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Euler's method

Learning goals

Uses the current point and derivative value to estimate the next point.
Advances the independent variable by the chosen step size in an Euler update.
Builds successive numerical approximations from an initial condition.
Explains how step size influences approximation accuracy.
Compares the numerical approximation with known or qualitative solution behavior when possible.

Difficulty: 1 – 5

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Euler's method 1

Jordan applies Euler's method to \(y'=\frac{1}{y}\), \(y(0)=0\). Explain why the first update cannot be made.

Hints

- Evaluate the differential equation at the initial point. - Check whether the denominator is zero. - Euler's method requires a defined current slope.

Solution

1. Euler's method first requires the slope at the initial point. 2. At \(y=0\), the expression \(\frac{1}{y}\) is undefined. 3. Therefore, the differential equation provides no initial slope, so the first Euler update cannot be made.

Answer

The slope \(\frac{1}{y}\) is undefined at the initial point because \(y=0\). Euler's method cannot begin without a defined initial slope.

Euler's method 2

Use one Euler step with \(h=0.2\) for \(y'=x+y\), \(y(0)=1\). Approximate \(y(0.2)\).

Hints

- Evaluate the derivative at the starting point. - Multiply that slope by the step size and add the change to the current y-value.

Solution

1. At the starting point \((0,1)\), the slope is \(0+1=1\). 2. One Euler update gives \(y(0.2)\approx1+0.2\cdot1=1.2\).

Answer

\(y(0.2)\approx1.2\)

Euler's method 3

Euler's method is about to take a step from the current point \((1,3)\) for \(y'=y-x\), using \(h=0.1\). Before doing the update, what slope must Euler's method use, and at which point is that slope evaluated?

Hints

- Euler's method samples the derivative before moving to the next x-value. - Substitute the current coordinates into \(y-x\).

Solution

1. Euler's method uses the derivative at the current point. 2. At \((1,3)\), the slope is \(3-1=2\).

Answer

Use slope \(2\), evaluated at the current point \((1,3)\).

Euler's method 4

A solution curve passes through \((2,5)\), and its slope there is \(y'(2)=-3\). Use the tangent-line idea behind Euler's method with step size \(h=0.1\) to estimate \(y(2.1)\).

Hints

- Treat the given derivative as the local tangent slope. - Multiply the slope by the horizontal step to estimate the change in \(y\).

Solution

1. Over a step of \(0.1\), the tangent-line change in \(y\) is approximately \(0.1\cdot(-3)=-0.3\). 2. Therefore, \(y(2.1)\approx5-0.3=4.7\).

Answer

\(y(2.1)\approx4.7\)

Euler's method 5

Starting at \(A\), use one Euler step of size \(h=0.25\). What \(y\)-value does it estimate at the next \(x\)-value?

Figure for problem 562986

Hints

- Read the rise and run of the short tangent segment. - Euler uses that local slope throughout the chosen small horizontal step. - The estimated change in height is slope times step size.

Solution

1. The marked point is \((0,1)\), and the tangent segment rises \(2\) units per unit of \(x\). 2. Euler's change in \(y\) is approximately \(0.25\cdot2=0.5\). 3. The next estimate is \(y(0.25)\approx1.5\).

Answer

\(y(0.25)\approx1.5\).

Euler's method 6

Starting at \(A\), read the tangent slope and take one Euler step with \(h=0.25\). Give the new point and say whether its height is higher or lower than at \(A\).

Figure for problem 562987

Hints

- A slope is vertical change divided by horizontal change. - Check whether the segment rises or falls from left to right. - Multiply the local slope by the stated step size, then add that change to the starting height.

Solution

1. The segment falls by \(0.8\) as \(x\) increases by \(0.8\), so the slope is \(-1\). 2. With \(A=(0,2)\), the step gives \(x_1=0.25\) and \(y_1=2+0.25(-1)=1.75\), so the height decreases.

Answer

Slope \(-1\); Euler point \((0.25,1.75)\), lower than \(A\).

Euler's method 7

An Euler step starts at \((1,2)\), where the slope is \(-3\). With step size \(h=0.2\), Sam says the next point is \((1.2,2.6)\). Is Sam correct? Find the next point and explain the mistake.

Hints

- The horizontal step is positive, but the slope is negative. What happens to the height? - Use the slope and the step size to find the vertical change. - Check the signs of both coordinate changes.

Solution

1. The new horizontal coordinate is \(1+0.2=1.2\). 2. The vertical change is \(h\cdot(-3)=0.2\cdot(-3)=-0.6\). 3. Thus the next point is \((1.2,1.4)\). Sam added \(0.6\) instead of subtracting it.

Answer

No. The next Euler point is \((1.2,1.4)\); a negative slope lowers the estimate.

Euler's method 8

A tank starts with \(5\,\text{g}\) of dye. The model is \(m'=10-0.2m\), with \(t\) in minutes and \(m\) in grams. Use one Euler step of \(h=0.5\) minute to estimate the dye mass.

Hints

- Evaluate the rate using the current dye mass. - Multiply that rate by the length of the time step. - Add the estimated change to the starting amount.

Solution

1. At the start, \(m'(0)=10-0.2\cdot5=9\,\text{g/min}\). 2. One Euler step gives \(m(0.5)\approx5+0.5\cdot9=9.5\,\text{g}\).

Answer

\(m(0.5)\approx9.5\,\text{g}\).

Euler's method 9

Use Euler's method with \(h=0.2\) to approximate \(y(0.6)\) for \(y'=x+y\), \(y(0)=1\). Do not use an exact solution.

Hints

- Three Euler steps of size \(0.2\) move from \(x=0\) to \(x=0.6\). - At each current point use the slope \(x_n+y_n\). - Update with \(y_{n+1}=y_n+0.2(x_n+y_n)\) before advancing to the next slope.

Solution

1. At \((x_0,y_0)=(0,1)\), the slope is \(1\), so \(y_1=1+0.2\cdot1=1.2\). 2. At \((x_1,y_1)=(0.2,1.2)\), the slope is \(1.4\), so \(y_2=1.2+0.2\cdot1.4=1.48\). 3. At \((x_2,y_2)=(0.4,1.48)\), the slope is \(1.88\), so \(y_3=1.48+0.2\cdot1.88=1.856\). 4. Therefore, \(y(0.6)\approx1.856\).

Answer

\(y(0.6)\approx1.856\)

Euler's method 10

Starting from \(y(1)=2\), carry out the Euler updates with \(h=0.25\) needed to estimate \(y(2)\) for \(y'=x-y\). Round the final estimate to four decimal places.

Hints

- Four steps of size \(0.25\) are needed to move from \(x=1\) to \(x=2\). - Use the current-point slope \(x_n-y_n\) in every update. - Retain unrounded values through all four steps and round only the final estimate.

Solution

1. At \((x_0,y_0)=(1,2)\), the slope is \(-1\), so \(y_1=2+0.25\cdot(-1)=1.75\). 2. At \((x_1,y_1)=(1.25,1.75)\), the slope is \(-0.5\), so \(y_2=1.75+0.25\cdot(-0.5)=1.625\). 3. At \((x_2,y_2)=(1.5,1.625)\), the slope is \(-0.125\), so \(y_3=1.625+0.25\cdot(-0.125)=1.59375\). 4. At \((x_3,y_3)=(1.75,1.59375)\), the slope is \(0.15625\), so \(y_4=1.59375+0.25\cdot0.15625=1.6328125\). 5. Therefore, \(y(2)\approx1.6328\).

Answer

\(y(2)\approx1.6328\)

Euler's method 11

Build an Euler approximation for \(y'=xy\), \(y(0)=2\), using \(h=0.5\), and report \(y(1.5)\).

Hints

- Use three steps of length \(0.5\) to reach \(x=1.5\). - At each step compute the slope as the product \(x_ny_n\). - Use the updated \(y\)-value in the next product rather than reusing the previous slope.

Solution

1. At \((x_0,y_0)=(0,2)\), the slope is \(0\), so \(y_1=2+0.5\cdot0=2\). 2. At \((x_1,y_1)=(0.5,2)\), the slope is \(1\), so \(y_2=2+0.5\cdot1=2.5\). 3. At \((x_2,y_2)=(1,2.5)\), the slope is \(2.5\), so \(y_3=2.5+0.5\cdot2.5=3.75\). 4. Therefore, \(y(1.5)\approx3.75\).

Answer

\(y(1.5)\approx3.75\)

Euler's method 12

Apply the explicit Euler update repeatedly with \(h=0.1\) to the initial-value problem \(y'=y-x^2\), \(y(0)=1\), until you approximate \(y(0.4)\). Round the final estimate to four decimal places.

Hints

- Four updates of size \(0.1\) reach \(x=0.4\). - Evaluate \(y_n-x_n^2\) at the current point before each update. - Carry all intermediate digits and apply the four-decimal rounding only at the end.

Solution

1. At \((x_0,y_0)=(0,1)\), the slope is \(1\), so \(y_1=1+0.1\cdot1=1.1\). 2. At \((x_1,y_1)=(0.1,1.1)\), the slope is \(1.09\), so \(y_2=1.1+0.1\cdot1.09=1.209\). 3. At \((x_2,y_2)=(0.2,1.209)\), the slope is \(1.169\), so \(y_3=1.209+0.1\cdot1.169=1.3259\). 4. At \((x_3,y_3)=(0.3,1.3259)\), the slope is \(1.2359\), so \(y_4=1.3259+0.1\cdot1.2359=1.44949\). 5. Therefore, \(y(0.4)\approx1.4495\).

Answer

\(y(0.4)\approx1.4495\)

Euler's method 13

Use Euler's method with \(h=0.2\) to approximate \(y(1)\) for \(y'=\frac{x}{1+y}\), \(y(0)=1\). Do not use an exact solution. Round the final estimate to four decimal places.

Hints

- Five steps of size \(0.2\) are required to reach \(x=1\). - At each current point evaluate \(x_n/(1+y_n)\), checking that the denominator remains nonzero. - Use unrounded intermediate values before reporting the final four-decimal estimate.

Solution

1. At \((x_0,y_0)=(0,1)\), the slope is \(0\), so \(y_1=1+0.2\cdot0=1\). 2. At \((x_1,y_1)=(0.2,1)\), the slope is \(0.1\), so \(y_2=1+0.2\cdot0.1=1.02\). 3. At \((x_2,y_2)=(0.4,1.02)\), the slope is \(\frac{0.4}{2.02}\approx0.198019802\), so \(y_3\approx1.02+0.2\cdot0.198019802\approx1.059603960\). 4. At \((x_3,y_3)\approx(0.6,1.059603960)\), the slope is approximately \(0.291318142\), so \(y_4\approx1.117867589\). 5. At \((x_4,y_4)\approx(0.8,1.117867589)\), the slope is approximately \(0.377738440\), so \(y_5\approx1.193415277\). 6. Therefore, \(y(1)\approx1.1934\).

Answer

\(y(1)\approx1.1934\)

Euler's method 14

Apply the explicit Euler update repeatedly with \(h=0.25\) to the initial-value problem \(y'=\sin x-y\), \(y(0)=1\), until you approximate \(y(1)\). Round the final estimate to four decimal places.

Hints

- Four steps of size \(0.25\) reach \(x=1\). - Use \(\sin x_n-y_n\) as the slope, evaluating sine at the current \(x\)-value. - Keep extra trigonometric precision until the final rounding.

Solution

1. At \((x_0,y_0)=(0,1)\), the slope is \(-1\), so \(y_1=1+0.25\cdot(-1)=0.75\). 2. At \((x_1,y_1)=(0.25,0.75)\), the slope is \(\sin(0.25)-0.75\approx-0.502596041\), so \(y_2\approx0.624350990\). 3. At \((x_2,y_2)\approx(0.5,0.624350990)\), the slope is approximately \(-0.144925451\), so \(y_3\approx0.588119627\). 4. At \((x_3,y_3)\approx(0.75,0.588119627)\), the slope is approximately \(0.093519133\), so \(y_4\approx0.611499410\). 5. Therefore, \(y(1)\approx0.6115\).

Answer

\(y(1)\approx0.6115\)

Euler's method 15

Build an Euler approximation for \(y'=1+y^2\), \(y(0)=0\), using \(h=0.1\), and report \(y(0.4)\). Round the final estimate to four decimal places.

Hints

- Use four steps of length \(0.1\) to reach \(x=0.4\). - The slope is \(1+y_n^2\), so square the current approximation rather than the next one. - Retain unrounded intermediate values and round only the final result.

Solution

1. At \((x_0,y_0)=(0,0)\), the slope is \(1\), so \(y_1=0+0.1\cdot1=0.1\). 2. At \((x_1,y_1)=(0.1,0.1)\), the slope is \(1.01\), so \(y_2=0.1+0.1\cdot1.01=0.201\). 3. At \((x_2,y_2)=(0.2,0.201)\), the slope is \(1.040401\), so \(y_3=0.201+0.1\cdot1.040401=0.3050401\). 4. At \((x_3,y_3)=(0.3,0.3050401)\), the slope is \(1.09304946260801\), so \(y_4=0.3050401+0.1\cdot1.09304946260801=0.414345046260801\). 5. Therefore, \(y(0.4)\approx0.4143\).

Answer

\(y(0.4)\approx0.4143\)

Euler's method 16

The table is set up for Euler's method applied to \(y'=x+y\) with \(y(0)=1\) and \(h=0.5\). Complete the missing \(y_n\) and slope entries. <table><thead><tr><th>Step</th><th>\(x_n\)</th><th>\(y_n\)</th><th>\(f(x_n,y_n)\)</th></tr></thead><tbody><tr><td>\(0\)</td><td>\(0\)</td><td>\(1\)</td><td>?</td></tr><tr><td>\(1\)</td><td>\(0.5\)</td><td>?</td><td>?</td></tr><tr><td>\(2\)</td><td>\(1\)</td><td>?</td><td>?</td></tr><tr><td>\(3\)</td><td>\(1.5\)</td><td>?</td><td>—</td></tr></tbody></table>

Hints

- For the first row, evaluate \(x_0+y_0\) using the given initial values. - Use \(y_{n+1}=y_n+0.5f(x_n,y_n)\) to fill the next \(y\)-entry. - Each new slope must be computed from the newly completed row.

Solution

1. Apply \(y_{n+1}=y_n+h f(x_n,y_n)\) row by row. 2. The completed values are \(x_0=0, y_0=1, f=1\); \(x_1=0.5, y_1=1.5, f=2\); \(x_2=1, y_2=2.5, f=3.5\); \(x_3=1.5, y_3=4.25\).

Answer

\(x_0=0, y_0=1, f=1\); \(x_1=0.5, y_1=1.5, f=2\); \(x_2=1, y_2=2.5, f=3.5\); \(x_3=1.5, y_3=4.25\)

Euler's method 17

The table is set up for Euler's method applied to \(y'=1-y^2\) with \(y(0)=0.5\) and \(h=0.25\). Complete the missing \(y_n\) and slope entries. Round displayed entries to four decimal places as needed, but use unrounded values in later updates. <table><thead><tr><th>Step</th><th>\(x_n\)</th><th>\(y_n\)</th><th>\(f(x_n,y_n)\)</th></tr></thead><tbody><tr><td>\(0\)</td><td>\(0\)</td><td>\(0.5\)</td><td>?</td></tr><tr><td>\(1\)</td><td>\(0.25\)</td><td>?</td><td>?</td></tr><tr><td>\(2\)</td><td>\(0.5\)</td><td>?</td><td>?</td></tr><tr><td>\(3\)</td><td>\(0.75\)</td><td>?</td><td>?</td></tr><tr><td>\(4\)</td><td>\(1\)</td><td>?</td><td>—</td></tr></tbody></table>

Hints

- Use \(1-y_n^2\) to compute the slope in each completed row. - Advance with \(y_{n+1}=y_n+0.25f(x_n,y_n)\). - Display four decimals as requested but retain unrounded \(y_n\) values for later slopes.

Solution

1. Apply \(y_{n+1}=y_n+h f(x_n,y_n)\) row by row, retaining unrounded values for each subsequent update. 2. The completed displayed values are \(x_0=0, y_0=0.5, f=0.75\); \(x_1=0.25, y_1=0.6875, f\approx0.5273\); \(x_2=0.5, y_2\approx0.8193, f\approx0.3287\); \(x_3=0.75, y_3\approx0.9015, f\approx0.1873\); \(x_4=1, y_4\approx0.9483\).

Answer

\(x_0=0, y_0=0.5, f=0.75\); \(x_1=0.25, y_1=0.6875, f\approx0.5273\); \(x_2=0.5, y_2\approx0.8193, f\approx0.3287\); \(x_3=0.75, y_3\approx0.9015, f\approx0.1873\); \(x_4=1, y_4\approx0.9483\)

Euler's method 18

An Euler step for \(y'=x+y\) starts at \((1, 2)\) with \(h=0.3\). Maya uses the slope at \((1.3, 2)\) to compute the next value. Correct the step.

Hints

- Identify the current point before evaluating the slope. - Euler's update uses \(f(x_n,y_n)\). - Advance \(x\) only after using the current slope.

Solution

1. Euler's method uses the slope at the current point, not at the next \(x\)-value. 2. At \((1, 2)\), the slope is \(f(1, 2)=1+2=3\). 3. Therefore, \(y_1=2+0.3\cdot3=2.9\), and the new point is \((1.3, 2.9)\).

Answer

Use the current-point slope \(f(1, 2)=3\). Then \(y_1=2+0.3\cdot3=2.9\), so the new point is \((1.3, 2.9)\).

Euler's method 19

Use one explicit Euler step with a negative step size for \(y'=x-y\), starting from \(y(2)=1\), to estimate \(y(1.5)\).

Hints

- Compute the signed step size from the two input values. - Evaluate the slope at the current point. - Use the negative step size in the Euler update.

Solution

1. The step size is \(h=1.5-2=-0.5\). 2. At the current point \((2,1)\), the slope is \(2-1=1\). 3. Euler's update gives \(y(1.5)\approx1+(-0.5)\cdot1=0.5\).

Answer

\(y(1.5)\approx0.5\)

Euler's method 20

For \(y'=x^2+y\), an Euler table with \(h=0.2\) lists \((x_1,y_1)=(0.2,1.2)\) from \((0,1)\), then uses slope \(1\) again for the second step. Identify and correct the second slope and \(y_2\).

Hints

- Use the updated point for the next slope evaluation. - Substitute both current coordinates into the differential equation. - Apply one Euler update with the corrected slope.

Solution

1. The second slope must be evaluated at the current point \((0.2,1.2)\). 2. The corrected slope is \(0.2^2+1.2=1.24\). 3. Therefore, \(y_2=1.2+0.2\cdot1.24=1.448\).

Answer

The corrected second slope is \(1.24\), and \(y_2=1.448\).

Euler's method 21

Suppose \(y'=f(x,y)\), \(y(3)=5\), and \(f(3,5)=-4\). What one-step Euler approximation with \(h=0.1\) is determined without any other information? Explain why a second step is not determined.

Hints

- Use the supplied current slope for the first update. - Identify the new point after that update. - Determine which slope value a second update would require.

Solution

1. Use the given current slope in one Euler update: \(y(3.1)\approx5+0.1\cdot(-4)=4.6\). 2. A second step would require the new slope \(f(3.1,4.6)\). 3. Because that value is not given and the function \(f\) is otherwise unspecified, no second Euler update is determined.

Answer

\(y(3.1)\approx4.6\). A second step is not determined because it would require the unknown value \(f(3.1,4.6)\).

Euler's method 22

Euler's method is applied to \(y'=y\), starting from \(y(0)=4\), using two equal positive steps of size \(h\). The approximation after the second step is \(9\). Determine \(h\) and the x-coordinate of the final approximation.

Hints

- Simplify one Euler update for this differential equation. - Apply the same multiplicative update twice. - Use the positive-step condition when solving the resulting square equation.

Solution

1. Each Euler update multiplies the current value by \(1+h\), so after two steps \(y_2=4(1+h)^2\). 2. The condition \(4(1+h)^2=9\) gives \(1+h=\frac{3}{2}\), since \(h>0\). 3. Thus \(h=\frac{1}{2}\), and the final x-coordinate is \(2h=1\).

Answer

\(h=\frac{1}{2}\), and the final approximation is at \(x=1\).

Euler's method 23

Euler's method with step size \(h=0.75\) is applied to \(y'=4-2y\), starting from \(y(0)=0\). Compute the first four approximations after the initial value, and describe their behavior relative to the equilibrium solution.

Hints

- Simplify the Euler update into a recurrence before calculating values. - Find the equilibrium by setting the derivative equal to zero. - Compare each approximation's signed difference from the equilibrium.

Solution

1. The Euler recurrence is \(y_{n+1}=y_n+0.75(4-2y_n)=3-0.5y_n\). 2. Starting from \(y_0=0\), the approximations are \(y_1=3\), \(y_2=1.5\), \(y_3=2.25\), and \(y_4=1.875\). 3. The equilibrium is \(y=2\). The approximations alternate above and below \(2\), and their distance from \(2\) is halved at each step.

Answer

\(y_1=3\), \(y_2=1.5\), \(y_3=2.25\), and \(y_4=1.875\). They alternate around \(2\) and converge toward it.

Euler's method 24

Euler's method with step size \(h=1\) is applied to \(y'=\frac{1}{2-y}\), starting from \(y(0)=1\). Carry out the first update and explain why a second update cannot be completed.

Hints

- Evaluate the slope at the initial point before updating. - Check that the differential equation is defined at every newly approximated point. - A valid first step does not guarantee that the next slope exists.

Solution

1. At \((0,1)\), the slope is \(\frac{1}{2-1}=1\). 2. The first update gives \(y(1)\approx1+1\cdot1=2\). 3. The next slope would require evaluating \(\frac{1}{2-y}\) at \(y=2\), where the denominator is zero. Therefore, the second Euler update is undefined.

Answer

The first approximation is \(y(1)\approx2\). A second update cannot be made because the differential equation is undefined at the approximated point.

Euler's method 25

Euler's method with \(h=0.4\) is applied to \(y'=y^2\), starting from \(y(0)=0.3\), for two steps. a) Compute the result using unrounded intermediate values. b) Taylor rounds the first approximation to \(0.34\) before the second step. Compute Taylor's result and the difference caused by the premature rounding.

Hints

- Complete the first update before evaluating the next slope. - Keep all digits in one calculation and deliberately round in the other. - Compare the two final approximations by subtraction.

Solution

1. Using full precision, \(y_1=0.3+0.4\cdot(0.3)^2=0.336\). 2. Then \(y_2=0.336+0.4\cdot(0.336)^2=0.3811584\). 3. With premature rounding, the second update is \(0.34+0.4\cdot(0.34)^2=0.38624\). 4. The rounded-path result exceeds the full-precision result by \(0.38624-0.3811584=0.0050816\).

Answer

a) \(0.3811584\) b) \(0.38624\), which is \(0.0050816\) larger

Euler's method 26

For \(y'=x+y\), start from \(y(0)=1\). Use one Euler step with \(h=0.2\), then use one Euler step with \(h=-0.2\) from the new point. Find the returned approximation at \(x=0\), and explain why the two steps do not undo each other.

Hints

- Evaluate the slope separately at each starting point of a step. - A negative step changes the sign of the increment but not the point at which the slope is sampled. - Compare the forward and backward slopes.

Solution

1. The forward slope is \(0+1=1\), so \(y(0.2)\approx1+0.2\cdot1=1.2\). 2. At \((0.2,1.2)\), the slope is \(1.4\). The backward step gives \(1.2-0.2\cdot1.4=0.92\). 3. The steps do not undo each other because the backward step uses the slope at the new approximated point, not the original slope.

Answer

The returned approximation is \(0.92\), not \(1\), because the two steps use different slopes.

Euler's method 27

An Euler approximation uses step size \(h=0.25\) and starts from \(y_0=2\). The slopes used in three consecutive updates are \(1.2\), \(-0.4\), and an unknown value \(m\). The final approximation is \(y_3=2.5\). Determine \(m\).

Hints

- Add the three Euler increments rather than reconstructing unknown coordinates. - Express the final value as the initial value plus step size times the slope sum. - Solve the resulting linear equation for the missing slope.

Solution

1. The total Euler change is \(0.25\cdot(1.2-0.4+m)\). 2. Since \(2.5=2+0.25\cdot(0.8+m)\), the total slope sum must satisfy \(0.8+m=2\). 3. Therefore, \(m=1.2\).

Answer

\(m=1.2\).

Euler's method 28

One Euler step of positive size \(h\) is applied to \(y'=-y^2\), starting from \(y(0)=2\). Find all values of \(h\) for which the first Euler approximation is nonnegative. What happens at the largest allowable step size?

Hints

- Write the first Euler update as a function of the step size. - Translate nonnegative into an inequality. - Evaluate the update and the differential equation at the boundary value of the inequality.

Solution

1. The initial slope is \(-2^2=-4\), so the first approximation is \(y_1=2-4h\). 2. Requiring \(y_1\ge0\) gives \(2-4h\ge0\), so \(0<h\le\frac{1}{2}\). 3. At \(h=\frac{1}{2}\), the approximation lands exactly at \(y_1=0\), where the differential equation has slope \(0\).

Answer

\(0<h\le\frac{1}{2}\). At \(h=\frac{1}{2}\), the first approximation is \(0\), an equilibrium value of the differential equation.

Euler's method 29

Euler's method with a constant positive step size \(h\) is applied to \(y'=y(4-y)\), starting from \(y(0)=1\). The first approximation is \(y_1=3\). Determine \(h\), and then compute the second approximation \(y_2\) using the same step size.

Hints

- Use the first prescribed approximation to solve for the step size. - Reevaluate the slope at the new approximation. - Apply the same step size in the second update.

Solution

1. At \(y_0=1\), the slope is \(1\cdot(4-1)=3\). 2. The first update gives \(3=1+3h\), so \(h=\frac{2}{3}\). 3. At \(y_1=3\), the slope is \(3\cdot(4-3)=3\). 4. The second update is \(y_2=3+\frac{2}{3}\cdot3=5\).

Answer

\(h=\frac{2}{3}\), and \(y_2=5\).

Euler's method 30

Euler's method with step size \(h=0.5\) is applied to \(y'=y(2-y)\), starting from \(y(0)=3\). Priya correctly finds \(y_1=1.5\), then reuses the original slope \(-3\) and reports \(y_2=0\). a) Explain the error. b) Compute the correct value of \(y_2\). c) Describe how the two correct approximations move relative to the equilibrium \(y=2\).

Hints

- Identify the point at which the second Euler slope must be evaluated. - Substitute the updated approximation into \(y(2-y)\). - Compare each approximation with the equilibrium value before describing its motion.

Solution

1. Euler's method requires a new slope at each updated point; the original slope cannot be reused automatically. 2. At \(y_1=1.5\), the new slope is \(1.5\cdot(2-1.5)=0.75\). 3. Therefore, \(y_2=1.5+0.5\cdot0.75=1.875\). 4. The first step crosses from above the equilibrium \(y=2\) to below it. The second step moves upward toward the equilibrium from below.

Answer

a) The slope must be recomputed at \(y_1=1.5\). b) \(y_2=1.875\) c) The first step overshoots below \(2\), and the second moves back upward toward \(2\).

Euler's method 31

Use Euler's method with step size \(h=0.25\) to approximate \(y(1)\) for \(y'=x-y\), \(y(0)=0\). The exact solution is \(y=x-1+e^{-x}\). State whether the Euler approximation is an overestimate or an underestimate.

Hints

- Update both coordinates after every step. - Use the new point to evaluate the next slope. - Compare the final approximation with the supplied exact solution.

Solution

1. Starting at \((0,0)\), the four Euler approximations are \(y_1=0\), \(y_2=0.0625\), \(y_3=0.171875\), and \(y_4=0.31640625\). 2. Thus Euler's method gives \(y(1)\approx0.31640625\). 3. The exact value is \(y(1)=e^{-1}\approx0.367879\). 4. Since \(0.31640625<0.367879\), the Euler approximation is an underestimate.

Answer

\(y(1)\approx0.31640625\), an underestimate of the exact value \(e^{-1}\).

Euler's method 32

Euler's method is applied to \(y'=1+y^2\), starting from \(y(0)=0\). The first step has size \(0.2\). The second step has an unknown positive size \(h_2\), and the second approximation is \(y_2=0.512\). Determine \(h_2\) and the x-coordinate \(x_2\).

Hints

- Complete the first Euler update before using the unknown second step. - Evaluate the second slope at the current approximation. - The second x-coordinate is the sum of the two step sizes.

Solution

1. The initial slope is \(1\), so the first approximation is \(y_1=0+0.2\cdot1=0.2\) at \(x_1=0.2\). 2. The next slope is \(1+(0.2)^2=1.04\). 3. The second update gives \(0.512=0.2+1.04h_2\), so \(h_2=0.3\). 4. Therefore, \(x_2=0.2+0.3=0.5\).

Answer

\(h_2=0.3\), and \(x_2=0.5\).

Euler's method 33

Euler's method with step size \(h=0.5\) is applied to \(y'=-2xy\), starting from \(y(0)=1\). Compute the first four approximations \(y_1,y_2,y_3,y_4\), and explain why the later approximations stop changing.

Hints

- Track the x-coordinate as well as the approximation at every step. - Evaluate the slope at the current Euler point. - Notice what happens to the differential equation when an approximation becomes zero.

Solution

1. At \(x_0=0\), the slope is \(0\), so \(y_1=1\). 2. At \(x_1=0.5\), the slope is \(-1\), so \(y_2=1+0.5\cdot(-1)=0.5\). 3. At \(x_2=1\), the slope is \(-1\), so \(y_3=0.5+0.5\cdot(-1)=0\). 4. Once an approximation equals \(0\), the slope \(-2xy\) is \(0\), so \(y_4=0\) and every later approximation remains \(0\).

Answer

\(y_1=1\), \(y_2=0.5\), \(y_3=0\), and \(y_4=0\). The approximations remain at \(0\) because \(y=0\) is an equilibrium.

Euler's method 34

Euler's method with step size \(h=2\) is applied to \(y'=y(1-y)\), starting from \(y(0)=0.25\). Compute the first three approximations and describe their motion relative to the equilibrium \(y=1\).

Hints

- Evaluate the nonlinear slope at each current approximation. - Compare every new value with the equilibrium. - Use the slope sign above and below the equilibrium to interpret the motion.

Solution

1. \(y_1=0.25+2\cdot0.25\cdot0.75=0.625\). 2. \(y_2=0.625+2\cdot0.625\cdot0.375=1.09375\). 3. \(y_3=1.09375+2\cdot1.09375\cdot(-0.09375)=0.888671875\). 4. The approximations begin below the equilibrium, cross above it on the second step, and return below it on the third step. The large step size produces oscillation around \(y=1\).

Answer

\(y_1=0.625\), \(y_2=1.09375\), and \(y_3=0.888671875\). The approximations oscillate around \(y=1\).

Euler's method 35

One Euler step of size \(h>0\) is used for \(y'=2x+1\), \(y(0)=0\). Find all step sizes for which the absolute one-step error at \(x=h\) is at most \(0.04\).

Hints

- Compute the one-step Euler value symbolically. - Solve the differential equation exactly for comparison. - Translate the error tolerance into an inequality for \(h\).

Solution

1. The initial slope is \(1\), so Euler's method gives \(y_E(h)=h\). 2. The exact solution is \(y=x^2+x\), so \(y(h)=h^2+h\). 3. The absolute error is \(h^2\). 4. Requiring \(h^2\le0.04\) and \(h>0\) gives \(0<h\le0.2\).

Answer

\(0<h\le0.2\).

Euler's method 36

Euler's method with a constant positive step size \(h\) is applied to \(y'=x^2-y\), starting from \(y(0)=0\). The second approximation is \(\frac{1}{8}\). Determine \(h\) and compute the third approximation.

Hints

- The first slope is zero, so the first approximation stays at zero. - Express the second approximation as a power of \(h\). - Use the new x-coordinate and approximation for the third slope.

Solution

1. The initial slope is \(0\), so \(y_1=0\) at \(x_1=h\). 2. The next slope is \(h^2\), so \(y_2=h^3\). 3. Since \(h^3=\frac{1}{8}\) and \(h>0\), \(h=\frac{1}{2}\). 4. At \((x_2,y_2)=\left(1,\frac{1}{8}\right)\), the slope is \(1-\frac{1}{8}=\frac{7}{8}\). 5. Thus \(y_3=\frac{1}{8}+\frac{1}{2}\cdot\frac{7}{8}=\frac{9}{16}\).

Answer

\(h=\frac{1}{2}\), and \(y_3=\frac{9}{16}\).

Euler's method 37

a) What slope does the displayed segment through \(A\) have? b) If Euler's method starts at \(A\) and uses step size \(h=0.5\), what value does one step give for \(y(0.5)\)?

Figure for problem 556042

Hints

- Read the local tangent direction at the marked point rather than a nearby segment. - Euler's method follows that local slope for one horizontal step.

Solution

1. The segment through \(A\) rises \(1\) unit for each \(1\) unit of horizontal change, so its slope is \(1\). 2. One Euler step gives \(y(0.5)\approx1+0.5\cdot1=1.5\).

Answer

a) \(1\) b) \(y(0.5)\approx1.5\)

Euler's method 38

Use the slope field to make two Euler steps of size \(h=1\), starting at \(A\). Give both new Euler points. No formula for the differential equation is supplied.

Figure for problem 556058

Hints

- Find the slope of the segment at the current point before taking a step. - With \(h=1\), the vertical change equals the current slope. - Read a new slope after reaching the first estimated point.

Solution

1. At \(A=(0,0)\), the segment has slope \(1\). The first Euler step gives \(y_1=0+1\cdot1=1\), so the next point is \((1,1)\). 2. At \((1,1)\), the segment is horizontal, with slope \(0\). The second step gives \(y_2=1+1\cdot0=1\), so the next point is \((2,1)\).

Answer

First point: \((1,1)\) Second point: \((2,1)\)

Euler's method 39

The paths p and q represent two proposed Euler approximations from A for \(y'=x+y\), with step size \(h=1\). Which path correctly applies Euler's method for two steps? Explain what slope is used at the second step.

Figure for problem 562990

Hints

- Euler recomputes the slope at each new approximation point. - Check the first update from A. - Then evaluate \(x+y\) at the updated point, not at A.

Solution

1. Starting at \((0,1)\), the first slope is \(0+1=1\), giving \((1,2)\). 2. The new slope is \(1+2=3\), not the original slope. 3. The next height is \(2+1\cdot3=5\), so path p is correct.

Answer

Path p; the second step uses slope \(3\) at \((1,2)\).

Euler's method 40

For \(y'=2x\) with \(y(0)=0\), paths p and q are Euler approximations with different step sizes. The smooth curve c is the exact solution. Which path uses the smaller step, and which gives the closer estimate at \(x=2\)? Explain.

Figure for problem 562991

Hints

- Count the horizontal increments along each path. - A smaller step produces more vertices over the same interval. - Compare the final heights with the smooth solution curve.

Solution

1. Path p has two steps of size \(1\); its estimate at \(x=2\) is \(2\). 2. Path q has four steps of size \(0.5\); its estimate at \(x=2\) is \(3\). 3. The exact curve has height \(4\) at \(x=2\), so q is closer.

Answer

Path q uses the smaller step, \(h=0.5\), and is closer at \(x=2\) (estimate \(3\) versus exact \(4\)).

Euler's method 41

A model for the mass \(M(t)\) of medicine remaining in the body is \(M'=-0.1M\), with \(t\) in hours and \(M(0)=100\,\text{mg}\). Use Euler's method with two steps of size \(h=1\) hour to estimate the amount after \(2\) hours.

Hints

- Begin with the rate at the initial amount. - Update the amount after the first hour. - Recompute the rate at the new amount before the second hour.

Solution

1. The starting rate is \(-0.1\cdot100=-10\,\text{mg/h}\), so \(M_1=100-10=90\,\text{mg}\). 2. At the new amount the rate is \(-0.1\cdot90=-9\,\text{mg/h}\), so \(M_2=90-9=81\,\text{mg}\).

Answer

\(M(2)\approx81\,\text{mg}\) by Euler's method.

Euler's method 42

Starting at \(A\), make one Euler step with \(h=0.5\) and one with \(h=1\). Give the new point for each step size. For which step size does the field provide the slope needed for a second Euler step? Complete that second step and explain why the other step size cannot be continued from the displayed information alone.

Figure for problem 562993

Hints

- The two first steps use the same starting slope but different horizontal distances. - Locate each new point on the graph before trying to take another step. - What information does Euler's method need at a new point that this diagram may not show?

Solution

1. The displayed slope at \(A=(0,1)\) is \(1\). With \(h=0.5\), the first Euler point is \((0.5,1.5)\). With \(h=1\), it is \((1,2)\). 2. A field segment is centered at \((0.5,1.5)\) and has slope \(2\), so a second step with \(h=0.5\) gives \((1,2.5)\). 3. No field segment is centered at \((1,2)\). Without a formula or a rule for estimating missing slopes, a second \(h=1\) update cannot be determined from this field alone.

Answer

With \(h=0.5\): first point \((0.5,1.5)\), second point \((1,2.5)\). With \(h=1\): first point \((1,2)\); the second step is not determined by the displayed slopes.

Euler's method 43

For \(y'=x+y\) with \(y(0)=1\), use Euler's method to approximate \(y(1)\) twice: first with \(h=1\), then with \(h=0.5\). Compare the approximations without claiming which is exact.

Hints

- The step size \(1\) requires one update, while \(0.5\) requires two updates to reach the same endpoint. - Use the slope \(x_n+y_n\) at the current point in each computation. - Compare the two approximations numerically without labeling either one exact.

Solution

1. With \(h=1\), the initial slope is \(1\), so the single update gives \(y(1)\approx1+1\cdot1=2\). 2. With \(h=0.5\), the first update gives \(y(0.5)\approx1+0.5\cdot1=1.5\). 3. At \((0.5,1.5)\), the slope is \(2\), so the second update gives \(y(1)\approx1.5+0.5\cdot2=2.5\). 4. The approximations differ by \(2.5-2=0.5\). The smaller step uses more tangent-line updates.

Answer

With \(h=1\): \(2\) With \(h=0.5\): \(2.5\) Difference: \(0.5\)

Euler's method 44

For \(y'=x^2-y\) with \(y(0)=1\), use Euler's method to approximate \(y(0.8)\) twice: first with \(h=0.4\), then with \(h=0.2\). Compare the approximations without claiming which is exact. Report each result to four decimal places.

Hints

- Two steps of \(0.4\) and four steps of \(0.2\) both reach \(x=0.8\). - Use \(x_n^2-y_n\) at each current point in both Euler sequences. - Keep each computation separate and compare the two four-decimal endpoint estimates.

Solution

1. With \(h=0.4\), the two updates give \(y_1=1+0.4\cdot(-1)=0.6\) and \(y_2=0.6+0.4\cdot(-0.44)=0.424\). Thus, \(y(0.8)\approx0.4240\). 2. With \(h=0.2\), the four updates give \(y_1=0.8\), \(y_2=0.648\), \(y_3=0.5504\), and \(y_4=0.51232\). Thus, \(y(0.8)\approx0.5123\). 3. The approximations differ by \(0.51232-0.424=0.08832\approx0.0883\). The smaller step uses more tangent-line updates.

Answer

With \(h=0.4\): \(0.4240\) With \(h=0.2\): \(0.5123\) Difference: \(0.0883\)

Euler's method 45

For \(y'=ay+x\), \(y(0)=2\), one Euler step of size \(0.5\) gives \(y(0.5)=3\). Find \(a\).

Hints

- Evaluate the differential equation at the initial point. - Substitute that slope into the one-step Euler formula. - Solve the resulting equation for \(a\).

Solution

1. At the initial point \((0,2)\), the slope is \(a\cdot2+0=2a\). 2. Euler's update gives \(3=2+0.5\cdot2a=2+a\). 3. Therefore, \(a=1\).

Answer

\(a=1\)

Euler's method 46

An Euler approximation for \(y'=y\), \(y(0)=1\), uses \(h=0.25\). Without doing four separate recomputations, express the approximation to \(y(1)\) as a power and evaluate it.

Hints

- Rewrite one Euler update as a multiplier applied to the current value. - Determine how many equal steps reach \(x=1\). - Apply the same multiplier as a power.

Solution

1. Euler's update is \(y_{n+1}=y_n+0.25y_n=1.25y_n\). 2. Four steps are needed to move from \(x=0\) to \(x=1\). 3. Therefore, \(y(1)\approx1\cdot(1.25)^4=2.44140625\).

Answer

\(y(1)\approx(1.25)^4=2.44140625\)

Euler's method 47

Euler's method with step size \(h=0.25\) is applied to \(y'=ax+by\), starting from \(y(0)=1\). The first two approximations are \(y(0.25)=1.5\) and \(y(0.5)=2.5\). Determine \(a\) and \(b\).

Hints

- Write the Euler update separately for each given step. - The first update involves only one of the unknown parameters. - Use the newly approximated point when evaluating the second slope.

Solution

1. The first update gives \(1.5=1+0.25b\), so \(b=2\). 2. At \((0.25,1.5)\), the slope is \(0.25a+1.5b=0.25a+3\). 3. The second update gives \(2.5=1.5+0.25(0.25a+3)\). 4. Solving yields \(0.25a+3=4\), so \(a=4\).

Answer

\(a=4\) and \(b=2\).

Euler's method 48

Euler's method with step size \(h=0.2\) is applied to \(y'=\frac{1}{1+x^2}\), \(y(0)=0\). Complete the table, then give the approximation to \(y(0.6)\) rounded to four decimal places. <table> <tr><th>\(n\)</th><th>\(x_n\)</th><th>\(y_n\)</th><th>Slope used</th></tr> <tr><td>0</td><td>0</td><td>0</td><td>\(1\)</td></tr> <tr><td>1</td><td>0.2</td><td></td><td>\(\frac{25}{26}\)</td></tr> <tr><td>2</td><td>0.4</td><td></td><td>\(\frac{25}{29}\)</td></tr> <tr><td>3</td><td>0.6</td><td></td><td>—</td></tr> </table> Explain why the total Euler update is a left Riemann sum in this problem.

Hints

- Use \(y_{n+1}=y_n+h f(x_n,y_n)\) one row at a time. - Keep the fractions exact until the final row. - Compare each slope's x-coordinate with the left endpoint of its width-\(0.2\) subinterval.

Solution

1. The first update gives \(y_1=0+0.2\cdot1=\frac{1}{5}\). 2. The second update gives \(y_2=\frac{1}{5}+0.2\cdot\frac{25}{26}=\frac{51}{130}\). 3. The third update gives \(y_3=\frac{51}{130}+0.2\cdot\frac{25}{29}=\frac{2129}{3770}\approx0.5647\). 4. Because the derivative depends only on \(x\), each Euler increment is the subinterval width times the derivative evaluated at the left endpoint. Their sum is therefore a left Riemann sum for \(\int_0^{0.6}\frac{1}{1+x^2}\,dx\).

Answer

\(y_1=\frac{1}{5}\), \(y_2=\frac{51}{130}\), and \(y_3=\frac{2129}{3770}\approx0.5647\). The increments use left-endpoint values of \(\frac{1}{1+x^2}\), so their sum is a left Riemann sum.

Euler's method 49

Euler's method with step size \(h=0.5\) is applied to \(y'=2-y\). After three steps starting at \(x=0\), the approximation is \(y(1.5)=2.25\). Determine the initial value \(y(0)\).

Hints

- Simplify the Euler update into a recurrence involving only the current approximation. - Apply the recurrence symbolically several times rather than guessing the starting value. - Use the given final approximation to solve for the remaining unknown.

Solution

1. The Euler update is \(y_{n+1}=y_n+0.5(2-y_n)=1+0.5y_n\). 2. Repeated substitution gives \(y_1=1+0.5y_0\), \(y_2=1.5+0.25y_0\), and \(y_3=1.75+0.125y_0\). 3. Using \(y_3=2.25\), \(2.25=1.75+0.125y_0\), so \(y_0=4\).

Answer

\(y(0)=4\).

Euler's method 50

Euler's method with step size \(h=0.5\) is applied to \(y'=x-y\), starting at \(x=0\) with an unknown value \(y(0)=a\). After two steps, the approximation at \(x=1\) equals the starting value \(a\). Determine \(a\).

Hints

- Keep the unknown starting value symbolic through both updates. - Evaluate the slope at the newly approximated point for the second step. - Use the stated return condition only after expressing the second approximation in terms of the start.

Solution

1. The first update is \(y_1=a+0.5(0-a)=\frac{a}{2}\). 2. The second update is \(y_2=\frac{a}{2}+0.5\left(0.5-\frac{a}{2}\right)=\frac{1}{4}+\frac{a}{4}\). 3. The condition \(y_2=a\) gives \(\frac{1}{4}+\frac{a}{4}=a\), so \(a=\frac{1}{3}\).

Answer

\(a=\frac{1}{3}\).

Euler's method 51

Euler's method is applied to \(y'=x+y\), starting from \(y(0)=1\), using two equal positive steps of size \(h\). The approximation after the second step is \(2\). Determine \(h\) and the x-coordinate of the final approximation.

Hints

- Keep the step size symbolic through both Euler updates. - Recalculate the second slope at the first approximated point. - Use the positive root of the resulting quadratic equation.

Solution

1. The first update gives \(y_1=1+h\). 2. At \((h,1+h)\), the slope is \(1+2h\), so \(y_2=1+h+h(1+2h)=1+2h+2h^2\). 3. Setting \(y_2=2\) gives \(2h^2+2h-1=0\). 4. The positive solution is \(h=\frac{\sqrt{3}-1}{2}\), and the final x-coordinate is \(2h=\sqrt{3}-1\).

Answer

\(h=\frac{\sqrt{3}-1}{2}\), and the final x-coordinate is \(\sqrt{3}-1\).

Euler's method 52

Euler's method is applied to \(y'=1-y\), starting from \(y(0)=4\), using a constant positive step size \(h\). After three steps, the approximation is \(1.375\). Determine \(h\).

Hints

- Rewrite the Euler update relative to the equilibrium value. - Look for a repeated multiplicative factor from one step to the next. - Use the given third approximation to solve for that factor.

Solution

1. The Euler update is \(y_{n+1}=y_n+h(1-y_n)\). 2. Subtracting the equilibrium value gives \(y_{n+1}-1=(1-h)(y_n-1)\). 3. Therefore, \(y_3-1=(1-h)^3(y_0-1)\), so \(0.375=3(1-h)^3\). 4. Thus \((1-h)^3=\frac{1}{8}\), giving \(1-h=\frac{1}{2}\) and \(h=\frac{1}{2}\).

Answer

\(h=\frac{1}{2}\).

Euler's method 53

Euler's method with step size \(h=0.2\) is applied to an autonomous linear differential equation \(y'=a+by\). Starting from \(y_0=1\), the first two approximations are \(y_1=1.4\) and \(y_2=1.24\). Determine \(a\) and \(b\), then compute \(y_3\).

Hints

- Convert each known Euler update into an equation for the unknown slope parameters. - Use the current approximation, not the starting value, in the second equation. - After finding the differential equation, perform one more update.

Solution

1. The first update gives \(1.4=1+0.2(a+b)\), so \(a+b=2\). 2. The second update gives \(1.24=1.4+0.2(a+1.4b)\), so \(a+1.4b=-0.8\). 3. Subtracting the equations gives \(0.4b=-2.8\), so \(b=-7\) and \(a=9\). 4. At \(y_2=1.24\), the slope is \(9-7\cdot1.24=0.32\). 5. Thus \(y_3=1.24+0.2\cdot0.32=1.304\).

Answer

\(a=9\), \(b=-7\), and \(y_3=1.304\).

Euler's method 54

Euler's method with step size \(h=0.25\) is applied twice to \(y'=x+y\), starting from \(y(0)=a\). The second approximation is \(2\). Determine \(a\).

Hints

- Keep the unknown initial value symbolic through both updates. - Include the new x-coordinate in the second slope. - Solve the final linear equation for the starting value.

Solution

1. The first update is \(y_1=a+0.25a=\frac{5}{4}a\). 2. At \(x_1=0.25\), the slope is \(0.25+\frac{5}{4}a\). 3. The second update is \(y_2=\frac{5}{4}a+\frac{1}{4}\left(\frac{1}{4}+\frac{5}{4}a\right)=\frac{25}{16}a+\frac{1}{16}\). 4. Setting \(y_2=2\) gives \(25a+1=32\), so \(a=\frac{31}{25}\).

Answer

\(a=\frac{31}{25}\).

Euler's method 55

Euler's method with \(n\) equal steps is used to approximate \(y(1)\) for \(y'=x\), \(y(0)=0\). Find the smallest positive integer \(n\) for which the Euler approximation has absolute error less than \(0.01\).

Hints

- Express the Euler approximation as a left Riemann sum. - Use the formula for the sum of the first \(n-1\) integers. - Compare the simplified approximation with the exact integral value.

Solution

1. The step size is \(h=\frac{1}{n}\), and Euler's method gives the left sum \(y_n=h\sum_{j=0}^{n-1}jh\). 2. Thus \(y_n=\frac{1}{n^2}\cdot\frac{n(n-1)}{2}=\frac{n-1}{2n}\). 3. The exact value is \(y(1)=\frac{1}{2}\). 4. The error is \(\frac{1}{2}-\frac{n-1}{2n}=\frac{1}{2n}\). 5. Requiring \(\frac{1}{2n}<0.01\) gives \(n>50\), so the smallest integer is \(51\).

Answer

\(n=51\).

Euler's method 56

Euler's method is used for \(y'=x^2\), \(y(0)=2\), with successive x-values \(0\), \(0.3\), \(0.8\), and \(1\). Find the Euler approximation to \(y(1)\), then compare it with the exact value.

Hints

- Determine each step size from consecutive x-values. - Use the slope at the left endpoint of each step. - Integrate the differential equation to obtain the exact comparison value.

Solution

1. The unequal step sizes are \(0.3\), \(0.5\), and \(0.2\). 2. The Euler approximation is \(2+0.3\cdot0^2+0.5\cdot0.3^2+0.2\cdot0.8^2=2.173\). 3. The exact solution is \(y=2+\frac{x^3}{3}\), so \(y(1)=\frac{7}{3}\approx2.333\). 4. The Euler value is an underestimate by \(\frac{7}{3}-2.173\approx0.1603\).

Answer

Euler approximation: \(2.173\). Exact value: \(\frac{7}{3}\approx2.333\). The approximation is low by about \(0.1603\).

Euler's method 57

Euler's method with a constant positive step size \(h\) is applied twice to \(y'=-3y\), starting from \(y(0)=2\). The second approximation is \(0.5\). Find all possible values of \(h\), and state whether each produces monotone or alternating approximations around the equilibrium.

Hints

- Rewrite the Euler update as multiplication by one constant factor. - Use the two-step condition to determine the possible factors. - The sign of the factor determines whether approximations alternate.

Solution

1. Each Euler step multiplies the approximation by \(1-3h\). 2. Thus \(0.5=2(1-3h)^2\), so \((1-3h)^2=\frac{1}{4}\). 3. If \(1-3h=\frac{1}{2}\), then \(h=\frac{1}{6}\). The factor is positive, so the approximations approach \(0\) without alternating signs. 4. If \(1-3h=-\frac{1}{2}\), then \(h=\frac{1}{2}\). The factor is negative, so the approximations alternate signs around \(0\).

Answer

\(h=\frac{1}{6}\) gives monotone sign-preserving decay, and \(h=\frac{1}{2}\) gives alternating decay.

Euler's method 58

Euler's method with a constant positive step size \(h\) is applied to \(y'=2-y\), starting from \(y(0)=5\). Find the step size for which each approximation is reflected to the opposite side of the equilibrium \(y=2\) at the same distance. Compute the first two approximations.

Hints

- Rewrite the update in terms of distance from equilibrium. - A reflection changes the sign of that distance without changing its magnitude. - Use the resulting step size in two updates.

Solution

1. Relative to equilibrium, the Euler update is \(y_{n+1}-2=(1-h)(y_n-2)\). 2. Reflection to the opposite side at the same distance requires the multiplier \(-1\), so \(1-h=-1\) and \(h=2\). 3. Starting from \(y_0=5\), \(y_1=5+2\cdot(2-5)=-1\). 4. Then \(y_2=-1+2\cdot(2-(-1))=5\).

Answer

\(h=2\), with \(y_1=-1\) and \(y_2=5\). The approximations alternate forever at equal distances from \(y=2\).

Euler's method 59

Euler's method approximates \(y(1)\) for \(y'=x\), \(y(0)=0\), using exactly two positive steps. The first step has size \(a\), and the second has size \(1-a\), where \(0<a<1\). Find the value of \(a\) that makes the Euler approximation as large as possible, and find that maximum approximation.

Hints

- The first left-endpoint slope is zero. - Express the second contribution using both step sizes. - Maximize the resulting quadratic on the allowed interval.

Solution

1. The first slope is \(0\), so after the first step the approximation is still \(0\) at \(x=a\). 2. The second slope is \(a\), and the second step has size \(1-a\). 3. Thus the final approximation is \(E(a)=a(1-a)\). 4. Since \(E'(a)=1-2a\) and \(E''(a)=-2<0\), the maximum on \((0,1)\) occurs at \(a=\frac{1}{2}\). 5. The maximum approximation is \(E\left(\frac{1}{2}\right)=\frac{1}{4}\).

Answer

\(a=\frac{1}{2}\), and the maximum Euler approximation is \(\frac{1}{4}\).

Euler's method 60

Euler's method with a constant positive step size \(h\) is applied to \(y'=x-y\), starting from \(y(0)=1\). The first and second Euler approximations are equal. Determine \(h\) and compute the first three approximations.

Hints

- Write the first approximation as a function of \(h\). - Equal consecutive approximations mean the intervening Euler slope is zero. - Use the resulting step size for one additional update.

Solution

1. The first approximation is \(y_1=1-h\), at \(x_1=h\). 2. Equality \(y_2=y_1\) requires the second-step slope to be zero: \(x_1-y_1=h-(1-h)=2h-1=0\). 3. Thus \(h=\frac{1}{2}\). 4. Then \(y_1=\frac{1}{2}\) and \(y_2=\frac{1}{2}\). 5. At \(x_2=1\), the slope is \(1-\frac{1}{2}=\frac{1}{2}\), so \(y_3=\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}=\frac{3}{4}\).

Answer

\(h=\frac{1}{2}\), with \(y_1=\frac{1}{2}\), \(y_2=\frac{1}{2}\), and \(y_3=\frac{3}{4}\).

Euler's method 61

Euler's method is applied to \(y'=6-2y\) with one positive step of size \(h\). Find the unique step size for which the first Euler approximation is independent of the starting value \(y_0\). What value does the method produce after that step, and what happens on later steps with the same size?

Hints

- Write the one-step update as a linear expression in the starting value. - Make the coefficient of the starting value equal to zero. - Check the differential equation at the resulting approximation.

Solution

1. The first update is \(y_1=y_0+h(6-2y_0)=(1-2h)y_0+6h\). 2. Independence from \(y_0\) requires \(1-2h=0\), so \(h=\frac{1}{2}\). 3. Then \(y_1=6\cdot\frac{1}{2}=3\), the equilibrium value. 4. At \(y=3\), the slope is \(6-2\cdot3=0\), so every later approximation remains \(3\).

Answer

\(h=\frac{1}{2}\). The first approximation is \(3\) for every starting value, and all later approximations remain \(3\).

Euler's method 62

Euler's method with step size \(h=0.5\) is applied to \(y'=y^2\), \(y(0)=1\), with the intention of approximating \(y(1)\). Compute the two Euler updates, then explain why the final number is not a valid approximation to \(y(1)\).

Hints

- Carry out the Euler updates exactly as stated. - Solve the separable initial-value problem independently. - Check whether the exact solution exists at the intended endpoint.

Solution

1. The first update is \(y_1=1+0.5\cdot1^2=1.5\). 2. The second update is \(y_2=1.5+0.5\cdot1.5^2=2.625\). 3. The exact solution of the initial-value problem is \(y=\frac{1}{1-x}\). 4. This solution becomes unbounded at \(x=1\), so there is no finite value \(y(1)\). 5. Therefore, the algorithmic output \(2.625\) cannot be interpreted as an approximation to the nonexistent finite endpoint value.

Answer

The Euler updates are \(y_1=1.5\) and \(y_2=2.625\). The number \(2.625\) is not a valid approximation to \(y(1)\) because the exact solution \(\frac{1}{1-x}\) blows up at \(x=1\).

Euler's method 63

The graph compares the exact solution c of \(y'=y\), \(y(0)=1\), with an Euler path p using \(h=0.5\). Find the Euler estimate at \(x=1\), calculate its absolute error to three decimal places, and explain why the Euler path lies below the exact curve.

Figure for problem 562994

Hints

- Use the two straight segments to understand how Euler builds its estimate. - Compare the final Euler height with the exact exponential height. - Think about how an upward-bending curve sits relative to its tangent lines.

Solution

1. The Euler steps give \(y_1=1+0.5\cdot1=1.5\) and \(y_2=1.5+0.5\cdot1.5=2.25\). 2. The exact solution is \(y=e^x\), so \(y(1)=e\approx2.718282\). 3. The absolute error is \(e-2.25\approx0.468\). 4. Since \(y''=e^x>0\), the exact curve bends upward above each forward tangent step, so Euler underestimates.

Answer

Euler estimate: \(2.25\). Absolute error: \(\approx0.468\). The upward-curving exact solution lies above the tangent-based Euler segments.

Euler's method 64

A chemical concentration follows \(C'=-kC^2\), where \(C\) is in \(\text{mg/L}\), \(t\) is in days, and \(k=1\,\frac{\text{L}}{\text{mg}\cdot\text{day}}\). Initially \(C(0)=4\,\text{mg/L}\). A student uses one Euler step of \(h=0.3\) day. Find the estimate and explain the problem with it. What is the largest positive step size that would keep the first Euler estimate nonnegative? The exact solution is \(C(t)=\frac{4}{1+4t}\). Use it to check whether the concentration is still positive after \(0.3\) day.

Hints

- Evaluate the starting rate before multiplying by the time step. - Ask whether a concentration can be negative. - For the step-size limit, require the next estimate to stay at least zero. - The exact solution is supplied so you can compare without solving the differential equation.

Solution

1. The initial rate is \(C'(0)=-1\cdot4^2=-16\,\frac{\text{mg}}{\text{L}\cdot\text{day}}\). 2. Euler with \(h=0.3\) gives \(C_1=4-0.3\cdot16=-0.8\,\text{mg/L}\), which is not physically meaningful. 3. The first Euler value is nonnegative only when \(4-16h\ge0\), hence \(0<h\le0.25\) day. 4. The exact model gives \(C(0.3)=4/2.2\approx1.82\,\text{mg/L}\), showing that a large Euler step can give a qualitatively wrong estimate.

Answer

Euler estimate: \(-0.8\,\text{mg/L}\), which is impossible for a concentration. The largest permitted first step is \(h=0.25\) day; the exact concentration at \(0.3\) day is about \(1.82\,\text{mg/L}\).

Euler's method 65

Euler's method is applied to \(y'=\frac{y}{x}\), starting from \((1,3)\), using any sequence of positive step sizes that keeps \(x>0\). Prove that every Euler point lies exactly on the solution \(y=3x\).

Hints

- Check the slope along the proposed exact line. - Assume one Euler point lies on the line and examine the next update. - Use induction to extend the result to every step.

Solution

1. The initial point satisfies \(y_0=3x_0\). 2. Suppose an Euler point satisfies \(y_n=3x_n\). Then its slope is \(\frac{y_n}{x_n}=3\). 3. For a step of size \(h_n\), Euler's method gives \(y_{n+1}=y_n+3h_n=3x_n+3h_n=3(x_n+h_n)=3x_{n+1}\). 4. By induction, every Euler point lies on \(y=3x\), which is the exact solution through \((1,3)\).

Answer

Every Euler point satisfies \(y_n=3x_n\), so Euler's method is exact for this initial-value problem for any allowed step sequence.

Euler's method 66

Use Euler's method with step size \(h=0.5\) to approximate \(y(1)\) for \(y'=\frac{1}{y}\), \(y(0)=1\). Compare the result with the exact solution and explain the error direction using concavity.

Hints

- Reevaluate the reciprocal slope after the first update. - Solve the differential equation exactly for comparison. - Determine the sign of the second derivative from the differential equation.

Solution

1. The first approximation is \(y_1=1+0.5\cdot1=1.5\). 2. The second approximation is \(y_2=1.5+0.5\cdot\frac{1}{1.5}=\frac{11}{6}\approx1.8333\). 3. The exact solution is \(y=\sqrt{2x+1}\), so \(y(1)=\sqrt{3}\approx1.7321\). 4. Since \(y''=-\frac{1}{y^3}<0\), the solution is concave down. Left-endpoint tangent steps lie above the curve, so Euler's method overestimates.

Answer

Euler approximation: \(\frac{11}{6}\approx1.8333\). Exact value: \(\sqrt{3}\approx1.7321\). The Euler value is an overestimate.

Euler's method 67

One Euler step of positive size \(h\) is applied to \(y'=x-y\), starting from \(y(0)=1\). The exact solution is \(y=x-1+2e^{-x}\). Prove that the Euler approximation at \(x=h\) is an underestimate for every \(h>0\).

Hints

- Write both the one-step approximation and the exact value at the same x-coordinate. - Simplify their difference. - Use the standard tangent-line inequality for the exponential function.

Solution

1. The initial slope is \(-1\), so Euler's method gives \(y_E(h)=1-h\). 2. The exact value is \(y(h)=h-1+2e^{-h}\). 3. Their difference is \(y(h)-y_E(h)=2(h-1+e^{-h})\). 4. For \(h>0\), the strict inequality \(e^{-h}>1-h\) gives \(h-1+e^{-h}>0\). 5. Therefore, \(y(h)>y_E(h)\) for every positive step size.

Answer

The one-step Euler value \(1-h\) is strictly less than the exact value for every \(h>0\).

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