Setting up some data structures for the iterations t0:=0.; tn:=1.; n :=10; h := (tn-t0)/n; xvals := Vector(n+1): tvals := Vector(n+1): xvals[1]:=x0: tvals[1]:=t0:

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RK4 using vectors for i from 1 to n do x := xvals[i]: t := tvals[i]: k1:= f(t,x): # left hand slope k2:= f(t+h/2,x+h*k1/2): # midpoint slope: first approx. k3:= f(t+h/2,x+h*k2/2): # midpoint slope: second approx. k4:= f(t+h,x+h*k3): # right hand slope approx. k:=(k1+2*k2+2*k3+k4)/6: # Simpson's integration rule xvals[i+1]:= x + h*k: # improved Euler update tvals[i+1]:= t+h: # increase x end do: This is the approximation to x(1) print(tvals[n+1],xvals[n+1],xtrue(tvals[n+1]));

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Setting up some data structures for the iterations t0:=0.; tn:=1.; n :=20; h := (tn-t0)/n; xvals := Vector(n+1): tvals := Vector(n+1): xvals[1]:=x0: tvals[1]:=t0:

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RK4 using vectors for i from 1 to n do x := xvals[i]: t := tvals[i]: k1:= f(t,x): # left hand slope k2:= f(t+h/2,x+h*k1/2): # midpoint slope: first approx. k3:= f(t+h/2,x+h*k2/2): # midpoint slope: second approx. k4:= f(t+h,x+h*k3): # right hand slope approx. k:=(k1+2*k2+2*k3+k4)/6: # Simpson's integration rule xvals[i+1]:= x + h*k: # improved Euler update tvals[i+1]:= t+h: # increase x end do: This is the approximation to x(1) print(tvals[n+1],xvals[n+1],xtrue(tvals[n+1]));

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This part of the code confirms there is a typo in the books xtrue. unassign('x'); unassign('t'); sys:={diff(x(t),t)=x(t)+2*y(t), diff(y(t),t)=x(t)+exp(-t)};

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dsolve(sys union {x(0)=0,y(0)=0});

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xtrue(t)[1]- (-(2/9)*exp(-t)+(2/9)*exp(2*t)-(2/3)*t*exp(-t));

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xtrue(t)[2] - (-(1/9)*exp(-t)+(1/9)*exp(2*t)+(2/3)*t*exp(-t));

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