June 18, 2026 - Calculus 2 - Separable Differential Equations and Applications of Integration

Evaluation of Equilibrium Solutions and Initial Conditions

• When solving a differential equation of the form $y' = f(y)g(x)$, the first step is to identify equilibrium solutions by setting the function of $y$ equal to zero ($y' = 0$) and factoring.

• Equilibrium solutions represent horizontal lines where the rate of change is zero for all values of $x$.

• Once identifies, equilibrium solutions must be checked against the provided initial condition.

• In the provided example, the equilibrium solutions are $y = 4$ and $y = -2$. The initial condition is $y(0) = 3$. Since neither equilibrium solution satisfies the initial condition ($y = 4 \neq 3$ and $y = -2 \neq 3$), the solver must proceed with separation of variables.

Solving Separable Differential Equations via Integration

• Rewrite Notation: Change the derivative notation from $y'$ to $\frac{dy}{dx}$ to facilitate the separation of variables.

• Separation of Variables: Group all terms involving $y$ with $dy$ and all terms involving $x$ with $dx$.

  • Example setup: $\frac{dy}{(y-4)(y+2)} = (2x + 1)dx$

• Integration of the X-component: Integrating a polynomial like $(2x + 1)$ results in $x^2 + x + C$.

• Integration of the Y-component (Partial Fraction Decomposition): If the $y$ function is rational (\text{degree of numerator} < \text{degree of denominator}), use partial fraction decomposition.

  • Setup: $\frac{1}{(y-4)(y+2)} = \frac{A}{y-4} + \frac{B}{y+2}$

  • Clearing denominators: $1 = A(y+2) + B(y-4)$

  • Solving for constants:

    • Let $y = -2 \implies 1 = -6B \implies B = -\frac{1}{6}$

    • Let $y = 4 \implies 1 = 6A \implies A = \frac{1}{6}$

  • The integral becomes: $\frac{1}{6} \int \frac{1}{y-4} \,dy - \frac{1}{6} \int \frac{1}{y+2} \,dy = \frac{1}{6} (\ln|y-4| - \ln|y+2|)$

Algebraic Manipulation and Solving for y

• Simplifying Logarithms: Combine the natural logs using the quotient rule: $\frac{1}{6} \ln \left| \frac{y-4}{y+2} \right| = x^2 + x + C$.

• Removing Coefficients: Multiply both sides by 6 to clear the fraction: $\ln \left| \frac{y-4}{y+2} \right| = 6x^2 + 6x + 6C$. Note that $6C$ remains a constant $C$.

• Exponential Form: Rewrite the equation in exponential form to isolate the rational function: $\frac{y-4}{y+2} = e^{6x^2 + 6x + C} = Ce^{6x^2 + 6x}$.

• Solving for the Constant C: Use the initial condition $y(0) = 3$ immediately to find the value of the constant.

  • $\frac{3-4}{3+2} = Ce^0 \implies -\frac{1}{5} = C$

• Isolating y: Multiply by the denominator, distribute, and group all $y$ terms on one side to factor and solve.

  • Example: $y-4 = C(y+2)e^{f(x)}$ (Proceed to solve for $y$).

• Caveat on Constants: Do not let a constant "eat" a coefficient if you are solving for specific values on both sides of an equation (e.g., if there are $C$ terms on both sides). This would change the calculation result leading to an incorrect particular solution (e.g., getting $7$ versus $\frac{7}{2}$).

Theoretical Concept: The "Quantum Mechanics" of Differential Equations

• A general solution to a differential equation represents a family of functions. These are all potential solutions that could describe the system.

• This is conceptually similar to a wave function in quantum mechanics, which describes a range of possible states for a particle.

• Applying an initial condition is analogous to taking a measurement in quantum mechanics: it "collapses the wave function" from a range of possibilities into a single, specific solution.

Newton's Law of Cooling

• Definition: The rate of change of the temperature of an object is proportional to the difference between its current temperature ($T$) and the temperature of its surroundings ($T_s$).

• Mathematical Formula: $\frac{dT}{dt} = k(T - T_s)$

  • $T(t)$: Temperature at time $t$.

  • $T_s$: Temperature of the surroundings (ambient temperature).

  • $k$: Proportionality constant (heat capacity constant associated with the specific object).

• Derivation to General Solution:

  • Separation: $\frac{dT}{T - T_s} = k \,dt$

  • Integration: $\ln|T - T_s| = kt + C$

  • Exponential Form: $T - T_s = Ce^{kt}$

  • Final Form: $T(t) = T_s + Ce^{kt}$

Case Study: Pizza Cooling Problem

• Data Points:

  • Initial temperature (at $t = 0$): $350^{\circ}F$

  • Room temperature ($T_s$): $75^{\circ}F$

  • Temperature after 5 minutes ($t = 5$): $340^{\circ}F$

  • Goal: Find time $t$ when temperature is $300^{\circ}F$

• Step 1: Solve for C:

  • $350 = 75 + Ce^{k(0)} \implies 350 = 75 + C \implies C = 275$

• Step 2: Solve for k:

  • $340 = 75 + 275e^{5k}$

  • $265 = 275e^{5k} \implies \frac{265}{275} = e^{5k}$

  • $k = \frac{\ln\left(\frac{265}{275}\right)}{5}$

• Step 3: Solve for t at 300 Degrees:

  • $300 = 75 + 275e^{kt}$

  • $225 = 275e^{kt} \implies \frac{225}{275} = e^{kt}$

  • $t = \frac{\ln\left(\frac{225}{275}\right)}{k}$

  • Substituting $k$: $t = \frac{5 \ln\left(\frac{225}{275}\right)}{\ln\left(\frac{265}{275}\right)} \approx 27.0 \, \text{minutes}$

Identifying the Order of Differential Equations

• The order of a differential equation is determined by the highest derivative present in the equation, not the power to which a derivative is raised.

• Examples:

  • $y^{(4)} + y''' + y^2 = 0$ is a 4th order equation.

  • $(y'')^6 + y' = x$ is a 2nd order equation (the power 6 is irrelevant to the order).

  • $\frac{d^3y}{dx^3}$ represents the 3rd derivative; the notation indicates three "d"s on top and three "dx"s on the bottom as a discrete value.

Mass and Density of One-Dimensional Objects

• The mass ($m$) of a linear object (like a rod or wire) is the integral of its linear density function $\rho(x)$.

• Linear Density Formula: $m = \int_a^b \rho(x) \,dx$

• Example Problem: A wire 8 feet long with density function $\rho(x) = x^2 + 4x$.

  • Setup: $m = \int_0^8 (x^2 + 4x) \,dx$

  • Antiderivative: $\left[ \frac{x^3}{3} + 2x^2 \right]_0^8$

  • Calculation: $m = \frac{512}{3} + 128 = \frac{512 + 384}{3} = \frac{896}{3} \, \text{units of mass}$

Mass of Circular Objects (Radial Density)

• For a circular disk where density changes based on the distance from the center, we use radial density.

• The mass is calculated by integrating the area of thin concentric rings. Imagine unravelling a circle of radius $x$ into a wire of length equal to its circumference: $2\pi x$.

• Radial Mass Formula: $m = \int_0^R 2\pi x \rho(x) \,dx$

• Example Setup: Disk with radius $R$ and $\rho(x) = x^3 - 4x + 5$.

  • $m = 2\pi \int_0^R (x^4 - 4x^2 + 5x) \,dx$

Surface Area of Revolution (Around the Y-axis)

• Standard Formula (X-axis): $S = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} \,dx$

• Y-axis Revolution: To revolve around the y-axis, it is often easier to convert the function and bounds into terms of $y$.

  • Formula: $S = 2\pi \int_c^d f(y) \sqrt{1 + [f'(y)]^2} \,dy$

• Example Problem: $y = 36 - 4x^2$ from $x = 0$ to $x = 3$ revolved around the y-axis.

  • Isolate x: $4x^2 = 36 - y \implies x^2 = \frac{36-y}{4} \implies x = \sqrt{\frac{36-y}{4}} = \frac{1}{2} \sqrt{36-y}$.

  • Find New Bounds:

    • If $x = 0, y = 36 - 4(0) = 36$.

    • If $x = 3, y = 36 - 4(9) = 0$.

  • Find the Derivative: $f'(y) = \frac{d}{dy} \left[ \frac{1}{2}(36-y)^{1/2} \right] = \frac{1}{2} \cdot \frac{1}{2} (36-y)^{-1/2} \cdot (-1) = -\frac{1}{4\sqrt{36-y}}$.

  • Setup Integral: $S = 2\pi \int_0^{36} \frac{1}{2} \sqrt{36-y} \sqrt{1 + \left(-\frac{1}{4\sqrt{36-y}}\right)^2} \,dy$

  • Simplify: Distribute the $\sqrt{36-y}$ into the second radical to simplify the expression before integrating. This results in $\sqrt{36 - y + \frac{1}{16}}$.

Questions & Discussion

• Question: At what point do you use shortcuts when solving for $y$, such as keeping parts as $f(x)$?

• Response: You can do this as soon as the variables are separate. If the function on the right is a single term, it's easy to keep. If it's a long multivariable string, replacing it with a placeholder like $f(x)$ saves time during the algebraic rearrangement of $y$.

• Question: When finding the constant $C$ for square roots, do we need the plus or minus?

• Response: Normally, taking a square root requires $\pm$. However, the initial condition will dictate which sign is correct. For example, if the initial $y$ value is positive, you use the positive root. Note that in logarithmic solutions, the constant $C$ often "absorbs" the plus/minus from an absolute value, but this does not happen automatically with square roots unless there is a coefficient $C$ in front of the radical.