Lecture 33: Separable Differential Equations and Applications

First Order Differential Equations: Separable Equations

• Lecture focuses on separable differential equations and their engineering applications.

First Order Differential Equations

• A first-order differential equation can be written as: $\frac{dy}{dx} = f(x, y)$, where $f(x, y)$ is a function depending on both $x$ and $y$.
• $y$ is the unknown solution, implicit in the equation.
• The set of all solutions forms the general solution, a family of functions.
• Imposing the condition $y(x<em>0) = y</em>0$ turns the differential equation into an initial value problem (IVP).

Separable First Order Differential Equations

• Separable differential equations have the form: $\frac{dy}{dx} = f(x)g(y)$.
• Separation of variables leads to: $\frac{dy}{g(y)} = f(x)dx$.
• Solving involves integrating both sides: $\int \frac{1}{g(y)} dy = \int f(x) dx + C$.

Examples of Separable Differential Equations

Example 1

• Given differential equation: $\frac{dy}{dx} = k(y - y<em>0)$, where $k$ and $y</em>0$ are constants.
• Separation gives: $\frac{dy}{y - y_0} = k dx$.
• Integrating both sides: $\int \frac{dy}{y - y<em>0} = \int k dx$ yields $\ln|y - y</em>0| = kx + c$.
• Taking exponentials: $y - y_0 = e^{kx + c} = e^{kx}e^c$.
• Rename $e^c = A$: $y = y_0 + Ae^{kx}$, where $A$ is an arbitrary constant.

Example 2 (Corrected)

• Differential equation: $\frac{dy}{dx} = -\frac{x}{y}$.
• Separating variables: $y dy = -x dx$.
• Integrating: $\int y dy = -\int x dx + C$ leads to $\frac{y^2}{2} = -\frac{x^2}{2} + C$.
• Which implies $y^2 + x^2 = 2C$.
• Impose initial condition $y(0) = r<em>0$ (with r0 > 0).
• Since $y = \sqrt{2C - x^2}$, then $y(0) = \sqrt{2C} = r<em>0$, so $r</em>0^2 = 2C$.
• The solution to the IVP is $y = \sqrt{r_0^2 - x^2}$.
  • Solution holds for positive $r<em>0$, a similar analysis can be performed for negative $r</em>0$.

Example 3

• Differential equation: $\frac{dy}{dx} = y(x + 1)$.
• Separating variables: $\frac{dy}{y} = (x + 1) dx$.
• Integrating: $\int \frac{dy}{y} = \int (x + 1) dx$ yields $\ln|y| = \frac{x^2}{2} + x + C$.
• Taking exponentials: $y = e^{\frac{x^2}{2} + x + C} = e^{\frac{x^2}{2} + x} e^C$.
• Let $A = e^C$: $y = A e^{\frac{x^2}{2} + x}$.

Application 1: R-C Circuit

• Consider a circuit with resistance $R$, capacitance $C$, and voltage source $V$.
• $I(t)$ is the current, and $Q(t)$ is the charge on the capacitor.
• Experimental law: $IR + \frac{Q}{C} = V$.
• Also, the current is the derivative of the charge with respect to time: $I = \frac{dQ}{dt}$.
• Substituting, we get: $R \frac{dQ}{dt} + \frac{Q}{C} = V$, a first-order separable differential equation.
• Given: $R = 50 \Omega$, $C = 2$ Farads, $V = 12$ Volts.
• The equation becomes: $50 \frac{dQ}{dt} + \frac{Q}{2} = 12$.
• Solving for $\frac{dQ}{dt}$: $50 \frac{dQ}{dt} = 12 - \frac{Q}{2} = \frac{24 - Q}{2}$.
• Separating variables: $\frac{dQ}{24 - Q} = \frac{dt}{100}$.
• Integrating: $\int \frac{dQ}{24 - Q} = \int \frac{dt}{100}$ gives $- \ln|24 - Q| = \frac{t}{100} + C$.
• Multiplying by -1, we have $\ln|24 - Q| = -\frac{t}{100} - C$.
• Exponential on both sides: 24 - Q = $e^{-\frac{t}{100} - C}$.
• Which becomes: $24 - Q = e^{-\frac{t}{100}}e^{-C}$.
• Which then becomes: $Q = 24 - e^{-\frac{t}{100}}e^{-C}$.
• Initial condition: $Q(0) = 0$.
• This means $0 = 24 - e^{0}e^{-C}$ which means that $e^{-C} = 24$.
• Therefore, $Q(t) = 24 - 24e^{-\frac{t}{100}}$.
• Final charge: $\lim<em>{t \to \infty} Q(t) = \lim</em>{t \to \infty} (24 - 24e^{-\frac{t}{100}}) = 24$ Coulombs.
• Time to reach 90% of total charge: $0.9 \times 24 = 21.6$.
• Solve $21.6 = 24 - 24e^{-\frac{t}{100}}$ for $t$.
• $21.6 - 24 = -24e^{-\frac{t}{100}}$
• $-2.4 = -24e^{-\frac{t}{100}}$.
• $0.1 = e^{-\frac{t}{100}}$.
• $\ln(0.1) = -\frac{t}{100}$.
• $t = -100 \ln(0.1) = -100 \ln(10^{-1}) = 100\ln(10) \approx 230.3$ seconds.
• Absolute value clarification: Removing the absolute value implies 24 - Q > 0, meaning Q < 24, consistent with the physical constraint.

Application 2: Dilution of Salt Solution

• Tank with 600 liters of water and 200 grams of salt at $t = 0$.
• Brine solution pumped in at 3 liters/minute with 4 grams of salt/liter.
• Mixed solution pumped out at 3 liters/minute.
• Problem: Find the salt concentration of the pumped-out solution as a function of time.
• Definitions:
  • $S(t)$ = Amount of salt in the tank at time $t$.
  • $\frac{dS}{dt}$ = Rate of change of salt in the tank.
• Rate of change equals to Rate of salt entering - Rate of salt leaving
• Rate of salt entering: 3 liters/minute * 4 grams/liter = 12 grams/minute.
• Rate of salt leaving: $(\frac{S(t)}{600}) \cdot 3 = \frac{3S(t)}{600}$ grams/minute.
• Differential equation: $\frac{dS}{dt} = 12 - \frac{3S}{600}$.
• Initial condition: $S(0) = 200$ grams.
• To solve the IVP all that is required is to:
  1. Separate variables
  2. Integrate both sides to get general solution
  3. Apply initial conditions to get specific solution
• Amount of salt in the tank at time t (solution after process), $S(t)=2400-2200e^{-t/200}$
• Concentration of salt leaving the tank: $\frac{S(t)}{600} = \frac{2400 - 2200e^{-\frac{t}{200}}}{600} = 4 - \frac{11}{3}e^{-\frac{t}{200}}$.
• Asymptotic behavior can be seen over time, concentration gets closer to 4.