Differential equations 1

1. Integrate both sides with respect to $x$:

   $$y = \int x\,dx = \frac{x^2}{2} + C$$
2. $$r = -\int \sin(\pi t)\,dt = \frac{\cos(\pi t)}{\pi} + C$$
3. $$y = \int bx^2\,dx = \frac{bx^3}{3} + C$$
4. Separate the variables:

   $$\int \frac{dy}{y} = \int \frac{3}{x-4}\,dx \quad\Rightarrow\quad \ln|y| = 3\ln|x-4| + D$$

   So, $y = A(x-4)^3$, where $A = \pm e^D$.
5. The variables are already separated:

   $$\int u\,du = \int (v+6)\,dv \quad\Rightarrow\quad \frac{u^2}{2} = \frac{v^2}{2} + 6v + D$$

   So, $u^2 = v^2 + 12v + C$, or $u = \pm\sqrt{v^2 + 12v + C}$.
6. Separate the variables, as in example 3 of the lecture:

   $$\int 7y\,dy = \int x e^{-x}\,dx$$

   Integrating the right-hand side by parts, with $u = x$ and $v' = e^{-x}$, gives $-xe^{-x} - e^{-x}$, so

   $$\frac{7y^2}{2} = -e^{-x}(x+1) + D \quad\Rightarrow\quad y^2 = -\frac{2}{7}e^{-x}(x+1) + C$$

1. Separate the variables and integrate:

   $$y = \int \frac{dx}{x} = \ln x + C$$

   (We can write $\ln x$ rather than $\ln|x|$, as the solution starts at $x = 2$.) $y(2) = 0$ gives $C = -\ln 2$, so

   $$y = \ln x - \ln 2 = \ln\frac{x}{2}$$
2. Separate the variables:

   $$\int \frac{dy}{y} = \int dx \quad\Rightarrow\quad \ln y = x + C \quad\Rightarrow\quad y = Ae^x$$

   $y(0) = 1$ gives $A = 1$, so $y = e^x$.

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import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt


def dydx(x, y):
    return 1 / x


x = np.linspace(2, 12, 1000)
solution = solve_ivp(dydx, [x[0], x[-1]], [0], t_eval=x, rtol=1e-8)
y_exact = np.log(x / 2)
print('largest difference =', np.max(np.abs(solution.y[0] - y_exact)))

plt.plot(x, y_exact, label='exact')
plt.plot(x, solution.y[0], '--', label='numerical')
plt.xlabel('x')
plt.ylabel('y')
plt.legend()

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largest difference = 2.4351254179766357e-06

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def dydx(x, y):
    return y


x = np.linspace(0, 5, 1000)
solution = solve_ivp(dydx, [x[0], x[-1]], [1], t_eval=x, rtol=1e-8)
y_exact = np.exp(x)
print('largest relative difference =', np.max(np.abs(solution.y[0] - y_exact) / y_exact))

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largest relative difference = 2.0386919889137807e-07

1. Let $n$ be the number of bacteria at time $t$ hours. The rate of increase is proportional to $n$, so

   $$\frac{dn}{dt} = kn$$

   where $k$ is a positive constant. Separating the variables, as for bacterial growth in the lecture, gives $n = n_0 e^{kt}$, since $n = n_0$ at $t = 0$. At $t = 2$, $n = n_1$, so

   $$n_1 = n_0 e^{2k} \quad\Rightarrow\quad k = \frac{1}{2}\ln\frac{n_1}{n_0}$$

   So, the number present after $t$ hours is

   $$n = n_0 e^{\frac{t}{2}\ln(n_1/n_0)} = n_0\left(\frac{n_1}{n_0}\right)^{t/2}$$
2. The number has tripled when $n = 3n_0$, so $e^{kt} = 3$, and

   $$t = \frac{\ln 3}{k} = \frac{2\ln 3}{\ln(n_1/n_0)}\text{ hours}$$
3. The curve starts at $n = n_0$ when $t = 0$ and grows exponentially, more and more steeply, passing through $n_1$ at $t = 2$:

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   import numpy as np
   from matplotlib import pyplot as plt
   
   n0 = 1
   n1 = 2  # for example
   t = np.linspace(0, 10, 200)
   
   plt.plot(t, n0 * (n1 / n0)**(t / 2))
   plt.xlabel('t (hours)')
   plt.ylabel('n')

4. The model assumes that every bacterium divides at the same constant rate, whatever the size of the population. This means there are unlimited nutrients and space, no bacteria die, and nothing else (such as temperature) changes. In practice, growth slows as resources run out, so the model only describes the early stages of growth.

1. Separate the variables:

   $$\int y^2\,dy = \int \frac{2x}{3}\,dx \quad\Rightarrow\quad \frac{y^3}{3} = \frac{x^2}{3} + C \quad\Rightarrow\quad y^3 = x^2 + 3C$$

   $y(\sqrt{2}) = 1$ gives $1 = 2 + 3C$, so $3C = -1$, and

   $$y = \left(x^2 - 1\right)^{1/3}$$
2. Integrate both sides:

   $$y = \beta\int\frac{dx}{x} = \beta\ln x + C$$

   $y(1) = 0$ gives $C = 0$, so $y = \beta\ln x$. Then $y(e^3) = 1$ gives $\beta\ln(e^3) = 3\beta = 1$, so $\beta = 1/3$.
3. Integrate both sides term by term:

   $$y = a_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + \frac{a_3x^4}{4} + \frac{a_4x^5}{5} + C$$

   $y(0) = \pi$ gives $C = \pi$, so

   $$y = \pi + a_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + \frac{a_3x^4}{4} + \frac{a_4x^5}{5}$$

1. $A$ is used up at a rate proportional to $m_t$, so

   $$\frac{dm_t}{dt} = -km_t$$

   where $k$ is a positive constant. Separating the variables gives $m_t = 130e^{-kt}$, since $m_0 = 130\,$g. After 150 seconds, 50g has been transformed, so $m_{150} = 80\,$g:

   $$80 = 130e^{-150k} \quad\Rightarrow\quad k = \frac{1}{150}\ln\frac{130}{80} \approx 0.00324\,\text{s}^{-1}$$
2. After 300 seconds,

   $$m_{300} = 130e^{-300k} = 130\left(e^{-150k}\right)^2 = 130\left(\frac{80}{130}\right)^2 = \frac{640}{13} \approx 49.2\,\text{g}$$

   So, the mass transformed is $130 - 49.2 \approx 80.8\,$g. (The first 150 seconds transforms 50g, but the next 150 seconds only about 30.8g, because there is less $A$ left.)
3. $m_t$ starts at 130g and decays exponentially towards zero, falling quickly at first and then more and more slowly:

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   import numpy as np
   from matplotlib import pyplot as plt
   
   k = np.log(130 / 80) / 150
   t = np.linspace(0, 1500, 200)
   
   plt.plot(t, 130 * np.exp(-k * t))
   plt.xlabel('t (s)')
   plt.ylabel('mass of A (g)')

1. The rate of decrease of $T$ is proportional to $T - T_s$, so

   $$\frac{dT}{dt} = -k(T - T_s)$$

   where $k$ is a positive constant.
2. Separate the variables ($T_s$ is a constant):

   $$\int\frac{dT}{T - T_s} = -\int k\,dt \quad\Rightarrow\quad \ln(T - T_s) = -kt + C$$

   At $t = 0$, $T = T_0$, so $C = \ln(T_0 - T_s)$, and

   $$\ln\frac{T - T_s}{T_0 - T_s} = -kt \quad\Rightarrow\quad T - T_s = (T_0 - T_s)\,e^{-kt}$$

   The exponent $kt$ must have no units, so $k$ has units of $\text{time}^{-1}$, for example $\text{min}^{-1}$.
3. Here $T_0 = 65$, $T_s = 22$ and $T = 59$ at $t = 2$ minutes:

   $$59 - 22 = (65 - 22)e^{-2k} \quad\Rightarrow\quad k = \frac{1}{2}\ln\frac{43}{37} \approx 0.0751\,\text{min}^{-1}$$

   The glycerol reaches $29^{\circ}$C when

   $$29 - 22 = 43e^{-kt} \quad\Rightarrow\quad t = \frac{1}{k}\ln\frac{43}{7} \approx 24.2\,\text{minutes}$$

   So, the glycerol can be added after about 24 minutes.
4. Taking logs of the solution gives

   $$\ln(T - T_s) = \ln(T_0 - T_s) - kt$$

   So, plotting $\ln(T - 22)$ against $t$ gives a straight line, with intercept $\ln 43$ and gradient $-k$. We can read off the time for any temperature, or extend the line, by using a ruler:

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   import numpy as np
   from matplotlib import pyplot as plt
   
   k = np.log(43 / 37) / 2
   t = np.linspace(0, 30, 100)
   T = 22 + 43 * np.exp(-k * t)
   
   plt.plot(t, np.log(T - 22))
   plt.xlabel('t (minutes)')
   plt.ylabel('ln(T - 22)')

5. No. The constant $k$ depends on the body that is cooling: its mass, heat capacity and surface area, and the container it is in. Once the glycerol is mixed with the protein sample, it is a different body, so it will cool with a different constant.

1. Since $T = I/I_o$, the absorbance is $A = \log_{10}(1/T) = -\log_{10}T$. The Beer-Lambert law then gives

   $$-\log_{10}T = \epsilon cl \quad\Rightarrow\quad T = 10^{-\epsilon cl}$$
2. The transmitted intensity is $I = I_oT = I_o10^{-\epsilon cl}$, so

   $$Q = I_o - I = I_o\left(1 - 10^{-\epsilon cl}\right)$$

   $Q$ is zero when $l = 0$, rises steeply at first, then levels off towards $I_o$ as $l$ increases, as almost all of the light is absorbed:

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   import numpy as np
   from matplotlib import pyplot as plt
   
   I_o = 1
   eps_c = 1  # epsilon times c, for example
   l = np.linspace(0, 3, 200)
   
   plt.plot(l, I_o * (1 - 10**(-eps_c * l)))
   plt.axhline(I_o, linestyle='--', color='grey')
   plt.xlabel('l')
   plt.ylabel('Q')

3. The rate at which the intensity is attenuated is $dQ/dl$. Using $\frac{d}{dl}a^{-kl} = -k\ln(a)\,a^{-kl}$ (lecture 5),

   $$\frac{dQ}{dl} = I_o\,\epsilon c\ln(10)\,10^{-\epsilon cl}$$

   The initial rate is at $l = 0$, where $10^{0} = 1$:

   $$\begin{aligned} \left.\frac{dQ}{dl}\right|_{l=0} &= I_o\,\epsilon c\ln 10 \\ &= 200 \times 15.7\times 10^3 \times 10\times 10^{-6} \times \ln 10 \approx 72.3\,\text{W/cm} \end{aligned}$$

   (The units of $\epsilon c$ are $\text{M}^{-1}\text{cm}^{-1}\times\text{M} = \text{cm}^{-1}$.)

   1. The initial rate is proportional to $c$, so doubling the concentration doubles it.
   2. The initial rate is proportional to $I_o$, so doubling the incident intensity doubles it.
   3. The initial rate does not depend on the length of the cell, so it does not change (though more light is absorbed in total).