Separable Differential Equations and Applications Flashcards

First-Order Separable Differential Equations

• Definition of Separable Differential Equations: A first-order differential equation is categorized as separable if it consists of a single derivative (the first derivative) and can be expressed in the mathematical form:     $y' = f(x) \times g(y)$

• Notation and Legal Mathematics:

  • In the standard notation $y' = \frac{dy}{dx}$, the term is viewed as an operation indicating that the derivative of function $y$ has been performed with respect to $x$.

  • In the context of separable equations, we treat $dy$ and $dx$ as differentials. These represent infinitesimal changes in the variables $y$ and $x$, respectively.

  • Technically, one cannot simply "break up" the fraction $\frac{dy}{dx}$; however, for the purposes of solving these specific types of equations, the operation allows for the separation of these differentials through a process that yields valid results.

• General Solving Strategy:

  1. Replace $y'$ with $\frac{dy}{dx}$.

  2. Separate the variables so that all terms involving $y$ (including $dy$) are on one side of the equation and all terms involving $x$ (including $dx$) are on the other.

  3. Integrate both sides of the equation.

  4. Solve for the variable $y$ to find the general solution.

Equilibrium Solutions

• Definition: Equilibrium solutions are functions where $y$ is equal to a constant (i.e., $y = k$), implying that the derivative $y'$ is zero. For these solutions, the value of $x$ can be anything, and the function appears as a horizontal line on a direction field.

• Method of Identification: To find equilibrium solutions for an equation of the form $y' = f(x) \times g(y)$, set the function of $y$ equal to zero ($g(y) = 0$) and solve for $y$.

• Example Case: In the equation $y' = (x - 4)(y^2 - 4)$, the equilibrium solutions are found by solving $y^2 - 4 = 0$, which yields:

  • $y = 2$

  • $y = -2$

  • Testing these in the original equation results in $0 = (x - 4)(0)$, confirming they are valid (though often "uninteresting") solutions.

Detailed Calculus Process: Rational Separable Equations

• Problem: Solve $y' = (x - 4)(y^2 - 4)$.

• Step 1: Separation of Variables:     $\frac{dy}{dx} = (x - 4)(y^2 - 4)$     $\frac{1}{y^2 - 4} \, dy = (x - 4) \, dx$

• Step 2: Integration:

  • The right side is a simple polynomial: $\int (x - 4) \, dx = \frac{1}{2}x^2 - 4x + C_1$.

  • The left side, $\int \frac{1}{y^2 - 4} \, dy$, requires Partial Fraction Decomposition because it is a rational function where the degree of the numerator (0) is less than the degree of the denominator (2).

• Step 3: Partial Fraction Decomposition:

  • Factor the denominator: $(y - 2)(y + 2)$.

  • Set up the equation: $\frac{1}{y^2 - 4} = \frac{A}{y - 2} + \frac{B}{y + 2}$.

  • Multiply through by the denominator: $1 = A(y + 2) + B(y - 2)$.

  • Solve for $A$ and $B$ by plugging in values for $y$:

    • If $y = 2$: $1 = A(4) \implies A = \frac{1}{4}$.

    • If $y = -2$: $1 = B(-4) \implies B = -\frac{1}{4}$.

  • The integral becomes: $\int (\frac{1}{4(y - 2)} - \frac{1}{4(y + 2)}) \, dy = \frac{1}{4} \ln|y - 2| - \frac{1}{4} \ln|y + 2| + C$.

• Step 4: Combining and Solving for y:

  • Combine the logarithms: $\frac{1}{4} \ln|\frac{y - 2}{y + 2}| = \frac{1}{2}x^2 - 4x + C_1$.

  • Multiply by 4 to clear the fraction: $\ln|\frac{y - 2}{y + 2}| = 2x^2 - 16x + 4C_1$. (Note: $4C_1$ is still just a constant, labeled $C_2$).

  • Exponentiate both sides with base $e$: $|\frac{y - 2}{y + 2}| = e^{2x^2 - 16x + C_2}$.

  • Using laws of exponents, rewrite as: $|\frac{y - 2}{y + 2}| = e^{C_2} \times e^{2x^2 - 16x}$. Since $e^{C_2}$ is a constant, it is replaced by a coefficient $C$.

  • The absolute value is absorbed into the constant coefficient $C$, as $C$ can be positive or negative.

  • Solution setup for simplified handling: Let $f(x) = C \times e^{2x^2 - 16x}$. Then $\frac{y - 2}{y + 2} = f(x)$.

  • Clear the denominator: $y - 2 = y \times f(x) + 2 \times f(x)$.

  • Isolate $y$ terms: $y - y \times f(x) = 2 \times f(x) + 2$.

  • Factor $y$: $y(1 - f(x)) = 2(f(x) + 1)$.

  • Final general form: $y = \frac{2(f(x) + 1)}{1 - f(x)}$, where $f(x) = C \times e^{2x^2 - 16x}$.

Basic Separable Equation Example

• Problem: Solve $y' = 6x^2 - 4x$.

• Process:

  1. Assume $g(y) = 1$. No equilibrium solutions exist because $1$ is never $0$.

  2. $\frac{dy}{dx} = 6x^2 - 4x \implies dy = (6x^2 - 4x) \, dx$.

  3. Integrate: $\int dy = \int (6x^2 - 4x) \, dx$.

  4. $y = \frac{6x^3}{3} - \frac{4x^2}{2} + C$.

  5. Simplified: $y = 2x^3 - 2x^2 + C$.

Trigonometric Separable Equations and Extraneous Solutions

• Problem: Solve $y' = \frac{\sec(y) + \tan(y)}{\cos(y)}$.

• Equilibrium Solution Check:

  • Setting the numerator equal to zero: $\sec(y) + \tan(y) = 0 \implies \frac{1}{\cos(y)} + \frac{\sin(y)}{\cos(y)} = 0$.

  • This implies $1 + \sin(y) = 0 \implies \sin(y) = -1$. This occurs at $y = \frac{3\pi}{2} + 2\pi n$.

  • However, at these values, $\cos(y) = 0$. Since the denominator cannot be zero, these are extraneous solutions and are not valid equilibrium solutions.

• Integration via U-Substitution:

  1. Separate: $\frac{dy}{\sec(y) + \tan(y)} = \frac{dx}{\cos(y)}$ is difficult; instead, use $\cos(y) \, dy = (\sec(y) + \tan(y)) \, dx$.

  2. Wait, rewrite the original derivative for easier integration: $\frac{dy}{dx} = \frac{1 + \sin(y)}{\cos^2(y)}$.

  3. Separate: $\frac{\cos^2(y)}{1 + \sin(y)} \, dy = dx$.

  4. Integrating LHS using $\cos^2(y) = 1 - \sin^2(y) = (1 - \sin(y))(1 + \sin(y))$:

     • The expression simplifies to $\int (1 - \sin(y)) \, dy = \int dx$.

  5. Alternatively, the speaker proposes u-substitution: Let $u = 1 + \sin(y)$, then $du = \cos(y) \, dy$.

  6. The specific problem demonstrated led to $\ln|1 + \sin(y)| = x + C$.

  7. Solving for $y$: $1 + \sin(y) = C \times e^x \implies \sin(y) = C \times e^x - 1 \implies y = \arcsin(C \times e^x - 1)$.

Factoring to Achieve Separability

• Concept: Some differential equations do not appear separable until they are factored, often via grouping.

• Problem: Solve $y' = xy - 2y + 3x - 6$.

• Factoring Step:

  • Group terms: $(xy - 2y) + (3x - 6)$.

  • Factor out common terms: $y(x - 2) + 3(x - 2)$.

  • Factor out the shared binomial: $(y + 3)(x - 2)$.

• Equilibrium Solution: $y + 3 = 0 \implies y = -3$.

• Solving:

  1. $\frac{1}{y + 3} \, dy = (x - 2) \, dx$.

  2. Integrate: $\ln|y + 3| = \frac{1}{2}x^2 - 2x + C$.

  3. Exponentiate: $y + 3 = C \times e^{0.5x^2 - 2x}$.

  4. Final equation: $y = C \times e^{0.5x^2 - 2x} - 3$.

Initial Value Problems (IVP) and Particular Solutions

• Initial Conditions: A particular solution determines the exact value of the constant $C$ based on a given point $(x_0, y_0)$.

• Example: Solve $y' = (x - 4)(y^2 - 4)$ with $y(0) = 3$.

• Process:

  1. Start with the general solution derived earlier: $y = \frac{2(f(x) + 1)}{1 - f(x)}$ where $f(x) = C \times e^{2x^2 - 16x}$.

  2. Plug in $x = 0$ and $y = 3$:

     • $f(0) = C \times e^{0} = C$.

     • $3 = \frac{2(C + 1)}{1 - C}$.

  3. Solve for $C$:

     • $3(1 - C) = 2(C + 1)$.

     • $3 - 3C = 2C + 2$.

     • $1 = 5C \implies C = \frac{1}{5}$.

  4. Replace $C$ in the function: $f(x) = \frac{1}{5} \times e^{2x^2 - 16x}$.

  5. Final Particular Solution: $y = \frac{2(\frac{1}{5}e^{2x^2 - 16x} + 1)}{1 - \frac{1}{5}e^{2x^2 - 16x}}$. Multiplying by 5 to clean up the complex fraction gives: $y = \frac{2e^{2x^2 - 16x} + 10}{5 - e^{2x^2 - 16x}}$.

Word Problem: Mixed Tank Brine Solution (The "Brine" Problem)

• Scenario Description:

  • "Brine is a mixture of salt and water."

  • Tank Volume: $100\,L$ (remains constant because inflow rate equals outflow rate).

  • Initial Salt Content: $4\,kg$ at $t = 0$.

  • Inflow: $2\,L/min$ brine containing a concentration of $0.5\,kg/L$ of salt.

  • Outflow: $2\,L/min$ of the mixed solution is drained from the bottom.

• Variable Definition: Let $S(t)$ be the amount of salt in kilograms in the tank as a function of time $t$.

• Setting up the Differential Equation:

  • $\frac{dS}{dt} = \text{Inflow Rate of Salt} - \text{Outflow Rate of Salt}$.

  • Inflow Rate of Salt: $2\,L/min \times 0.5\,kg/L = 1\,kg/min$.

  • Outflow Rate of Salt: $2\,L/min \times \frac{S(t)}{100\,L} = \frac{S(t)}{50}\,kg/min$.

  • Equation: $\frac{dS}{dt} = 1 - \frac{S}{50}$.

• Solving the Equation:

  1. Combine common denominator: $\frac{dS}{dt} = \frac{50 - S}{50}$.

  2. Separate: $\frac{dS}{50 - S} = \frac{1}{50} \, dt$.

  3. Integrate: $\int \frac{1}{50 - S} \, dS = \int \frac{1}{50} \, dt \implies -\ln|50 - S| = \frac{t}{50} + C$.

  4. Multiply by $-1$: $\ln|50 - S| = -\frac{t}{50} + C$.

  5. Exponentiate: $50 - S = C \times e^{-t/50}$.

  6. Solve for $S$: $S(t) = 50 - C \times e^{-t/50}$.

• Applying Initial Condition $S(0) = 4$:

  • $4 = 50 - C \times e^0 \implies 4 = 50 - C \implies C = 46$.

• Final Salt Function: $S(t) = 50 - 46e^{-t/50}$.

• Limiting Amount of Salt: As time goes to infinity ($t \rightarrow \infty$), the term $e^{-t/50}$ goes to $0$. Thus, the limiting amount of salt is $50\,kg$. This means the tank eventually reaches the concentration of the incoming brine solution ($0.5\,kg/L \times 100\,L = 50\,kg$).

Questions & Discussion

• Question: Does the absolute value change anything when integrating for natural logs?

  • Answer: In these problems, it is typically absorbed into the constant coefficient $C$. When we exponentiate both sides, the constant $e^{C}$ becomes a coefficient that can account for both positive and negative results.

• Question: How does raising everything to the power of $e$ work? Is $e$ the base for everything?

  • Answer: Yes, $e$ is the base of the natural log. Applying it to both sides cancels the natural log, turning one side into the argument of the log and the other side into an exponent of $e$.

• Question: Looking at Line 2 of the factoring problem (Example 4), how did you pull that out?

  • Answer: It is factoring by grouping. By pulling a $y$ out of the first two terms and a $3$ out of the second two, we see that the binomial $(x - 2)$ is being multiplied by both, allowing us to factor it out entirely.

• Question: Why do we only have one constant $C$?

  • Answer: If we have constants on both sides (e.g., $C_1$ and $C_2$), we can simply subtract one from the other to create a single resulting constant. Therefore, we generally only write it on the side with the independent variable $x$.

• Question on Tank Problem: Could you explain the outflow concentration again?

  • Answer: The outflow rate is the volume of water leaving ($2\,L/min$) multiplied by the concentration of salt. Since the salt is mixed, its concentration is the total current amount of salt ($S(t)$) divided by the total volume ($100\,L$). This yields the rate at which mass is leaving the system.

Class Information and Attendance

• Planned Topics: Newton's Law of Cooling (Word Problems) and more IVP particular solutions.

• Upcoming Quiz: Will cover finding the order of a differential equation and a non-word problem separable IVP.

• Attendance proper nouns identified: Amber, Jordan, Matthew, Ethan, Kevin Davis, Bowen, Steen, Alan, Celeste, Kevin, Antonio, Jaden, Carson, Kate, Lindsay, Corbin, Olivia, Brian, Sam, Susie, Dylan, Jamar.