- Solve differential equations by separating variables
- Apply separation of variables to real-world problems
Separation of Variables
The Main Idea
Think of separation of variables like untangling a knot—you separate the [latex]x[/latex] stuff from the [latex]y[/latex] stuff, then deal with each side independently.
When Can You Use This Method? Your equation must be separable: [latex]y' = f(x) \cdot g(y)[/latex]. The right side breaks into one function of [latex]x[/latex] times one function of [latex]y[/latex].
Quick Recognition Test:
- [latex]y' = (x^2 - 4)(3y + 2)[/latex] → Separable ✓
- [latex]y' = xy + 2x[/latex] → Separable (factor as [latex]x(y + 2)[/latex]) ✓
- [latex]y' = x + y^2[/latex] → Not separable (can’t factor) ✗
Problem-Solving Strategy:
- Check for constant solutions: Set [latex]g(y) = 0[/latex] to find equilibrium points
- Separate variables: Rearrange to [latex]\frac{dy}{g(y)} = f(x)dx[/latex]
- Integrate both sides: [latex]\int \frac{dy}{g(y)} = \int f(x)dx[/latex]
- Solve for [latex]y[/latex] (if possible—sometimes you’re stuck with implicit solutions)
- Apply initial conditions to find specific constants
You’re essentially moving all the [latex]y[/latex] terms to one side and all the [latex]x[/latex] terms to the other, then integrating each side separately.
Don’t forget those constant solutions from Step 1! They’re often missed but represent important equilibrium states.
Use the method of separation of variables to find a general solution to the differential equation [latex]y^{\prime} =2xy+3y - 4x - 6[/latex].
Find the solution to the initial-value problem
using the method of separation of variables.
Applications of Separation of Variables
The Main Idea
Separation of variables isn’t just an abstract math technique—it’s the key to solving problems you encounter in labs, kitchens, and anywhere things mix, flow, or change temperature.
The Two Big Categories:
Tank/Mixing Problems: Track how concentrations change over time
- Setup: [latex]\frac{du}{dt} = \text{INFLOW RATE} - \text{OUTFLOW RATE}[/latex]
- Key insight: Inflow brings in new stuff, outflow removes mixed stuff
- Common pattern: Salt solution flowing into a tank while mixed solution flows out
Cooling/Heating Problems: Newton’s Law of Cooling
- Setup: [latex]\frac{dT}{dt} = k(T - T_s)[/latex] where [latex]T_s[/latex] is ambient temperature
- Key insight: Temperature change rate depends on the temperature difference
- Common pattern: Hot coffee cooling to room temperature, pizza cooling after leaving the oven
Problem-Solving Strategy:
- Step 1: Identify what’s changing (salt amount, temperature, etc.)
- Step 2: Set up the rate equation based on physical principles
- Step 3: Apply separation of variables
- Step 4: Use given information to find constants
- Step 5: Answer the specific question asked
Units matter! Make sure rates, concentrations, and times all work together.
A tank contains [latex]3[/latex] kilograms of salt dissolved in [latex]75[/latex] liters of water. A salt solution of [latex]0.4\text{kg salt/L}[/latex] is pumped into the tank at a rate of [latex]6\text{L/min}[/latex] and is drained at the same rate. Solve for the salt concentration at time [latex]t[/latex]. Assume the tank is well mixed at all times.
A cake is removed from the oven after baking thoroughly, and the temperature of the oven is [latex]450^\circ\text{F}\text{.}[/latex] The temperature of the kitchen is [latex]70^\circ\text{F}[/latex], and after [latex]10[/latex] minutes the temperature of the cake is [latex]430^\circ\text{F}\text{.}[/latex]
- Write the appropriate initial-value problem to describe this situation.
- Solve the initial-value problem for [latex]T\left(t\right)[/latex].
- How long will it take until the temperature of the cake is within [latex]5^\circ\text{F}[/latex] of room temperature?