Euler's method: accuracy and contexts Problem set B One question at a time

\(\dfrac{dy}{dx}=6x^2\) and \(y=2\) when \(x=0\).

1. (a) Use Euler's method with \(h=1\) to estimate \(y(2)\) and \(y(4)\).

   • The gradient depends on x only: \(y_{n+1}=y_n+1\times6x_n^2\)
   • Add the 4 steps in a table: \(y(2)\approx8,\ y(4)\approx86\)

2. (b) Use Euler's method with \(h=0.5\) to estimate \(y(2)\) and \(y(4)\).

   • Same rule with \(h=0.5\): \(y(2)\approx12.5,\ y(4)\approx107\)

3. (c) Solve the differential equation exactly. Hence say which of your four estimates is furthest from the true value.

   • \(y=\int6x^2\,dx=2x^3+2\)
   • True values: \(y(2)=18,\ y(4)=130\)
   • Errors, in the order above: 10, 44, 5.5, 23
   • Furthest: the \(h=1\) estimate of \(y(4)\)

\(\dfrac{dy}{dx}=0.25y\) and \(y=8\) when \(x=0\).

1. (a) Use Euler's method with \(h=1\) to estimate \(y(4)\).

   • \(y_{n+1}=y_n+1(0.25y_n)=1.25y_n\)
   • y: \(8, 10, 12.5, 15.625, 19.53\)

2. (b) Solve the differential equation exactly and find \(y(4)\).

   • \(\int\dfrac1y\,dy=\int0.25\,dx\Rightarrow\ln y=0.25x+c\)
   • \(y=8e^{0.25x}\)
   • \(y(4)=21.746\ldots\approx21.7\)

3. (c) Find the percentage error of your estimate.

   • \(\dfrac{|19.53-21.75|}{21.75}\times100=10.2\%\)

4. (d) Explain how the estimate could be made more accurate.

   • A smaller step means shorter tangent segments, so the estimate follows the curve more closely.

\(\dfrac{dy}{dx}=\dfrac{x^2}{1+y}\) and \(y=1\) when \(x=1\). Use Euler's method with \(h=0.25\) to estimate \(y(2)\).

• Start at \(x_0=1,\ y_0=1\); the rule is \(y_{n+1}=y_n+0.25\,f(x_n,y_n)\)
• \(y_1=1.125,\ y_2=1.309\)
• \(y_{4}=1.852\ \text{at}\ x=2\)
• Answer: \(y(2)\approx1.85\)

\(\dfrac{dy}{dx}=2y^2\) and \(y=0.25\) when \(x=0\). The graph shows the exact solution and the Euler polyline with \(h=0.25\).

1. (a) Use Euler's method with \(h=0.25\) to estimate \(y(1)\).

   • \(y_{n+1}=y_n+0.25\,f(y_n)\)
   • y: \(0.25, 0.2812, 0.3208, 0.3723, 0.4415\)

2. (b) Solve the differential equation exactly and find \(y(1)\).

   • Separate the variables, integrate and use the start value.
   • \(y=\dfrac{1}{4-2x}\)
   • \(y(1)=0.5\)

3. (c) Use the shape of the exact solution to explain why the Euler estimate is smaller than the true value.

   • The second derivative of the solution is positive, so the curve is concave up.
   • A tangent to a concave up curve lies below it. Every Euler step ends below the curve and the error builds up.

\(\dfrac{dy}{dx}=\dfrac{3}{x+y+1}\) and \(y=2\) when \(x=0\). Use Euler's method with \(h=0.5\) and technology to estimate \(y(10)\).

• There are 20 steps, so use a spreadsheet: x in column A, y in column B.
• First rows: \(y_1=2.5,\ y_2=2.875\)
• Last row: \(x=10,\ y=5.709\)
• Answer: \(y(10)\approx5.71\)

The depth \(d\) cm of snow on a pitch, \(t\) days after the thaw begins, is modelled by \(\dfrac{dd}{dt}=-0.3d-0.2t\). At the start \(d=12\). Use Euler's method with a step length you choose to estimate the time, to the nearest day, when the pitch is clear of snow.

• Choose \(h=1\): \(d_{n+1}=d_n+1\times(-0.3d_n-0.2t_n)\)
• \(d\): \(12, 8.4, 5.68, 3.58, 1.9, 0.532, -0.627\)
• \(d(5)=0.532>0,\ d(6)=-0.627<0\): the sign changes between 5 and 6.
• A step of 0.5 gives the same answer to the nearest day: 6 days.

The amounts of solution \(x\) litres and \(y\) litres in two connected tanks, \(t\) hours after a pump starts, are modelled by \(\dfrac{dx}{dt}=0.2y-0.3x\) and \(\dfrac{dy}{dt}=0.3x-0.2y-0.1t\). At the start \(x=50\) and \(y=20\). Use Euler's method with \(h=0.25\) to estimate \(x\) and \(y\) after 1 hour.

• Both updates use the old values: \(x_{n+1}=x_n+h\,f_1,\ y_{n+1}=y_n+h\,f_2\)
• Step 1: \(x_1=47.25,\ y_1=22.75\)
• After 4 steps: \(x_{4}=40.895,\ y_{4}=29.068\)
• Answer: \(x\approx40.9,\ y\approx29.1\)

A train's speed is measured every 6 seconds as it brakes. The table shows the speeds.

1. (a) Use Euler's method to estimate the distance travelled in the first 30 seconds.

   • Distance in each interval is about \(h\,v\), with the speed at the start of the interval.
   • \(s=6(30+24+18+14+11)=582\text{ m}\)

2. (b) Is your estimate too large or too small? Give a reason.

   • The speed is decreasing, so the speed at the start of each interval is higher than the speed during it.
   • The estimate is too large.

\(\dfrac{dy}{dx}=3\sin x-0.6y\) with \(y=2\) when \(x=0\) (\(x\) in radians).

1. (a) Use Euler's method with \(h=0.1\) and technology to sketch the solution for \(0\le x\le6\).

   • Spreadsheet: 60 steps; next y = y + 0.1 × (3 sin(x) − 0.6y), using the row above. Plot column B against column A.
   • Starts at \((0,2)\), then a minimum of about 1.73 at \(x=0.4\), then a maximum of about 3.53 at \(x=2.4\), then a minimum of about −2.51 at \(x=5.8\), and ends at \(y=-2.46\) when \(x=6\).

2. (b) Hence estimate, to 1 decimal place, the largest value of \(y\) in this interval.

   • The largest value in column B is 3.533 at \(x=2.4\)
   • To 1 decimal place: \(y_{\max}\approx3.5\)