Euler's method: accuracy and contexts Problem set A Worked solutions

\(\dfrac{dy}{dx}=3x^2\) and \(y=1\) when \(x=0\).

1. (a) Use Euler's method with \(h=0.5\) to estimate \(y(1)\) and \(y(2)\).

   • The gradient depends on x only: \(y_{n+1}=y_n+0.5\times3x_n^2\)
   • Add the 4 steps in a table: \(y(1)=1.375,\ y(2)=6.25\)

2. (b) Use Euler's method with \(h=0.25\) to estimate \(y(1)\) and \(y(2)\).

   • Same rule with \(h=0.25\): \(y(1)=1.65625,\ y(2)=7.5625\)

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

   • \(y=\int3x^2\,dx=x^3+1\)
   • True values: \(y(1)=2,\ y(2)=9\)
   • Errors, in the order above: 0.625, 2.75, 0.344, 1.44
   • Furthest: the \(h=0.5\) estimate of \(y(2)\)

\(\dfrac{dy}{dx}=-0.5y\) and \(y=6\) when \(x=0\).

1. (a) Use Euler's method with \(h=0.5\) to estimate \(y(2)\).

   • \(y_{n+1}=y_n+0.5(-0.5y_n)=0.75y_n\)
   • y: \(6, 4.5, 3.375, 2.531, 1.898\)

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

   • \(\int\dfrac1y\,dy=\int-0.5\,dx\Rightarrow\ln y=-0.5x+c\)
   • \(y=6e^{-0.5x}\)
   • \(y(2)=2.207\)

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

   • \(\dfrac{|1.898-2.207|}{2.207}\times100=14\%\)

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.

\(\dfrac{dy}{dx}=xy^2\) and \(y=0.4\) when \(x=1\). Use Euler's method with \(h=0.2\) to estimate \(y(2)\).

• Start at \(x_0=1,\ y_0=0.4\); the rule is \(y_{n+1}=y_n+0.2\,f(x_n,y_n)\)
• \(y_1=0.432,\ y_2=0.4768\)
• \(y_{5}=0.7786\ \text{at}\ x=2\)
• Answer: \(y(2)\approx0.779\)

\(\dfrac{dy}{dx}=y^2\) and \(y=0.5\) 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: \(0.5, 0.5625, 0.6416, 0.7445, 0.8831\)

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

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

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.

\(\dfrac{dy}{dx}=\dfrac{1}{x+y}\) and \(y=1\) when \(x=0\). Use Euler's method with \(h=0.5\) and technology to estimate \(y(8)\).

• There are 16 steps, so use a spreadsheet: x in column A, y in column B.
• First rows: \(y_1=1.5,\ y_2=1.75\)
• Last row: \(x=8,\ y=2.969\)
• Answer: \(y(8)\approx2.97\)

During a drought, the volume \(V\) million litres of water in a reservoir, \(t\) weeks after the rains stop, is modelled by \(\dfrac{dV}{dt}=-0.25\sqrt{V}-0.1t\). At the start \(V=12\). Use Euler's method with a step length you choose to estimate the time, to the nearest week, when the reservoir is empty.

• Choose \(h=1\): \(V_{n+1}=V_n+1\times(-0.25\sqrt{V_n}-0.1t_n)\)
• \(V\): \(12, 11.1, 10.2, 9.2, 8.14, 7.03, 5.87, 4.66, 3.42, 2.16, 0.892, -0.344\)
• \(V(10)=0.892>0,\ V(11)=-0.344<0\): the sign changes between 10 and 11.
• A step of 0.5 gives the same answer to the nearest week: 11 weeks.

The numbers of voles \(x\) and owls \(y\) in a wood, \(t\) years after a survey starts, are modelled by \(\dfrac{dx}{dt}=0.8x-0.02xy\) and \(\dfrac{dy}{dt}=0.005xy-0.4y\). At the start there are 100 voles and 30 owls. Use Euler's method with \(h=0.1\) to estimate \(x\) and \(y\) after 0.5 years.

• Both updates use the old values: \(x_{n+1}=x_n+h\,f_1,\ y_{n+1}=y_n+h\,f_2\)
• Step 1: \(x_1=102,\ y_1=30.3\)
• After 5 steps: \(x_{5}=109.69,\ y_{5}=31.839\)
• Answer: \(x\approx110,\ y\approx31.8\)

A sprinter's speed is measured every second after the start of a race. The table shows the speeds.

1. (a) Use Euler's method to estimate the distance travelled in the first 5 seconds.

   • Distance in each interval is about \(h\,v\), with the speed at the start of the interval.
   • \(s=1(0+5.8+8.1+9.4+10)=33.3\text{ m}\)

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

   • The speed is increasing, so the speed at the start of each interval is lower than the speed during it.
   • The estimate is too small.

\(\dfrac{dy}{dx}=2\sin x-0.5y\) with \(y=3\) when \(x=0\) (\(x\) in radians).

1. (a) Use Euler's method with \(h=0.1\) and technology to sketch the solution for \(0\le x\le8\).

   • Spreadsheet: 80 steps; next y = y + 0.1 × (2 sin(x) − 0.5y), using the row above. Plot column B against column A.
   • Starts at \((0,3)\), then a minimum of about 2.47 at \(x=0.7\), then a maximum of about 3.1 at \(x=2.3\), then a minimum of about −1.6 at \(x=5.9\), and ends at \(y=1.06\) when \(x=8\).

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

   • The largest value in column B is 3.103 at \(x=2.3\)
   • To 1 decimal place: \(y_{\max}\approx3.1\)