Differential Equations — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for differential equations.

UT-1: Separable Equations — Lost Equilibrium Solutions

Question:

(a) Solve $\dfrac{dy}{dx} = y^2 - 1$ with the general solution.

(b) A student separates variables and writes $\dfrac{dy}{y^2 - 1} = dx$, integrates, and arrives at the general solution $\dfrac{1}{2}\ln\!\left\lvert\dfrac{y - 1}{y + 1}\right\rvert = x + C$. They claim this covers all solutions. Is this correct?

(c) Find the particular solution with $y(0) = 1$.

[Difficulty: hard. Tests the common error of losing equilibrium solutions when dividing by $g(y)$.]

Solution:

(a) First check equilibrium solutions: $y^2 - 1 = 0 \implies y = 1$ or $y = -1$.

For $y \neq \pm 1$, separate variables using partial fractions:

$\frac{1}{y^2 - 1} = \frac{1}{(y - 1)(y + 1)} = \frac{1}{2(y - 1)} - \frac{1}{2(y + 1)}$

$\int \frac{1}{y^2 - 1}\,dy = \int dx$

$\frac{1}{2}\ln\!\left\lvert\frac{y - 1}{y + 1}\right\rvert = x + C$

$\left\lvert\frac{y - 1}{y + 1}\right\rvert = e^{2(x + C)} = Ae^{2x}$

where $A = e^{2C} \gt 0$. Including the equilibrium solutions, the general solution is:

$y = 1, \quad y = -1, \quad \text{or} \quad \frac{y - 1}{y + 1} = \pm Ae^{2x}$

(b) The student's solution is incomplete because they lost the equilibrium solutions $y = 1$ and $y = -1$. By dividing by $y^2 - 1$, the student implicitly assumed $y^2 - 1 \neq 0$. The equilibrium solutions must be stated separately and are not captured by the formula $\frac{1}{2}\ln\!\left\lvert\frac{y - 1}{y + 1}\right\rvert = x + C$.

(c) The particular solution with $y(0) = 1$ is simply the equilibrium solution $y = 1$ for all $x$. This can be verified: $\frac{dy}{dx} = 0$ and $y^2 - 1 = 1 - 1 = 0$. Confirmed.

Note that if the student tries to use their formula: $\frac{1}{2}\ln\!\left\lvert\frac{0}{2}\right\rvert = 0 + C \implies \frac{1}{2}\ln 0$, which is undefined. This shows the equilibrium solution $y = 1$ cannot be obtained from the separated formula.

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UT-2: Integrating Factor — Sign Error

Question:

Solve $\dfrac{dy}{dx} = 2y + e^{3x}$ with $y(0) = 1$.

(a) Find the general solution using an integrating factor.

(b) A student rewrites the equation as $\dfrac{dy}{dx} - 2y = e^{3x}$, computes the integrating factor as $\mu = e^{-2x}$, and gets the wrong answer. Identify the error in their working:

Their working: $\dfrac{d}{dx}(ye^{-2x}) = e^{3x} \cdot e^{-2x} = e^x$.

So $ye^{-2x} = e^x + C \implies y = e^{3x} + Ce^{2x}$.

With $y(0) = 1$: $1 = 1 + C \implies C = 0$, so $y = e^{3x}$.

Verify: $\dfrac{dy}{dx} = 3e^{3x}$ and $2y + e^{3x} = 2e^{3x} + e^{3x} = 3e^{3x}$. Confirmed.

(c) Was the student actually wrong? If not, explain why.

[Difficulty: hard. Tests the integrating factor method and careful verification of solutions.]

Solution:

(a) Rewrite in standard form: $\dfrac{dy}{dx} - 2y = e^{3x}$.

Integrating factor: $\mu = e^{\int -2\,dx} = e^{-2x}$.

Multiply through: $e^{-2x}\dfrac{dy}{dx} - 2e^{-2x}y = e^{3x} \cdot e^{-2x} = e^x$.

The left side is $\dfrac{d}{dx}(ye^{-2x})$:

$\frac{d}{dx}(ye^{-2x}) = e^x$

$ye^{-2x} = e^x + C$

$y = e^{3x} + Ce^{2x}$

With $y(0) = 1$: $1 = 1 + C \implies C = 0$.

Particular solution: $y = e^{3x}$.

(b) The student's working is actually correct. Despite the framing of the question, the student:

1. Correctly identified the standard form.
2. Correctly computed the integrating factor $\mu = e^{-2x}$.
3. Correctly applied the method.
4. Correctly found the particular solution.
5. Correctly verified it.

(c) The student was not wrong. This question is designed to test whether the student can recognise that a seemingly suspicious answer is actually correct when verified. The key lesson is to always verify solutions by substitution, rather than relying on intuition about whether an answer "looks right."

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UT-3: Second Order — Repeated Root Error

Question:

Solve $y'' + 4y' + 4y = 0$ with $y(0) = 1$ and $y'(0) = 0$.

(a) Find the general solution.

(b) A student writes the general solution as $y = Ae^{-2x} + Be^{-2x} = (A + B)e^{-2x} = Ce^{-2x}$ and then uses the initial conditions to find $C = 1$. Explain why this is wrong.

[Difficulty: hard. Tests the repeated root case of the characteristic equation.]

Solution:

(a) Characteristic equation: $\lambda^2 + 4\lambda + 4 = 0 \implies (\lambda + 2)^2 = 0$.

Repeated root: $\lambda = -2$ (algebraic multiplicity 2).

The general solution for a repeated root is:

$y = (A + Bx)e^{-2x}$

Note the factor of $x$ in the second term. This is essential.

(b) The student's error is treating the repeated root $\lambda = -2$ as two independent solutions $e^{-2x}$ and $e^{-2x}$. These are the same function, so they are linearly dependent. The general solution requires two linearly independent solutions.

For a repeated root $\lambda$, the two independent solutions are $e^{\lambda x}$ and $xe^{\lambda x}$. The factor of $x$ is derived from the method of reduction of order or from the Taylor expansion perspective: when the characteristic equation has a repeated root, the second solution involves the derivative of $e^{\lambda x}$ with respect to $\lambda$.

Using the correct general solution with $y(0) = 1$ and $y'(0) = 0$:

$y = (A + Bx)e^{-2x}, \quad y' = Be^{-2x} + (A + Bx)(-2)e^{-2x} = e^{-2x}(B - 2A - 2Bx)$

$y(0) = A = 1$

$y'(0) = B - 2A = B - 2 = 0 \implies B = 2$

$y = (1 + 2x)e^{-2x}$

The student's answer $y = e^{-2x}$ does not satisfy $y'(0) = 0$ since $y' = -2e^{-2x}$ and $y'(0) = -2 \neq 0$.

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Integration Tests

Tests synthesis of differential equations with other topics.

IT-1: Newton's Law of Cooling with Exact Values (with Logarithms)

Question:

A body at temperature $95\degree\mathrm{C}$ is placed in a room at constant temperature $20\degree\mathrm{C}$. After $10$ minutes, the body's temperature is $60\degree\mathrm{C}$.

(a) Find the temperature of the body as a function of time.

(b) Determine how long it takes for the body to cool to $30\degree\mathrm{C}$. Give your answer in exact form.

(c) Show that the body temperature approaches $20\degree\mathrm{C}$ as $t \to \infty$.

[Difficulty: hard. Combines separable DEs, exponential modelling, and logarithm manipulation.]

Solution:

(a) Newton's law of cooling: $\dfrac{dT}{dt} = -k(T - 20)$, where $k \gt 0$.

Separate variables: $\dfrac{dT}{T - 20} = -k\,dt$.

$\ln(T - 20) = -kt + C$

$T - 20 = e^{-kt + C} = Ae^{-kt}$

$T(t) = 20 + Ae^{-kt}$

Initial condition: $T(0) = 95 \implies A = 75$.

$T(t) = 20 + 75e^{-kt}$

At $t = 10$: $T(10) = 60 \implies 60 = 20 + 75e^{-10k} \implies 75e^{-10k} = 40 \implies e^{-10k} = \frac{8}{15}$.

$-10k = \ln\!\left(\frac{8}{15}\right) \implies k = \frac{1}{10}\ln\!\left(\frac{15}{8}\right)$

$T(t) = 20 + 75\left(\frac{8}{15}\right)^{t/10}$

(b) Set $T(t) = 30$:

$30 = 20 + 75\left(\frac{8}{15}\right)^{t/10} \implies 75\left(\frac{8}{15}\right)^{t/10} = 10$

$\left(\frac{8}{15}\right)^{t/10} = \frac{2}{15}$

$\frac{t}{10}\ln\!\left(\frac{8}{15}\right) = \ln\!\left(\frac{2}{15}\right)$

$t = \frac{10\ln\!\left(\frac{2}{15}\right)}{\ln\!\left(\frac{8}{15}\right)} = \frac{10\ln\!\left(\frac{2}{15}\right)}{\ln 8 - \ln 15} = \frac{10\ln\!\left(\frac{2}{15}\right)}{3\ln 2 - \ln 3 - \ln 5}$

Numerically: $t \approx \frac{10 \times (-2.015)}{0.628} \approx 32.1$ minutes.

(c) As $t \to \infty$, since $\frac{8}{15} \lt 1$, we have $\left(\frac{8}{15}\right)^{t/10} \to 0$.

Therefore $T(t) \to 20 + 75 \times 0 = 20\degree\mathrm{C}$.

This confirms that the body temperature asymptotically approaches the ambient temperature, as expected from Newton's law of cooling.

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IT-2: Euler's Method — Comparing with Exact Solution (with Number and Algebra)

Question:

Use Euler's method with step size $h = 0.5$ to approximate $y(2)$ for the initial value problem:

$\frac{dy}{dx} = \frac{x}{y}, \quad y(1) = 2$

(a) Complete the Euler's method table.

(b) Find the exact solution and compute the exact value of $y(2)$.

(c) Compute the percentage error of the Euler approximation.

[Difficulty: hard. Combines numerical methods, separable DEs, and error analysis.]

Solution:

(a)

$n$	$x_n$	$y_n$	$f(x_n, y_n) = \dfrac{x_n}{y_n}$
0	1.0	2.000	0.500
1	1.5	2.250	0.667
2	2.0	2.583	---

$y_1 = 2.000 + 0.5 \times 0.500 = 2.250$

$y_2 = 2.250 + 0.5 \times \frac{1.5}{2.250} = 2.250 + 0.5 \times 0.667 = 2.250 + 0.333 = 2.583$

Euler approximation: $y(2) \approx 2.583$.

(b) Separate variables: $y\,dy = x\,dx$.

$\int y\,dy = \int x\,dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C$

With $y(1) = 2$: $\frac{4}{2} = \frac{1}{2} + C \implies C = \frac{3}{2}$.

$y^2 = x^2 + 3 \implies y = \sqrt{x^2 + 3}$

(Taking the positive root since $y(1) = 2 \gt 0$.)

$y(2) = \sqrt{4 + 3} = \sqrt{7} \approx 2.646$

(c)

$\text{Percentage error} = \frac{\lvert 2.583 - \sqrt{7} \rvert}{\sqrt{7}} \times 100\% = \frac{2.646 - 2.583}{2.646} \times 100\% \approx 2.4\%$