Differential Equations Quiz: Improved Euler Heuns Method

What this quiz covers

This quiz focuses on Improved Euler Heuns Method, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Use linear approximation $\sin(0.0524) \approx 0.0524$ to avoid computational complexity
2. Calculate $\sin(0.0524)$ exactly using series expansion to maintain method accuracy
3. Use calculator or numerical evaluation of $\sin(0.0524)$ as the method requires (correct answer)
4. Replace $\sin(xy)$ with $xy$ approximation since the argument is small

Explanation: Heun's method requires accurate evaluation of the function $f(x,y)$ at both predictor and corrector stages. Using approximations (choices A, D) or unnecessary series expansions (choice B) defeats the purpose of the numerical method. Standard numerical evaluation is appropriate and necessary. Choice A introduces additional error. Choice B is unnecessarily complex. Choice D changes the differential equation being solved.

Question 2

1. $0$ (correct answer)
2. $h^2$
3. Order $O(h^2)$ but non-zero
4. Order $O(h^3)$ but non-zero

Explanation: Heun's method has a local truncation error of $O(h^3)$, meaning it can find the exact value for polynomial solutions of degree 2 or less in a single step. First, find the exact solution to the IVP by integration: $y(x) = \int (2x - 3) dx = x^2 - 3x + C$. Using $y(1)=5$, we have $5 = 1^2 - 3(1) + C \implies 5 = -2 + C \implies C=7$. The exact solution is $y(x) = x^2 - 3x + 7$. The exact value at $x=1+h$ is $y(1+h) = (1+h)^2 - 3(1+h) + 7 = (1+2h+h^2) - (3+3h) + 7 = h^2 - h + 5$. Now, apply one step of Heun's method from $(x_0, y_0) = (1,5)$: $f(x,y) = 2x-3$. $m_1 = f(1, 5) = 2(1) - 3 = -1$. Predictor: $y_1^* = y_0 + h \cdot m_1 = 5 + h(-1) = 5-h$. Corrector slope: $m_2 = f(1+h, 5-h) = 2(1+h) - 3 = 2+2h-3 = 2h-1$. Approximation: $y_1 = y_0 + \frac{h}{2}(m_1 + m_2) = 5 + \frac{h}{2}(-1 + 2h-1) = 5 + \frac{h}{2}(2h-2) = 5 + h^2 - h$. The error is the exact value minus the approximation: $(h^2 - h + 5) - (5 + h^2 - h) = 0$. The approximation is exact.

Question 3

1. Euler's Method
2. Improved Euler (Heun's) Method (correct answer)
3. Midpoint Method
4. Classic Fourth-Order Runge-Kutta Method

Explanation: Let's analyze the given formulas. The first calculation is $k_1 = h \cdot f(x_n, y_n)$. This represents the increment in $y$ according to Euler's method. The term $y_n+k_1$ is therefore $y_n + h \cdot f(x_n, y_n)$, which is the predictor step in Heun's method, often denoted $y_{n+1}^*$. Substituting this into the second equation: $y_{n+1} = y_n + \frac{1}{2} (h f(x_n, y_n) + h f(x_n+h, y_{n+1}^*))$ Factoring out $h/2$: $y_{n+1} = y_n + \frac{h}{2} [f(x_n, y_n) + f(x_n+h, y_{n+1}^*)]$. This is the standard formula for the Improved Euler (Heun's) method, which averages the slope at the beginning of the interval, $f(x_n, y_n)$, and the estimated slope at the end of the interval, $f(x_n+h, y_{n+1}^*)$.

Question 4

1. $1$
2. $4$
3. $2$
4. $5$ (correct answer)

Explanation: When you encounter Heun's method problems, remember that this predictor-corrector technique uses two slope calculations: an initial predictor slope and a corrector slope based on the predicted y-value. Given the differential equation $y' = x + y^2$ with $y(0) = 1$ and step size $h = 1$, you need to find the corrector slope $m_2 = f(x_1, y_1^*)$ where the student already found $y_1^* = 2$. First, identify your coordinates. Starting at $(x_0, y_0) = (0, 1)$, after one step of size $h = 1$, you reach $x_1 = 0 + 1 = 1$. The corrector slope is simply the function $f(x,y) = x + y^2$ evaluated at the point $(x_1, y_1^*) = (1, 2)$. Substituting these values: $m_2 = f(1, 2) = 1 + 2^2 = 1 + 4 = 5$. This confirms answer D is correct. Looking at the wrong answers: A) $1$ represents only the $x$ component, ignoring the $y^2$ term entirely. B) $4$ captures only the $y^2 = 2^2$ component while forgetting to add the $x = 1$ term. C) $2$ likely comes from mistakenly using the predicted $y$-value itself rather than evaluating the function. The key strategy for Heun's method problems is to carefully track your coordinates and systematically substitute into the given differential equation. Always double-check that you're evaluating $f(x,y)$ at the correct point—many errors stem from using wrong coordinates or forgetting terms in the function.

Question 5

1. $1.088$
2. $1.472$
3. $1.428$
4. $1.278$ (correct answer)

Explanation: This problem requires two steps of Heun's method. Let $f(x,y)=4-y^2$. Step 1: Approximate $y(0.2)$ from $(x_0, y_0) = (0,0)$ with $h=0.2$. $m_1 = f(0,0) = 4 - 0^2 = 4$. $y_1^* = y_0 + h m_1 = 0 + 0.2(4) = 0.8$. $m_2 = f(x_1, y_1^*) = f(0.2, 0.8) = 4 - (0.8)^2 = 4 - 0.64 = 3.36$. $y_1 = y_0 + \frac{h}{2}(m_1+m_2) = 0 + \frac{0.2}{2}(4+3.36) = 0.1(7.36) = 0.736$. So $y(0.2) \approx 0.736$. Step 2: Approximate $y(0.4)$ from $(x_1, y_1) = (0.2, 0.736)$ with $h=0.2$. $m_1' = f(0.2, 0.736) = 4 - (0.736)^2 = 4 - 0.541696 = 3.458304$. $y_2^* = y_1 + h m_1' = 0.736 + 0.2(3.458304) = 0.736 + 0.6916608 = 1.4276608$. $m_2' = f(x_2, y_2^*) = f(0.4, 1.4276608) = 4 - (1.4276608)^2 \approx 4 - 2.03821 = 1.96179$. $y_2 = y_1 + \frac{h}{2}(m_1'+m_2') \approx 0.736 + \frac{0.2}{2}(3.4583 + 1.9618) = 0.736 + 0.1(5.4201) = 0.736 + 0.54201 = 1.27801$. The closest answer is $1.278$.

Question 6

1. $1.500$
2. $1.550$ (correct answer)
3. $1.625$
4. $1.750$

Explanation: We apply one step of Heun's method with $f(x,y) = \frac{2x}{y+1}$, $(x_0, y_0) = (1,1)$, and $h=0.5$.

1. Calculate the predictor slope $m_1$: $m_1 = f(1,1) = \frac{2(1)}{1+1} = 1$.
2. Calculate the predictor value $y_1^*$: $y_1^* = y_0 + h \cdot m_1 = 1 + 0.5(1) = 1.5$.
3. Calculate the corrector slope $m_2$ at $(x_1, y_1^*) = (1.5, 1.5)$: $m_2 = f(1.5, 1.5) = \frac{2(1.5)}{1.5+1} = \frac{3}{2.5} = \frac{6}{5} = 1.2$.
4. Calculate the final approximation $y_1$: $y_1 = y_0 + \frac{h}{2}(m_1+m_2) = 1 + \frac{0.5}{2}(1+1.2) = 1 + 0.25(2.2) = 1 + 0.55 = 1.55$.

Question 7

1. $h^0$
2. $h^1$
3. $h^2$ (correct answer)
4. $h^3$

Explanation: The order of a numerical method is related to its local truncation error. The local truncation error is defined as the difference between the true solution and the numerical approximation after one step, assuming the previous value was exact. For a method to have a local truncation error of order $O(h^{p+1})$, its Taylor series expansion must match the true solution's Taylor series expansion up to and including the term of order $h^p$. Since Heun's method has a local truncation error of $O(h^3)$, its expansion must match the true solution's expansion up to the $h^2$ term. The mismatch first occurs at the $h^3$ term, which defines the order of the local error. Therefore, the expansions match for terms with powers $h^0, h^1$, and $h^2$.

Question 8

1. $-0.89$ (correct answer)
2. $-1.00$
3. $-0.99$
4. $-0.945$

Explanation: The Improved Euler (Heun's) method consists of a predictor step and a corrector step. Given $f(x,y) = x^2 - y$, $x_0=0$, $y_0=1$, and $h=0.1$.

1. Calculate the predictor slope at the initial point: $m_1 = f(x_0, y_0) = f(0, 1) = 0^2 - 1 = -1$.
2. Use the predictor slope to find a preliminary estimate for $y_1$ (the predictor value $y_1^*$): $y_1^* = y_0 + h \cdot f(x_0, y_0) = 1 + 0.1 \cdot (-1) = 0.9$.
3. Calculate the corrector slope $m_2$ at the new point $(x_1, y_1^*) = (0.1, 0.9)$: $m_2 = f(x_1, y_1^*) = f(0.1, 0.9) = (0.1)^2 - 0.9 = 0.01 - 0.9 = -0.89$.

Question 9

1. $1.0$
2. $1.5$
3. $1.75$
4. $1.4$ (correct answer)

Explanation: When you encounter Heun's method problems, remember that this is a two-stage Runge-Kutta method that uses both the slope at the current point and a predicted slope at the next point to make its approximation. For Heun's method with $y' = f(x,y)$, the formula is: $y_1 = y_0 + \frac{h}{2}[f(x_0, y_0) + f(x_0 + h, y_0 + hf(x_0, y_0))]$ Given $y' = 2y - 4x$ with $y(0) = y_0$, $h = 0.5$, and $y_1 = 3$: First, calculate $f(0, y_0) = 2y_0 - 4(0) = 2y_0$ Next, find the predictor: $y_0 + hf(0, y_0) = y_0 + 0.5(2y_0) = 2y_0$ Then calculate $f(0.5, 2y_0) = 2(2y_0) - 4(0.5) = 4y_0 - 2$ Applying Heun's formula: $3 = y_0 + \frac{0.5}{2}[2y_0 + (4y_0 - 2)]$ $3 = y_0 + 0.25[6y_0 - 2]$ $3 = y_0 + 1.5y_0 - 0.5$ $3 = 2.5y_0 - 0.5$ $3.5 = 2.5y_0$ $y_0 = 1.4$ Choice A (1.0) would give $y_1 = 2$, not 3. Choice B (1.5) results in $y_1 = 3.25$. Choice C (1.75) yields $y_1 = 3.625$. These all stem from computational errors in applying Heun's two-stage process. The answer is D. Study tip: Always work backwards systematically when given the final result in numerical methods problems. Set up the complete formula first, then substitute known values and solve algebraically.

Question 10

1. $y' = x^2$
2. $y' = y$
3. $y' = -10$ (correct answer)
4. $y' = x+y$

Explanation: Let $y_1^E$ be the Euler approximation and $y_1^H$ be the Heun's approximation. $y_1^E = y_0 + h f(x_0, y_0)$. $y_1^H = y_0 + \frac{h}{2} [f(x_0, y_0) + f(x_0+h, y_1^*)]$, where $y_1^* = y_0 + h f(x_0, y_0) = y_1^E$. For the two approximations to be equal, we must have: $y_0 + h f(x_0, y_0) = y_0 + \frac{h}{2} [f(x_0, y_0) + f(x_0+h, y_1^E)]$ $h f(x_0, y_0) = \frac{h}{2} [f(x_0, y_0) + f(x_0+h, y_1^E)]$ $2 f(x_0, y_0) = f(x_0, y_0) + f(x_0+h, y_1^E)$ $f(x_0, y_0) = f(x_0+h, y_1^E)$. This condition means that the slope function $f(x,y)$ must have the same value at the initial point $(x_0, y_0)$ and at the predicted next point $(x_0+h, y_1^E)$. This is only guaranteed to be true for any $(x_0, y_0)$ and $h$ if the function $f(x,y)$ is a constant. For the choice $y'=-10$, we have $f(x,y) = -10$. Thus, $f(x_0, y_0) = -10$ and $f(x_0+h, y_1^E) = -10$. The condition is met.

Question 11

1. The slope at the beginning of the interval and an estimated slope at the end of the interval. (correct answer)
2. The slope at the beginning of the interval and an estimated slope at the midpoint of the interval.
3. The exact slopes of the true solution curve at the beginning and the end of the interval.
4. The slope given by the forward Euler method and the slope given by the backward Euler method.

Explanation: Heun's method first calculates the slope at the current point $(x_n, y_n)$, which is $m_1 = f(x_n, y_n)$. It then uses this slope to make a preliminary estimate (a 'prediction') of the value at the end of the interval, $y_{n+1}^* = y_n + h \cdot m_1$. It then calculates a second slope, $m_2 = f(x_{n+1}, y_{n+1}^*)$, which is an estimate for the slope at the end of the interval. The final approximation is made using the average of these two slopes. Therefore, the two slopes are the slope at the beginning of the interval and an estimated slope at the end.

Question 12

1. $0.1$
2. $0.2$ (correct answer)
3. $0.5$
4. $1.0$

Explanation: We need to find the condition on $h$ such that $|y_1| < |y_0|$. Here, $y_0=1$. First, let's compute $y_1$ in terms of $h$. $f(x,y) = -10y$, $(x_0, y_0) = (0,1)$. $m_1 = f(0,1) = -10(1) = -10$. $y_1^* = y_0 + h m_1 = 1 - 10h$. $m_2 = f(h, y_1^*) = -10(1-10h) = -10 + 100h$. $y_1 = y_0 + \frac{h}{2}(m_1 + m_2) = 1 + \frac{h}{2}(-10 - 10 + 100h) = 1 + \frac{h}{2}(-20 + 100h) = 1 - 10h + 50h^2$. The condition for stability is $|y_1| < 1$, which translates to $-1 < 1 - 10h + 50h^2 < 1$. First inequality: $1 - 10h + 50h^2 < 1 \implies 50h^2 - 10h < 0 \implies 10h(5h-1) < 0$. Since $h>0$, we need $5h-1 < 0$, which gives $h < 1/5$ or $h < 0.2$. Second inequality: $-1 < 1 - 10h + 50h^2 \implies 0 < 2 - 10h + 50h^2$. The quadratic $50h^2 - 10h + 2$ has discriminant $\Delta = (-10)^2 - 4(50)(2) = 100 - 400 = -300 < 0$. Since the leading coefficient is positive, the quadratic is always positive. Thus, this inequality holds for all $h$. Combining the conditions, the step size must satisfy $h < 0.2$.

Question 13

1. $y(0.3) \approx 1.368$ (correct answer)
2. $y(0.3) \approx 1.377$
3. $y(0.3) \approx 1.386$
4. $y(0.3) \approx 1.395$

Explanation: Heun's method has global error $O(h^2)$. The original error was $1.423 - 1.350 = 0.073$. When step size halves, error reduces by factor of $4$: new error $\approx 0.073/4 = 0.018$. So $y(0.3) \approx 1.350 + 0.018 = 1.368$. Choice B uses factor of $2$ reduction. Choice C uses $\sqrt{4}$ factor. Choice D assumes error reduces linearly with step size.

Question 14

1. $y(0.4) \approx 1.5824$
2. $y(0.4) \approx 1.5936$ (correct answer)
3. $y(0.4) \approx 1.6048$
4. $y(0.4) \approx 1.6160$

Explanation: Heun's method uses predictor-corrector: First step: $y_1^* = 1 + 0.2(0 + 1) = 1.2$, then $y_1 = 1 + 0.1[(0 + 1) + (0.2 + 1.2)] = 1.24$. Second step: $y_2^* = 1.24 + 0.2(0.2 + 1.24) = 1.528$, then $y_2 = 1.24 + 0.1[(0.2 + 1.24) + (0.4 + 1.528)] = 1.5936$. Choice A uses only the predictor value. Choice C incorrectly adds an extra correction term. Choice D uses forward Euler throughout.

Question 15

1. $y_1^* = 2 + 0.25[-2(2) + 3(1)]$ (correct answer)
2. $y_1^* = 2 + 0.25[-2(2) + 3(1.25)]$
3. $y_1^* = 2 + 0.125[-2(2) + 3(1)] + 0.125[-2(2) + 3(1.25)]$
4. $y_1^* = 2 + 0.25\{[-2(2) + 3(1)] + [-2(2) + 3(1.25)]\}$

Explanation: In Heun's method, the predictor step uses only the slope at the initial point: $y_1^* = y_0 + h \cdot f(x_0, y_0)$. Here, $f(1,2) = -2(2) + 3(1) = -1$, so $y_1^* = 2 + 0.25(-1)$. Choice B incorrectly uses the endpoint $x$-value in the predictor. Choice C represents the corrector step formula. Choice D attempts to average slopes in the predictor step, which is incorrect procedure.

Question 16

1. Student B's result is closer to the true solution than Student A's result
2. Student B's result equals Student A's result since extra correction has no effect
3. Student B's result has larger error than Student A's due to over-correction instability (correct answer)
4. Student B inadvertently implements a higher-order method with better accuracy

Explanation: Applying the corrector formula multiple times in Heun's method leads to over-correction and can cause instability, especially when the step size is not very small. The method is designed for one predictor and one corrector step. Multiple corrections don't improve accuracy and typically introduce additional error. Choice A is incorrect as over-correction worsens accuracy. Choice B is wrong since extra correction does change the result. Choice D misunderstands that this doesn't create a higher-order method.

Question 17

1. The analytical solution contains a transcendental function that cannot be exactly represented
2. Heun's method accumulates truncation error from the trapezoidal approximation over the integration interval (correct answer)
3. The step size $h = 0.4$ is too large for the exponential decay component of the solution
4. Round-off errors in computing $\frac{1}{1+x^2}$ compound through the numerical iterations

Explanation: The primary source of error in Heun's method is the local truncation error $O(h^3)$ at each step, which accumulates to global error $O(h^2)$. This comes from the trapezoidal rule approximation inherent in the method. Choice A is irrelevant to numerical vs analytical comparison. Choice C misidentifies the error source. Choice D overestimates the impact of round-off error compared to truncation error for this step size.

Question 18

1. The predictor step linearizes the nonlinear relationship between $x$ and $y$
2. The corrector step over-compensates for the curvature in the solution curve
3. The method fails to account for the changing rate of change of the slope function
4. The relatively large step size amplifies truncation errors from the trapezoidal approximation (correct answer)

Explanation: Heun's method has local truncation error $O(h^3)$, and the primary error source with $h = 0.4$ is that this step size is relatively large for the curvature of the solution. The trapezoidal rule approximation in the corrector step introduces significant error when $h$ is not small enough. Choice A misidentifies the linearization issue. Choice B incorrectly suggests over-compensation. Choice C is vague and doesn't identify the primary error mechanism.