6.P — Worked Problems: Differential Equations

Twelve problems, one for each method, each solved to the end and checked by substitution. Every formula used is derived in the chapter that introduces it.

Problem 1 — A separable equation with an initial condition

Solve \dfrac{dy}{dx} = \dfrac{2x}{y} with y(0) = 3.

Solution

Step 1 — separate, putting every y on the left and every x on the right:

y\,dy = 2x\,dx

Step 2 — integrate both sides. Each side gets its own constant, but they can be combined into one on the right:

\frac{y^2}{2} = x^2 + C

Step 3 — apply the initial condition now, before tidying, because it is easier while the equation is still simple. Put x=0, y=3:

\frac{9}{2} = 0 + C \quad \Rightarrow \quad C = 4.5

Step 4 — write the solution.

\frac{y^2}{2} = x^2 + 4.5 \quad \Rightarrow \quad y^2 = 2x^2+9 \quad \Rightarrow \quad y = \sqrt{2x^2+9}

The positive root is the right one because y(0) = 3 is positive, and the solution cannot jump across zero without the original equation breaking (it has y in a denominator).

Answer: y = \sqrt{2x^2+9}.

Check by substituting back. Differentiate:

\frac{dy}{dx} = \frac{4x}{2\sqrt{2x^2+9}} = \frac{2x}{\sqrt{2x^2+9}} = \frac{2x}{y} \; ✓

And y(0) = \sqrt9 = 3 ✓

What to notice. The solution is one branch of a hyperbola. The original equation is undefined where y=0, and the solution obligingly never goes there — the mathematics protects itself.

Problem 2 — A first-order linear equation

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

Solution

Step 1 — identify P and Q. The equation is already in the standard form y' + Py = Q, with

P(x) = 2, \qquad Q(x) = e^{-x}

Step 2 — compute the integrating factor.

\mu = e^{\int P\,dx} = e^{\int 2\,dx} = e^{2x}

Step 3 — multiply the whole equation by it.

e^{2x}y' + 2e^{2x}y = e^{2x}e^{-x} = e^{x}

Step 4 — recognise the left side as a single derivative, which is the entire point of the factor:

\frac{d}{dx}\left(e^{2x}y\right) = e^x

Confirm with the product rule: \frac{d}{dx}(e^{2x}y) = 2e^{2x}y + e^{2x}y' ✓ — the same as the left side above.

Step 5 — integrate both sides.

e^{2x}y = \int e^xdx = e^x + C

Step 6 — solve for y.

y = e^{-2x}\left(e^x+C\right) = e^{-x}+Ce^{-2x}

Step 7 — apply the initial condition.

1 = e^0 + Ce^0 = 1 + C \quad \Rightarrow \quad C = 0

Answer: y = e^{-x}.

Check. y' = -e^{-x}, so y'+2y = -e^{-x}+2e^{-x} = e^{-x} ✓ and y(0)=1 ✓

What to notice. The constant came out zero, which means this initial condition happened to sit exactly on the particular solution with no transient at all. Start at y(0) = 2 instead and you would get y = e^{-x}+e^{-2x}, where the second term is the transient dying away faster than the first. In the long run the e^{-2x} part is irrelevant — the behaviour is set by the slowest-decaying term, which is the general rule for every linear system.

Problem 3 — Newton's cooling with a real question

A body is found at 10:00 pm with a temperature of 27 °C in a room held at 20 °C. An hour later it is 25 °C. Assuming a normal body temperature of 37 °C at death, when did the person die?

Solution

Step 1 — write the model. With T_s = 20:

T(t) = 20 + \left(T_{\text{at }t=0}-20\right)e^{-kt}

Take t=0 at 10:00 pm, so the initial excess is 27-20 = 7:

T(t) = 20 + 7e^{-kt}

Step 2 — find k from the second measurement. At t=1, T = 25:

25 = 20+7e^{-k} \quad \Rightarrow \quad 5 = 7e^{-k} \quad \Rightarrow \quad e^{-k} = \frac57 = 0.714286

Take logarithms:

-k = \ln(0.714286) = -0.336472 \quad \Rightarrow \quad k = 0.336472 \text{ per hour}

Step 3 — run the model backwards to find death. We need the time t at which T = 37. That will come out negative, meaning before 10:00 pm, which is exactly what we expect.

37 = 20 + 7e^{-0.336472t}

17 = 7e^{-0.336472t} \quad \Rightarrow \quad e^{-0.336472t} = \frac{17}{7} = 2.428571

-0.336472\,t = \ln(2.428571) = 0.887303

t = \frac{0.887303}{-0.336472} = -2.637 \text{ hours}

Step 4 — convert. 2.637 hours is 2 hours and 38 minutes before 10:00 pm.

Answer: death occurred at about 7:22 pm.

Check by running forwards from that time. At t = -2.637: 20 + 7e^{0.887} = 20+7(2.4286) = 37 ✓

The honest caveat. This calculation is used in real investigations and it is known to be rough. The model assumes a constant room temperature, no clothing, no air movement, and that cooling began immediately — none of which is quite true, and the first hour after death behaves differently in any case. A forensic report gives a range of hours, not a time to the minute, and the mathematics is only part of the reasoning.

Problem 4 — The logistic equation with real numbers

A lake can support 1,000 fish. It is stocked with 50, and after one year there are 150. When will the population reach 900?

Solution

Step 1 — write the solution with the known constants. With K = 1000 and P_0 = 50:

\frac{K-P_0}{P_0} = \frac{950}{50} = 19

P(t) = \frac{1000}{1+19e^{-rt}}

Step 2 — find r from the one-year measurement.

150 = \frac{1000}{1+19e^{-r}}

Rearrange:

1+19e^{-r} = \frac{1000}{150} = 6.6667 \quad \Rightarrow \quad 19e^{-r} = 5.6667 \quad \Rightarrow \quad e^{-r} = 0.298246

r = -\ln(0.298246) = 1.20999 \approx 1.21 \text{ per year}

Step 3 — solve for the time when P = 900.

900 = \frac{1000}{1+19e^{-1.21t}}

1+19e^{-1.21t} = \frac{1000}{900} = 1.11111

19e^{-1.21t} = 0.11111 \quad \Rightarrow \quad e^{-1.21t} = 0.0058480

-1.21t = \ln(0.0058480) = -5.14166

t = \frac{5.14166}{1.21} = 4.25 \text{ years}

Answer: about 4 years and 3 months after stocking.

Step 4 — find the fastest-growing moment, which is at P = \frac K2 = 500:

500 = \frac{1000}{1+19e^{-1.21t}} \quad \Rightarrow \quad 19e^{-1.21t} = 1 \quad \Rightarrow \quad e^{-1.21t} = \frac{1}{19}

t = \frac{\ln 19}{1.21} = \frac{2.9444}{1.21} = 2.43 \text{ years}

At that moment the growth rate is rP(1-P/K) = 1.21(500)(0.5) = 302 fish per year, the highest it ever gets.

What to notice. Getting from 50 to 500 took 2.43 years; getting from 500 to 900 took only another 1.82. The population accelerates all the way to half capacity and decelerates after. If you had extrapolated from the first year's growth — tripling annually — you would have predicted 1,350 fish by year three and 4,050 by year four. The lake would never allow it, and the difference between those two forecasts is why the carrying capacity term matters so much in every real model.

Problem 5 — A second-order equation, three cases in one setting

Solve y'' + 5y' + 6y = 0, then y''+4y'+4y=0, then y''+2y'+5y=0, each with y(0)=1 and y'(0)=0.

Solution

Case A: y''+5y'+6y = 0.

Step 1 — the characteristic equation.

m^2+5m+6 = 0 \quad \Rightarrow \quad (m+2)(m+3) = 0 \quad \Rightarrow \quad m = -2,\ -3

Two distinct real roots, so

y = Ae^{-2x}+Be^{-3x}

Step 2 — apply the conditions. From y(0)=1:

A + B = 1

Differentiate: y' = -2Ae^{-2x}-3Be^{-3x}, and y'(0) = 0 gives

-2A - 3B = 0 \quad \Rightarrow \quad A = -\frac{3B}{2}

Substitute into the first: -\frac{3B}{2}+B = 1, so -\frac B2 = 1, giving B = -2 and A = 3.

\boxed{y = 3e^{-2x}-2e^{-3x}}

Case B: y''+4y'+4y=0.

m^2+4m+4 = (m+2)^2 = 0 \quad \Rightarrow \quad m = -2 \text{ twice}

A repeated root needs the extra factor of x:

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

From y(0)=1: A=1. Differentiate with the product rule:

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

At x=0: B - 2A = 0, so B = 2.

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

Case C: y''+2y'+5y=0.

m^2+2m+5 = 0 \quad \Rightarrow \quad m = \frac{-2\pm\sqrt{4-20}}{2} = \frac{-2\pm4i}{2} = -1\pm2i

Complex roots with real part -1 and imaginary part 2:

y = e^{-x}\left(A\cos2x+B\sin2x\right)

From y(0)=1: A = 1. Differentiating,

y' = -e^{-x}(A\cos2x+B\sin2x)+e^{-x}(-2A\sin2x+2B\cos2x)

At x=0: -A + 2B = 0, so B = \frac12.

\boxed{y = e^{-x}\left(\cos2x+\tfrac12\sin2x\right)}

What the three answers look like. All three start at height 1 with zero slope and settle to zero. Case A slides down smoothly. Case B does the same but is the borderline — the fastest possible return. Case C wobbles: it dips below zero, comes back up, and rings its way to rest. Case C is a system with too little friction, and the ringing is the visible sign of complex roots.

Problem 6 — A spring with numbers

A 2 kg mass hangs on a spring of stiffness 50 N/m with a damper of coefficient 8 N·s/m. Find the damping ratio, classify the motion, and write the displacement if it starts 0.1 m from rest with no initial velocity.

Solution

Step 1 — the natural frequency.

\omega_0 = \sqrt{\frac km} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \text{ rad/s}

Step 2 — the damping ratio.

\zeta = \frac{c}{2\sqrt{mk}} = \frac{8}{2\sqrt{100}} = \frac{8}{20} = 0.4

Since 0 < \zeta < 1, the motion is underdamped: it will oscillate with a shrinking amplitude.

Step 3 — the damped frequency.

\omega_d = \omega_0\sqrt{1-\zeta^2} = 5\sqrt{1-0.16} = 5\sqrt{0.84} = 5(0.916515) = 4.583 \text{ rad/s}

Note it is lower than 5 — the friction slows the wobble as well as shrinking it.

Step 4 — write the general solution.

x(t) = e^{-\zeta\omega_0t}\left(A\cos\omega_dt+B\sin\omega_dt\right) = e^{-2t}\left(A\cos4.583t + B\sin4.583t\right)

with \zeta\omega_0 = 0.4\times5 = 2.

Step 5 — apply the conditions. From x(0)=0.1: A = 0.1.

Differentiate:

\dot x = -2e^{-2t}(A\cos\omega_dt+B\sin\omega_dt) + e^{-2t}(-A\omega_d\sin\omega_dt+B\omega_d\cos\omega_dt)

At t=0:

0 = -2A + B\omega_d \quad \Rightarrow \quad B = \frac{2(0.1)}{4.583} = 0.04364

Answer:

x(t) = e^{-2t}\left(0.1\cos 4.583t + 0.04364\sin4.583t\right)

Step 6 — the practical numbers. The period of the wobble is \frac{2\pi}{4.583} = 1.371 s. The envelope e^{-2t} halves every \frac{\ln 2}{2} = 0.347 s, so after one full oscillation the amplitude has fallen by a factor of e^{-2(1.371)} = 0.064 — down to 6% of where it started. After two oscillations it is essentially at rest.

How much damping would be critical? \zeta=1 needs c = 2\sqrt{mk} = 20 N·s/m, two and a half times what we have. A car with \zeta = 0.4 would bounce noticeably after a speed bump, which is exactly the test a mechanic does by pushing down on the wing and watching whether it comes back once or three times.

Problem 7 — Resonance

An undamped system has \omega_0 = 10 rad/s and is driven by F_0\cos\omega t with F_0/m = 4. Compare the steady-state amplitude at \omega = 5, \omega = 9.5 and \omega = 9.9, then say what happens at \omega = 10.

Solution

Step 1 — the amplitude formula with no damping (\zeta = 0) simplifies to

A(\omega) = \frac{F_0/m}{\left|\omega_0^2-\omega^2\right|} = \frac{4}{\left|100-\omega^2\right|}

Step 2 — evaluate at each frequency.

At \omega=5: \;A = \frac{4}{|100-25|} = \frac{4}{75} = 0.0533

At \omega = 9.5: \;A = \frac{4}{|100-90.25|} = \frac{4}{9.75} = 0.410

At \omega = 9.9: \;A = \frac{4}{|100-98.01|} = \frac{4}{1.99} = 2.010

Step 3 — at \omega = 10 exactly. The denominator is zero and the formula gives infinity. What that really means is that the assumed steady-state form is wrong: the guess A\cos\omega t is already a solution of the unforced equation, so §8 of 6.1 — first order says to multiply by t. The true solution is

x(t) = \frac{F_0}{2m\omega_0}\,t\sin\omega_0 t

an oscillation whose amplitude grows linearly and without limit.

Answers: amplitudes 0.053, 0.41 and 2.01 — a 38-fold increase between \omega=5 and \omega=9.9 — and unbounded growth at \omega = 10.

What to take from the numbers. The amplitude at 9.9 is nearly five times the amplitude at 9.5, for a frequency change of only 4%. Resonance is sharp: the response is unremarkable everywhere except in a narrow band, and inside that band it is enormous. That is what makes it dangerous — a machine can run safely at a hundred speeds and destroy itself at one.

Why nothing actually reaches infinity. Real systems always have some damping, which caps the peak at A_{\max} \approx \frac{F_0/m}{2\zeta\omega_0^2}. With \zeta = 0.01 that is still 50 times the low-frequency response — enough to break most things. And long before infinity, the linear model itself fails: the spring stops obeying Hooke's law, or the material yields. The equation does not predict the collapse; it predicts that the model will stop applying, and the collapse happens in the gap.

Problem 8 — Non-homogeneous, by undetermined coefficients

Solve y'' - 3y' + 2y = 4x.

Solution

Step 1 — the complementary function. Solve the equation with zero on the right:

m^2-3m+2 = 0 \quad \Rightarrow \quad (m-1)(m-2)=0 \quad \Rightarrow \quad m = 1, 2

y_c = Ae^x + Be^{2x}

Step 2 — guess a particular solution shaped like the right-hand side. The right side is a degree-1 polynomial, so try the general degree-1 polynomial:

y_p = ax+b

Check it is not already part of y_c — it is not, since y_c is made of exponentials. So no extra factor of x is needed.

Step 3 — substitute. With y_p' = a and y_p''=0:

0 - 3a + 2(ax+b) = 4x

2ax + (2b-3a) = 4x + 0

Step 4 — match coefficients on each power of x.

The x terms: 2a = 4, so a = 2.

The constant terms: 2b - 3a = 0, so 2b = 6 and b = 3.

y_p = 2x+3

Step 5 — add them.

y = Ae^x + Be^{2x} + 2x + 3

Answer: y = Ae^x+Be^{2x}+2x+3.

Check the particular part alone, since that is where errors hide:

y_p'' - 3y_p' + 2y_p = 0 - 3(2) + 2(2x+3) = -6 + 4x + 6 = 4x \; ✓

Why the guess has to include the constant b. Guessing only y_p = ax would give -3a + 2ax = 4x, which cannot be satisfied for all x: it needs a=2 from the x terms and a=0 from the constant terms. Always guess the complete general form of that type, not just the term you can see on the right.

Problem 9 — Bernoulli

Solve y' + \dfrac{y}{x} = xy^2.

Solution

Step 1 — recognise the form. This is y' + Py = Qy^n with P = \frac1x, Q = x and n = 2. It is not linear because of the y^2.

Step 2 — divide by y^n = y^2.

y^{-2}y' + \frac{y^{-1}}{x} = x

Step 3 — substitute u = y^{1-n} = y^{-1}. Then by the chain rule

u' = -y^{-2}y' \quad \Rightarrow \quad y^{-2}y' = -u'

Step 4 — rewrite the whole equation in u.

-u' + \frac ux = x \quad \Rightarrow \quad u' - \frac ux = -x

This is now first-order linear, which is the entire purpose of the substitution.

Step 5 — integrating factor. With P = -\frac1x:

\mu = e^{\int -\frac1x dx} = e^{-\ln x} = \frac1x

Multiply through:

\frac{u'}{x}-\frac{u}{x^2} = -1 \quad \Rightarrow \quad \frac{d}{dx}\left(\frac ux\right) = -1

Step 6 — integrate.

\frac ux = -x + C \quad \Rightarrow \quad u = -x^2+Cx

Step 7 — undo the substitution. Since u = \frac1y:

\frac1y = Cx - x^2 \quad \Rightarrow \quad y = \frac{1}{Cx-x^2} = \frac{1}{x(C-x)}

Answer: y = \dfrac{1}{x(C-x)}.

Check with C=1, so y = \frac{1}{x-x^2}. Then

y' = -\frac{1-2x}{(x-x^2)^2}

and

y' + \frac yx = -\frac{1-2x}{(x-x^2)^2}+\frac{1}{x(x-x^2)} = \frac{-(1-2x) + (x - x^2)/x}{(x-x^2)^2}

The numerator is -1+2x+1-x = x, so the whole thing is \frac{x}{(x-x^2)^2} = xy^2 ✓

Problem 10 — A mixing problem

A 100-litre tank holds pure water. Brine containing 20 g of salt per litre flows in at 5 l/min, and the well-mixed solution flows out at the same rate. How much salt is in the tank after 30 minutes, and what is the long-run amount?

Solution

Step 1 — define the unknown. Let A(t) be the grams of salt in the tank at time t minutes. The volume stays at 100 litres because inflow equals outflow.

Step 2 — rate in.

20 \frac{\text{g}}{\text{l}} \times 5\frac{\text{l}}{\text{min}} = 100 \text{ g/min}

Step 3 — rate out. The outflowing liquid has whatever concentration the tank currently has, which is \frac{A}{100} g/l:

\frac{A}{100}\times 5 = \frac{A}{20} \text{ g/min}

Step 4 — the differential equation.

\frac{dA}{dt} = 100 - \frac{A}{20}

Step 5 — solve it. This is separable:

\frac{dA}{100 - \frac{A}{20}} = dt

Multiply top and bottom of the left by 20 to tidy:

\frac{20\,dA}{2000-A} = dt

Integrate, remembering that the derivative of 2000-A with respect to A is -1:

-20\ln|2000-A| = t + C

Step 6 — apply A(0)=0, since the tank starts as pure water:

-20\ln(2000) = C

Step 7 — solve for A.

-20\ln(2000-A) + 20\ln 2000 = t

20\ln\frac{2000}{2000-A} = t \quad \Rightarrow \quad \frac{2000}{2000-A} = e^{t/20}

2000 - A = 2000e^{-t/20} \quad \Rightarrow \quad A(t) = 2000\left(1-e^{-t/20}\right)

Step 8 — answer both questions.

At t=30:

A = 2000\left(1 - e^{-1.5}\right) = 2000(1-0.22313) = 2000(0.77687) = 1554 \text{ g}

As t\to\infty, e^{-t/20}\to0, so A\to 2000 g.

Answers: 1,554 g after 30 minutes, approaching 2,000 g in the long run.

Check the long-run answer without any calculus. Eventually the tank must reach the same concentration as the inflow, 20 g/l. In 100 litres that is 2,000 g ✓ — and this is worth doing first on any mixing problem, because it tells you the answer your algebra must produce.

The time constant. The exponent is -\frac{t}{20}, so the characteristic time is 20 minutes: after 20 minutes the tank is 63% of the way there, after 60 minutes 95%, after 100 minutes over 99%. That number, 20, is the tank volume divided by the flow rate — the time it takes to replace the tank's contents once.

Problem 11 — A system of two equations

Solve \begin{cases}\dot x = x + 2y\\ \dot y = 3x + 2y\end{cases} and describe the long-run behaviour.

Solution

Step 1 — write it as a matrix equation.

\dot{\mathbf{x}} = A\mathbf{x}, \qquad A = \begin{pmatrix}1&2\\3&2\end{pmatrix}

Step 2 — find the eigenvalues.

\det(A-\lambda I) = (1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda + 2 - 6 = \lambda^2-3\lambda-4

\lambda^2-3\lambda-4 = 0 \quad \Rightarrow \quad (\lambda-4)(\lambda+1) = 0 \quad \Rightarrow \quad \lambda = 4,\ -1

Check against trace and determinant: 4+(-1) = 3 = trace ✓, and 4\times(-1) = -4 = determinant ✓

Step 3 — the eigenvector for \lambda = 4.

(A-4I) = \begin{pmatrix}-3&2\\3&-2\end{pmatrix}

The first row says -3x+2y = 0, so y = \frac{3x}{2}. Taking x=2 gives \mathbf{v}_1 = (2,3).

Step 4 — the eigenvector for \lambda = -1.

(A+I) = \begin{pmatrix}2&2\\3&3\end{pmatrix}

Both rows say x + y = 0, so \mathbf{v}_2 = (1,-1).

Step 5 — write the general solution.

\begin{pmatrix}x\\y\end{pmatrix} = c_1e^{4t}\begin{pmatrix}2\\3\end{pmatrix}+c_2e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}

Answer, written out:

x(t) = 2c_1e^{4t}+c_2e^{-t}, \qquad y(t) = 3c_1e^{4t}-c_2e^{-t}

Step 6 — the long-run behaviour. One eigenvalue is positive and one negative, so by the table in 6.1 — first order §9 this is a saddle, and it is unstable.

Concretely: the e^{-t} part dies away and the e^{4t} part explodes. So unless c_1 is exactly zero, every trajectory eventually runs off to infinity along the direction (2,3) — the eigenvector belonging to the growing eigenvalue.

The one exception is a start exactly on the line y = -x, where c_1 = 0. Such a trajectory slides straight into the origin. But it is a knife edge: any deviation, however small, has a non-zero c_1 which then grows by a factor of e^4 = 55 every unit of time.

What a saddle means in practice. The system has a stable direction and an unstable one, and the stable one cannot be relied on because reaching it exactly is impossible. A pencil balanced on its point is the physical picture — there is a position where it stays, and no way to stay there.

Problem 12 — Everything at once: separation of variables on the heat equation

A metal rod of length L is held at 0 °C at both ends, starting from a temperature profile u(x,0) = \sin\frac{\pi x}{L}. Solve u_t = \alpha u_{xx} and describe what happens.

Solution

Step 1 — assume the solution factorises.

u(x,t) = X(x)T(t)

Step 2 — substitute into the equation. The time derivative only hits T, and the space derivatives only hit X:

XT' = \alpha X''T

Step 3 — separate the variables, dividing both sides by \alpha XT:

\frac{T'}{\alpha T} = \frac{X''}{X}

The left side depends only on t and the right only on x. Fix x and vary t: the right side cannot change, so the left cannot either. Both must equal the same constant, which we call -\lambda (the minus is a convention chosen because we expect decay, and it makes the algebra come out with sines rather than exponentials).

Step 4 — two ordinary equations.

X'' + \lambda X = 0, \qquad T' + \alpha\lambda T = 0

Step 5 — solve the space equation with its boundary conditions. The ends are held at zero, so X(0)=X(L)=0. For \lambda>0 the solution is

X(x) = A\cos\sqrt\lambda x + B\sin\sqrt\lambda x

X(0)=0 forces A=0. Then X(L) = B\sin\sqrt\lambda L = 0, and since B=0 would give nothing at all, we need

\sqrt\lambda L = n\pi \quad \Rightarrow \quad \lambda_n = \left(\frac{n\pi}{L}\right)^2, \quad n = 1,2,3,\ldots

This is the crucial step. The boundary conditions have picked out a discrete list of allowed values from a continuous range. Only certain shapes can exist on a rod with fixed ends, exactly as only certain notes can sound on a guitar string of a given length.

Step 6 — solve the time equation.

T' = -\alpha\lambda_n T \quad \Rightarrow \quad T = e^{-\alpha\lambda_n t}

Step 7 — match the initial condition. The starting profile is \sin\frac{\pi x}{L}, which is exactly the n=1 shape and nothing else. So only that one term is needed:

u(x,t) = \sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}

Answer: the temperature keeps its shape — a single hump — and its height decays exponentially with time constant \frac{L^2}{\alpha\pi^2}.

Check. u_t = -\frac{\alpha\pi^2}{L^2}u and u_{xx} = -\frac{\pi^2}{L^2}u, so \alpha u_{xx} = -\frac{\alpha\pi^2}{L^2}u = u_t ✓

Two things worth taking away.

The L^2 is the important one. Double the rod's length and it takes four times as long to cool. Heat diffuses in proportion to the square of the distance, which is why a thick steak takes far more than twice as long to cook as one half its thickness, and why thin fins are used on radiators and heat sinks.

What if the starting profile is not a single sine? Then write it as a sum of sines — one for each allowed n — and each term decays at its own rate e^{-\alpha n^2\pi^2t/L^2}. The high-n terms, which are the sharp wiggles, have an n^2 in their exponent and so die away far fastest. A rough temperature profile smooths itself out before it cools down. Writing an arbitrary function as a sum of sines is precisely the problem Fourier solved in order to do this, and it is Part 9.

Next: Part 7 — Probability & Statistics, where the quantities stop being determined and start being uncertain.