Hard Further Maths Numerical methods Questions

Challenging, exam-style Further Maths Numerical methods questions with worked solutions. Stretch yourself on the hardest mid-ordinate-rule, numerical-integration, simpsons-rule, eulers-method problems.

mid-ordinate-rulenumerical-integrationsimpsons-ruleeulers-methodnumerical-solution-of-differential-equationsimproved-euler
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The function y(x)y(x) satisfies dydx=x+y\frac{dy}{dx}=\sqrt{x+y} with y=1y=1 at x=0x=0. Using the improved Euler formula yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right) with step length h=0.1h=0.1, estimate the value of yy at x=0.5x=0.5, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote the improved Euler formula

    yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right)

    The gradients at the start and at the predicted end of the step are averaged, which is far more accurate than plain Euler.

  2. Write down the step length and the starting values

    h=0.1,x0=0,y0=1h=0.1,\quad x_{0}=0,\quad y_{0}=1

    55 steps of length 0.10.1 take xx from 00 to 0.50.5.

  3. List the xx-values used

    x0=0,x1=0.1,x2=0.2,x3=0.3,x4=0.4,x5=0.5x_{0}=0,\quad x_{1}=0.1,\quad x_{2}=0.2,\quad x_{3}=0.3,\quad x_{4}=0.4,\quad x_{5}=0.5

    Each step advances xx by h=0.1h=0.1.

  4. Find k1k_{1} for step 11

    k1=hf(0,1.00000000)=0.1×1.00000000=0.10000000k_{1}=hf\left(0,1.00000000\right)=0.1\times1.00000000=0.10000000

    k1k_{1} uses the gradient at the current point (0,1.00000000)\left(0,1.00000000\right).

  5. Find k2k_{2} for step 11

    k2=hf(0.1,1.10000000)=0.1×1.09544512=0.10954451k_{2}=hf\left(0.1,1.10000000\right)=0.1\times1.09544512=0.10954451

    k2k_{2} uses the gradient at the Euler-predicted point (0.1,1.10000000)\left(0.1,1.10000000\right).

  6. Average the two gradients to find y1y_{1}

    y1=1.00000000+12(0.10000000+0.10954451)=1.10477226y_{1}=1.00000000+\frac{1}{2}\left(0.10000000+0.10954451\right)=1.10477226

    The mean of k1k_{1} and k2k_{2} is added to y0y_{0}.

  7. Find k1k_{1} for step 22

    k1=hf(0.1,1.10477226)=0.1×1.09762118=0.10976212k_{1}=hf\left(0.1,1.10477226\right)=0.1\times1.09762118=0.10976212

    k1k_{1} uses the gradient at the current point (0.1,1.10477226)\left(0.1,1.10477226\right).

  8. Find k2k_{2} for step 22

    k2=hf(0.2,1.21453437)=0.1×1.18934199=0.11893420k_{2}=hf\left(0.2,1.21453437\right)=0.1\times1.18934199=0.11893420

    k2k_{2} uses the gradient at the Euler-predicted point (0.2,1.21453437)\left(0.2,1.21453437\right).

  9. Average the two gradients to find y2y_{2}

    y2=1.10477226+12(0.10976212+0.11893420)=1.21912041y_{2}=1.10477226+\frac{1}{2}\left(0.10976212+0.11893420\right)=1.21912041

    The mean of k1k_{1} and k2k_{2} is added to y1y_{1}.

  10. Find k1k_{1} for step 33

    k1=hf(0.2,1.21912041)=0.1×1.19126841=0.11912684k_{1}=hf\left(0.2,1.21912041\right)=0.1\times1.19126841=0.11912684

    k1k_{1} uses the gradient at the current point (0.2,1.21912041)\left(0.2,1.21912041\right).

  11. Find k2k_{2} for step 33

    k2=hf(0.3,1.33824725)=0.1×1.27994033=0.12799403k_{2}=hf\left(0.3,1.33824725\right)=0.1\times1.27994033=0.12799403

    k2k_{2} uses the gradient at the Euler-predicted point (0.3,1.33824725)\left(0.3,1.33824725\right).

  12. Average the two gradients to find y3y_{3}

    y3=1.21912041+12(0.11912684+0.12799403)=1.34268085y_{3}=1.21912041+\frac{1}{2}\left(0.11912684+0.12799403\right)=1.34268085

    The mean of k1k_{1} and k2k_{2} is added to y2y_{2}.

  13. Find k1k_{1} for step 44

    k1=hf(0.3,1.34268085)=0.1×1.28167112=0.12816711k_{1}=hf\left(0.3,1.34268085\right)=0.1\times1.28167112=0.12816711

    k1k_{1} uses the gradient at the current point (0.3,1.34268085)\left(0.3,1.34268085\right).

  14. Find k2k_{2} for step 44

    k2=hf(0.4,1.47084796)=0.1×1.36778944=0.13677894k_{2}=hf\left(0.4,1.47084796\right)=0.1\times1.36778944=0.13677894

    k2k_{2} uses the gradient at the Euler-predicted point (0.4,1.47084796)\left(0.4,1.47084796\right).

  15. Average the two gradients to find y4y_{4}

    y4=1.34268085+12(0.12816711+0.13677894)=1.47515388y_{4}=1.34268085+\frac{1}{2}\left(0.12816711+0.13677894\right)=1.47515388

    The mean of k1k_{1} and k2k_{2} is added to y3y_{3}.

  16. Find k1k_{1} for step 55

    k1=hf(0.4,1.47515388)=0.1×1.36936258=0.13693626k_{1}=hf\left(0.4,1.47515388\right)=0.1\times1.36936258=0.13693626

    k1k_{1} uses the gradient at the current point (0.4,1.47515388)\left(0.4,1.47515388\right).

  17. Find k2k_{2} for step 55

    k2=hf(0.5,1.61209014)=0.1×1.45330318=0.14533032k_{2}=hf\left(0.5,1.61209014\right)=0.1\times1.45330318=0.14533032

    k2k_{2} uses the gradient at the Euler-predicted point (0.5,1.61209014)\left(0.5,1.61209014\right).

  18. Average the two gradients to find y5y_{5}

    y5=1.47515388+12(0.13693626+0.14533032)=1.61628717y_{5}=1.47515388+\frac{1}{2}\left(0.13693626+0.14533032\right)=1.61628717

    The mean of k1k_{1} and k2k_{2} is added to y4y_{4}.

  19. Select the option that matches this estimate

    y(0.5)≈1.6163y\left(0.5\right)\approx1.6163

    The value after the final step, rounded to 44 decimal places.

Answer
1.61631.6163
Question 2
9 markschallenging
Use Simpson's rule with 88 strips to estimate ∫0411+x2 dx\int_{0}^{4}\frac{1}{1+x^{2}}\,dx, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote Simpson's rule

    ∫abf(x) dx≈h3(y0+yn+4(y1+y3+⋯ )+2(y2+y4+⋯ ))\int_{a}^{b}f(x)\,dx\approx\frac{h}{3}\left(y_{0}+y_{n}+4\left(y_{1}+y_{3}+\cdots\right)+2\left(y_{2}+y_{4}+\cdots\right)\right)

    The end ordinates have weight 11, the odd ordinates weight 44 and the remaining even ordinates weight 22.

  2. Find the strip width hh

    h=b−an=4−08=0.5h=\frac{b-a}{n}=\frac{4-0}{8}=0.5

    The interval is divided into 88 strips of equal width.

  3. Find the xx-values of the ordinates

    x0=0,x1=0.5,x2=1,x3=1.5,x4=2,x5=2.5,x6=3,x7=3.5,x8=4x_{0}=0,\quad x_{1}=0.5,\quad x_{2}=1,\quad x_{3}=1.5,\quad x_{4}=2,\quad x_{5}=2.5,\quad x_{6}=3,\quad x_{7}=3.5,\quad x_{8}=4

    There are 88 strips, so there are 99 ordinates xi=a+ihx_{i}=a+ih.

  4. Evaluate the ordinate at x0=0x_{0}=0

    y0=11+(0)2=1.00000000y_{0}=\frac{1}{1+\left(0\right)^{2}}=1.00000000

    Substituting x0=0x_{0}=0 into 11+x2\frac{1}{1+x^{2}}.

  5. Evaluate the ordinate at x1=0.5x_{1}=0.5

    y1=11+(0.5)2=0.80000000y_{1}=\frac{1}{1+\left(0.5\right)^{2}}=0.80000000

    Substituting x1=0.5x_{1}=0.5 into 11+x2\frac{1}{1+x^{2}}.

  6. Evaluate the ordinate at x2=1x_{2}=1

    y2=11+(1)2=0.50000000y_{2}=\frac{1}{1+\left(1\right)^{2}}=0.50000000

    Substituting x2=1x_{2}=1 into 11+x2\frac{1}{1+x^{2}}.

  7. Evaluate the ordinate at x3=1.5x_{3}=1.5

    y3=11+(1.5)2=0.30769231y_{3}=\frac{1}{1+\left(1.5\right)^{2}}=0.30769231

    Substituting x3=1.5x_{3}=1.5 into 11+x2\frac{1}{1+x^{2}}.

  8. Evaluate the ordinate at x4=2x_{4}=2

    y4=11+(2)2=0.20000000y_{4}=\frac{1}{1+\left(2\right)^{2}}=0.20000000

    Substituting x4=2x_{4}=2 into 11+x2\frac{1}{1+x^{2}}.

  9. Evaluate the ordinate at x5=2.5x_{5}=2.5

    y5=11+(2.5)2=0.13793103y_{5}=\frac{1}{1+\left(2.5\right)^{2}}=0.13793103

    Substituting x5=2.5x_{5}=2.5 into 11+x2\frac{1}{1+x^{2}}.

  10. Evaluate the ordinate at x6=3x_{6}=3

    y6=11+(3)2=0.10000000y_{6}=\frac{1}{1+\left(3\right)^{2}}=0.10000000

    Substituting x6=3x_{6}=3 into 11+x2\frac{1}{1+x^{2}}.

  11. Evaluate the ordinate at x7=3.5x_{7}=3.5

    y7=11+(3.5)2=0.07547170y_{7}=\frac{1}{1+\left(3.5\right)^{2}}=0.07547170

    Substituting x7=3.5x_{7}=3.5 into 11+x2\frac{1}{1+x^{2}}.

  12. Evaluate the ordinate at x8=4x_{8}=4

    y8=11+(4)2=0.05882353y_{8}=\frac{1}{1+\left(4\right)^{2}}=0.05882353

    Substituting x8=4x_{8}=4 into 11+x2\frac{1}{1+x^{2}}.

  13. Substitute the ordinates into Simpson's rule

    I≈0.53[1.00000000+0.05882353+4(0.80000000+0.30769231+0.13793103+0.07547170)+2(0.50000000+0.20000000+0.10000000)]I\approx\frac{0.5}{3}\left[1.00000000+0.05882353+4\left(0.80000000+0.30769231+0.13793103+0.07547170\right)+2\left(0.50000000+0.20000000+0.10000000\right)\right]

    The values are placed into the rule in order.

  14. Evaluate the bracket

    1.00000000+0.05882353+4(0.80000000+0.30769231+0.13793103+0.07547170)+2(0.50000000+0.20000000+0.10000000)=7.943203691.00000000+0.05882353+4\left(0.80000000+0.30769231+0.13793103+0.07547170\right)+2\left(0.50000000+0.20000000+0.10000000\right)=7.94320369

    Add the weighted ordinates together.

  15. Multiply by the factor outside the bracket

    I≈0.53×7.94320369=1.32386728I\approx\frac{0.5}{3}\times7.94320369=1.32386728

    This is the estimate before rounding.

  16. Select the option that matches this estimate

    I≈1.3239I\approx1.3239

    The question asks for 44 decimal places, so the estimate is rounded at the very end.

Answer
1.32391.3239
Question 3
9 markschallenging
Use the mid-ordinate rule with 88 strips to estimate ∫04ln⁡(1+x) dx\int_{0}^{4}\ln(1+x)\,dx, giving your answer to 44 decimal places. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Quote the mid-ordinate rule

    ∫abf(x) dx≈h(f(m1)+f(m2)+⋯+f(mn))\int_{a}^{b}f(x)\,dx\approx h\left(f(m_{1})+f(m_{2})+\cdots+f(m_{n})\right)

    Each strip is replaced by a rectangle whose height is the value of ff at the middle of that strip.

  2. Find the strip width hh

    h=b−an=4−08=0.5h=\frac{b-a}{n}=\frac{4-0}{8}=0.5

    The interval is divided into 88 strips of equal width.

  3. Find the mid-ordinates

    m1=0.25,m2=0.75,m3=1.25,m4=1.75,m5=2.25,m6=2.75,m7=3.25,m8=3.75m_{1}=0.25,\quad m_{2}=0.75,\quad m_{3}=1.25,\quad m_{4}=1.75,\quad m_{5}=2.25,\quad m_{6}=2.75,\quad m_{7}=3.25,\quad m_{8}=3.75

    The mid-ordinate of the ii-th strip is mi=a+(i−12)hm_{i}=a+\left(i-\frac{1}{2}\right)h.

  4. Evaluate the integrand at m1=0.25m_{1}=0.25

    f(m1)=ln⁡(1+(0.25))=0.22314355f\left(m_{1}\right)=\ln(1+\left(0.25\right))=0.22314355

    Substituting m1=0.25m_{1}=0.25 into ln⁡(1+x)\ln(1+x).

  5. Evaluate the integrand at m2=0.75m_{2}=0.75

    f(m2)=ln⁡(1+(0.75))=0.55961579f\left(m_{2}\right)=\ln(1+\left(0.75\right))=0.55961579

    Substituting m2=0.75m_{2}=0.75 into ln⁡(1+x)\ln(1+x).

  6. Evaluate the integrand at m3=1.25m_{3}=1.25

    f(m3)=ln⁡(1+(1.25))=0.81093022f\left(m_{3}\right)=\ln(1+\left(1.25\right))=0.81093022

    Substituting m3=1.25m_{3}=1.25 into ln⁡(1+x)\ln(1+x).

  7. Evaluate the integrand at m4=1.75m_{4}=1.75

    f(m4)=ln⁡(1+(1.75))=1.01160091f\left(m_{4}\right)=\ln(1+\left(1.75\right))=1.01160091

    Substituting m4=1.75m_{4}=1.75 into ln⁡(1+x)\ln(1+x).

  8. Evaluate the integrand at m5=2.25m_{5}=2.25

    f(m5)=ln⁡(1+(2.25))=1.17865500f\left(m_{5}\right)=\ln(1+\left(2.25\right))=1.17865500

    Substituting m5=2.25m_{5}=2.25 into ln⁡(1+x)\ln(1+x).

  9. Evaluate the integrand at m6=2.75m_{6}=2.75

    f(m6)=ln⁡(1+(2.75))=1.32175584f\left(m_{6}\right)=\ln(1+\left(2.75\right))=1.32175584

    Substituting m6=2.75m_{6}=2.75 into ln⁡(1+x)\ln(1+x).

  10. Evaluate the integrand at m7=3.25m_{7}=3.25

    f(m7)=ln⁡(1+(3.25))=1.44691898f\left(m_{7}\right)=\ln(1+\left(3.25\right))=1.44691898

    Substituting m7=3.25m_{7}=3.25 into ln⁡(1+x)\ln(1+x).

  11. Evaluate the integrand at m8=3.75m_{8}=3.75

    f(m8)=ln⁡(1+(3.75))=1.55814462f\left(m_{8}\right)=\ln(1+\left(3.75\right))=1.55814462

    Substituting m8=3.75m_{8}=3.75 into ln⁡(1+x)\ln(1+x).

  12. Substitute the mid-ordinate values into the rule

    I≈0.5(0.22314355+0.55961579+0.81093022+1.01160091+1.17865500+1.32175584+1.44691898+1.55814462)I\approx0.5\left(0.22314355+0.55961579+0.81093022+1.01160091+1.17865500+1.32175584+1.44691898+1.55814462\right)

    The values are placed into the rule in order.

  13. Evaluate the bracket

    0.22314355+0.55961579+0.81093022+1.01160091+1.17865500+1.32175584+1.44691898+1.55814462=8.110764900.22314355+0.55961579+0.81093022+1.01160091+1.17865500+1.32175584+1.44691898+1.55814462=8.11076490

    Add the weighted ordinates together.

  14. Multiply by the factor outside the bracket

    I≈0.5×8.11076490=4.05538245I\approx0.5\times8.11076490=4.05538245

    This is the estimate before rounding.

  15. Select the option that matches this estimate

    I≈4.0554I\approx4.0554

    The question asks for 44 decimal places, so the estimate is rounded at the very end.

Answer
4.05544.0554
Question 4
9 markschallenging
The function y(x)y(x) satisfies dydx=xy\frac{dy}{dx}=xy with y=1y=1 at x=1x=1. Using the improved Euler formula yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right) with step length h=0.1h=0.1, estimate the value of yy at x=1.4x=1.4, giving your answer to 44 decimal places.
Show worked solution

Worked solution

  1. Quote the improved Euler formula

    yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right)

    The gradients at the start and at the predicted end of the step are averaged, which is far more accurate than plain Euler.

  2. Write down the step length and the starting values

    h=0.1,x0=1,y0=1h=0.1,\quad x_{0}=1,\quad y_{0}=1

    44 steps of length 0.10.1 take xx from 11 to 1.41.4.

  3. List the xx-values used

    x0=1,x1=1.1,x2=1.2,x3=1.3,x4=1.4x_{0}=1,\quad x_{1}=1.1,\quad x_{2}=1.2,\quad x_{3}=1.3,\quad x_{4}=1.4

    Each step advances xx by h=0.1h=0.1.

  4. Find k1k_{1} for step 11

    k1=hf(1,1.00000000)=0.1×1.00000000=0.10000000k_{1}=hf\left(1,1.00000000\right)=0.1\times1.00000000=0.10000000

    k1k_{1} uses the gradient at the current point (1,1.00000000)\left(1,1.00000000\right).

  5. Find k2k_{2} for step 11

    k2=hf(1.1,1.10000000)=0.1×1.21000000=0.12100000k_{2}=hf\left(1.1,1.10000000\right)=0.1\times1.21000000=0.12100000

    k2k_{2} uses the gradient at the Euler-predicted point (1.1,1.10000000)\left(1.1,1.10000000\right).

  6. Average the two gradients to find y1y_{1}

    y1=1.00000000+12(0.10000000+0.12100000)=1.11050000y_{1}=1.00000000+\frac{1}{2}\left(0.10000000+0.12100000\right)=1.11050000

    The mean of k1k_{1} and k2k_{2} is added to y0y_{0}.

  7. Find k1k_{1} for step 22

    k1=hf(1.1,1.11050000)=0.1×1.22155000=0.12215500k_{1}=hf\left(1.1,1.11050000\right)=0.1\times1.22155000=0.12215500

    k1k_{1} uses the gradient at the current point (1.1,1.11050000)\left(1.1,1.11050000\right).

  8. Find k2k_{2} for step 22

    k2=hf(1.2,1.23265500)=0.1×1.47918600=0.14791860k_{2}=hf\left(1.2,1.23265500\right)=0.1\times1.47918600=0.14791860

    k2k_{2} uses the gradient at the Euler-predicted point (1.2,1.23265500)\left(1.2,1.23265500\right).

  9. Average the two gradients to find y2y_{2}

    y2=1.11050000+12(0.12215500+0.14791860)=1.24553680y_{2}=1.11050000+\frac{1}{2}\left(0.12215500+0.14791860\right)=1.24553680

    The mean of k1k_{1} and k2k_{2} is added to y1y_{1}.

  10. Find k1k_{1} for step 33

    k1=hf(1.2,1.24553680)=0.1×1.49464416=0.14946442k_{1}=hf\left(1.2,1.24553680\right)=0.1\times1.49464416=0.14946442

    k1k_{1} uses the gradient at the current point (1.2,1.24553680)\left(1.2,1.24553680\right).

  11. Find k2k_{2} for step 33

    k2=hf(1.3,1.39500122)=0.1×1.81350158=0.18135016k_{2}=hf\left(1.3,1.39500122\right)=0.1\times1.81350158=0.18135016

    k2k_{2} uses the gradient at the Euler-predicted point (1.3,1.39500122)\left(1.3,1.39500122\right).

  12. Average the two gradients to find y3y_{3}

    y3=1.24553680+12(0.14946442+0.18135016)=1.41094409y_{3}=1.24553680+\frac{1}{2}\left(0.14946442+0.18135016\right)=1.41094409

    The mean of k1k_{1} and k2k_{2} is added to y2y_{2}.

  13. Find k1k_{1} for step 44

    k1=hf(1.3,1.41094409)=0.1×1.83422731=0.18342273k_{1}=hf\left(1.3,1.41094409\right)=0.1\times1.83422731=0.18342273

    k1k_{1} uses the gradient at the current point (1.3,1.41094409)\left(1.3,1.41094409\right).

  14. Find k2k_{2} for step 44

    k2=hf(1.4,1.59436682)=0.1×2.23211355=0.22321135k_{2}=hf\left(1.4,1.59436682\right)=0.1\times2.23211355=0.22321135

    k2k_{2} uses the gradient at the Euler-predicted point (1.4,1.59436682)\left(1.4,1.59436682\right).

  15. Average the two gradients to find y4y_{4}

    y4=1.41094409+12(0.18342273+0.22321135)=1.61426113y_{4}=1.41094409+\frac{1}{2}\left(0.18342273+0.22321135\right)=1.61426113

    The mean of k1k_{1} and k2k_{2} is added to y3y_{3}.

  16. State the estimate to the required accuracy

    y(1.4)≈1.6143y\left(1.4\right)\approx1.6143

    The value after the final step, rounded to 44 decimal places.

Answer
1.61431.6143
Question 5
9 markschallenging
The function y(x)y(x) satisfies dydx=x2+y\frac{dy}{dx}=x^{2}+y with y=1y=1 at x=0x=0. Using the improved Euler formula yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right) with step length h=0.1h=0.1, estimate the value of yy at x=0.5x=0.5, giving your answer to 44 decimal places.
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Worked solution

  1. Quote the improved Euler formula

    yr+1=yr+12(k1+k2),k1=hf(xr,yr),k2=hf(xr+h,yr+k1)y_{r+1}=y_{r}+\frac{1}{2}\left(k_{1}+k_{2}\right),\quad k_{1}=hf\left(x_{r},y_{r}\right),\quad k_{2}=hf\left(x_{r}+h,y_{r}+k_{1}\right)

    The gradients at the start and at the predicted end of the step are averaged, which is far more accurate than plain Euler.

  2. Write down the step length and the starting values

    h=0.1,x0=0,y0=1h=0.1,\quad x_{0}=0,\quad y_{0}=1

    55 steps of length 0.10.1 take xx from 00 to 0.50.5.

  3. List the xx-values used

    x0=0,x1=0.1,x2=0.2,x3=0.3,x4=0.4,x5=0.5x_{0}=0,\quad x_{1}=0.1,\quad x_{2}=0.2,\quad x_{3}=0.3,\quad x_{4}=0.4,\quad x_{5}=0.5

    Each step advances xx by h=0.1h=0.1.

  4. Find k1k_{1} for step 11

    k1=hf(0,1.00000000)=0.1×1.00000000=0.10000000k_{1}=hf\left(0,1.00000000\right)=0.1\times1.00000000=0.10000000

    k1k_{1} uses the gradient at the current point (0,1.00000000)\left(0,1.00000000\right).

  5. Find k2k_{2} for step 11

    k2=hf(0.1,1.10000000)=0.1×1.11000000=0.11100000k_{2}=hf\left(0.1,1.10000000\right)=0.1\times1.11000000=0.11100000

    k2k_{2} uses the gradient at the Euler-predicted point (0.1,1.10000000)\left(0.1,1.10000000\right).

  6. Average the two gradients to find y1y_{1}

    y1=1.00000000+12(0.10000000+0.11100000)=1.10550000y_{1}=1.00000000+\frac{1}{2}\left(0.10000000+0.11100000\right)=1.10550000

    The mean of k1k_{1} and k2k_{2} is added to y0y_{0}.

  7. Find k1k_{1} for step 22

    k1=hf(0.1,1.10550000)=0.1×1.11550000=0.11155000k_{1}=hf\left(0.1,1.10550000\right)=0.1\times1.11550000=0.11155000

    k1k_{1} uses the gradient at the current point (0.1,1.10550000)\left(0.1,1.10550000\right).

  8. Find k2k_{2} for step 22

    k2=hf(0.2,1.21705000)=0.1×1.25705000=0.12570500k_{2}=hf\left(0.2,1.21705000\right)=0.1\times1.25705000=0.12570500

    k2k_{2} uses the gradient at the Euler-predicted point (0.2,1.21705000)\left(0.2,1.21705000\right).

  9. Average the two gradients to find y2y_{2}

    y2=1.10550000+12(0.11155000+0.12570500)=1.22412750y_{2}=1.10550000+\frac{1}{2}\left(0.11155000+0.12570500\right)=1.22412750

    The mean of k1k_{1} and k2k_{2} is added to y1y_{1}.

  10. Find k1k_{1} for step 33

    k1=hf(0.2,1.22412750)=0.1×1.26412750=0.12641275k_{1}=hf\left(0.2,1.22412750\right)=0.1\times1.26412750=0.12641275

    k1k_{1} uses the gradient at the current point (0.2,1.22412750)\left(0.2,1.22412750\right).

  11. Find k2k_{2} for step 33

    k2=hf(0.3,1.35054025)=0.1×1.44054025=0.14405403k_{2}=hf\left(0.3,1.35054025\right)=0.1\times1.44054025=0.14405403

    k2k_{2} uses the gradient at the Euler-predicted point (0.3,1.35054025)\left(0.3,1.35054025\right).

  12. Average the two gradients to find y3y_{3}

    y3=1.22412750+12(0.12641275+0.14405403)=1.35936089y_{3}=1.22412750+\frac{1}{2}\left(0.12641275+0.14405403\right)=1.35936089

    The mean of k1k_{1} and k2k_{2} is added to y2y_{2}.

  13. Find k1k_{1} for step 44

    k1=hf(0.3,1.35936089)=0.1×1.44936089=0.14493609k_{1}=hf\left(0.3,1.35936089\right)=0.1\times1.44936089=0.14493609

    k1k_{1} uses the gradient at the current point (0.3,1.35936089)\left(0.3,1.35936089\right).

  14. Find k2k_{2} for step 44

    k2=hf(0.4,1.50429698)=0.1×1.66429698=0.16642970k_{2}=hf\left(0.4,1.50429698\right)=0.1\times1.66429698=0.16642970

    k2k_{2} uses the gradient at the Euler-predicted point (0.4,1.50429698)\left(0.4,1.50429698\right).

  15. Average the two gradients to find y4y_{4}

    y4=1.35936089+12(0.14493609+0.16642970)=1.51504378y_{4}=1.35936089+\frac{1}{2}\left(0.14493609+0.16642970\right)=1.51504378

    The mean of k1k_{1} and k2k_{2} is added to y3y_{3}.

  16. Find k1k_{1} for step 55

    k1=hf(0.4,1.51504378)=0.1×1.67504378=0.16750438k_{1}=hf\left(0.4,1.51504378\right)=0.1\times1.67504378=0.16750438

    k1k_{1} uses the gradient at the current point (0.4,1.51504378)\left(0.4,1.51504378\right).

  17. Find k2k_{2} for step 55

    k2=hf(0.5,1.68254816)=0.1×1.93254816=0.19325482k_{2}=hf\left(0.5,1.68254816\right)=0.1\times1.93254816=0.19325482

    k2k_{2} uses the gradient at the Euler-predicted point (0.5,1.68254816)\left(0.5,1.68254816\right).

  18. Average the two gradients to find y5y_{5}

    y5=1.51504378+12(0.16750438+0.19325482)=1.69542338y_{5}=1.51504378+\frac{1}{2}\left(0.16750438+0.19325482\right)=1.69542338

    The mean of k1k_{1} and k2k_{2} is added to y4y_{4}.

  19. State the estimate to the required accuracy

    y(0.5)≈1.6954y\left(0.5\right)\approx1.6954

    The value after the final step, rounded to 44 decimal places.

Answer
1.69541.6954

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