Definition of Separable First-order ODEs
Separable First-order ODEs are ODEs that can be written in the form of
\begin{equation}
\frac{dy}{dx} = f(x) g(y) \tag{1}
\end{equation}
where \( f(x) \) \([ g(y) ]\) is some function of \( x \) \([ y ]\) only.
Solution Method: Separation of Variables
To solve this kind of separable ODEs, we heuristically move \( g(y) \) \([ dx ]\) and group it with \( dy \) \([ f(x) ]\). This involves treating the derivative \( dy/dx \) like a fraction informally and the rigorous justification is out of our scope. Subsequently, we integrate both sides together:
\begin{align}
\frac{dy}{dx} &= f(x) g(y) \\
\frac{dy}{g(y)} &= f(x) dx \\
\int\frac{dy}{g(y)} &= \int f(x) dx \tag{2} \\
\end{align}
This is known as the method of Separation of Variables.
Initial Condition
The full solution also requires an initial condition (I.C.) to determine the value of the integration constant that arises in the end. Let’s illustrate this with a small example: Solve
\begin{equation}
\frac{dy}{dx} = \frac{x^2}{y} \tag{3}
\end{equation}
with the initial condition \( y=1 \) when \( x=0 \), or succinctly written as \( y(0) = 1 \). Separation of Variables gives
\begin{align}
\frac{dy}{dx} &= \frac{x^2}{y} \\
\int y dy &= \int x^2 dx \\
\frac{1}{2}y^2 &= \frac{1}{3}x^3 + C \\
y &= \pm \sqrt{2}\sqrt{\frac{1}{3}x^3 + C} \tag{4}
\end{align}
where the constant \( +C \) emerges after the indefinite integration. Given the I.C. \( y(0) = 1 \), we will have to take the positive root in the solution, and substitution gives
\begin{align}
(1) &= \sqrt{2}\sqrt{\frac{1}{3}(0)^3 + C} \\
1 &= \sqrt{2C} \\
\Rightarrow C &= \frac{1}{2} \tag{5}
\end{align}
and therefore the full solution will be
\begin{equation}
y = \sqrt{2}\sqrt{\frac{1}{3}x^3 + \frac{1}{2}} = \sqrt{\frac{2}{3}x^3 + 1} \tag{6}
\end{equation}
\( x > (-3/2)^{1/3} \).
Examples
Radioactive Decay
The physical process of Radioactive Decay follows the ODE
\begin{equation}
\frac{dN}{dt} = -kN \tag{7}
\end{equation}
where \( N(t) \) is the number of radioactive atoms at time \( t \). \(k\) is the decay constant which dictates the probability of an individual isotope atom to decay per unit time and has the unit of \( \text{s}^{-1} \). So, on average, there will be \( k \cdot N \) radioactive atoms to decay over a unit time interval and this implies the form of the ODE, where the L.H.S. \( dN/dt\) represents the corresponding rate of change in the number of radioactive atoms.
Using Separation of Variables to solve (7) leads to
\begin{align}
\frac{dN}{dt} &= -kN \\
\int \frac{dN}{N} &= -\int k dt \\
\ln N &= -kt + C & \text{(Required \( N > 0 \))} \\
N &= e^{C} e^{-kt} \tag{8}
\end{align}
If the initial isotope concentration is \( N_0 \) when \( t = 0 \), then
\begin{align}
(N_0) &= e^{C} e^{-k(0)} \\
\Rightarrow e^{C} &= N_0 \\
\Rightarrow N &= N_0 e^{-kt} \tag{9}
\end{align}
The full solution (9) is thus an exponential decay. From this we can infer the time required for the radioactive isotope to halve in amount: Half-life \( \tau_{1/2} \).
\begin{align}
\frac{\frac{1}{2}N}{N} &= \frac{N_0 e^{-k(t+\tau_{1/2})}}{N_0 e^{-kt}} \\
\frac{1}{2} &= e^{-k\tau_{1/2}} \\
\tau_{1/2} &= \frac{\ln 2}{k} \tag{10}
\end{align}
Hence there exists an inverse relationship between half-life and the decay constant. For example, the carbon-14 isotope is known to have a half-life of about 5730 years, and the corresponding decay constant is therefore
\begin{equation}
k = \frac{\ln 2}{5730 \text{ yrs}} = \frac{0.693147}{1.81 \times 10^{11} \text{ s}} \approx 3.83 \times 10^{-12} \text{ s}^{-1}
\end{equation}
Exercise
Solve the following ODE
\begin{equation}
(\csc x) \frac{dy}{dx} = y+1
\end{equation}
with the I.C. \( y(0) = 2 \).
Answer
\begin{align}
(\csc x) \frac{dy}{dx} &= y+1 \\
\int \frac{dy}{y+1} &= \int \sin x dx \\
\ln |y+1| &= -\cos x + C \\
y &= \pm e^{C -\cos x} -1 = Ae^{-\cos x} -1
\end{align}
where \(A = \pm e^C\). Substituting the I.C. yields \( A = 3e \), and therefore the final solution is
\begin{equation}
y = 3e^{1 -\cos x} -1
\end{equation}
Leave a Reply