Separable Equations

Lecture 1.4: Separable Equations

Introduction

• Focuses on separable differential equations expressed as:
  • $\frac{dy}{dt} = f(t, y)$
• Definition: A first-order ordinary differential equation (ODE) is separable if it can be expressed as:
  • $f(t, y) = g(t) h(y)$
  • Where both functions depend solely on t and y respectively.

Key Examples of Separable Equations

Example 1: Basic Separable Equation

• Given:
  $\frac{dy}{dt} = ty + 2(1-t) - y$
• Simplifying $f(t, y)$ gives:
  $f(t, y) = ty + 2 - 2t - y$
• Factorization leads to:
  • $(t-1)(y-2)$
  • Thus $g(t) = (t-1)$ and $h(y) = (y-2)$.

Example 2: Exponential Function

• Given:
  $\frac{dy}{dt} = e^{t}$
• Here,
  • $f(t, y) = e^t$
• Note: This can be rewritten in a separable form, but not all ODEs are separable.

Steps to Solve Separable Differential Equations

Step 1: Identify Functions

• Write the equation as:
  • $\frac{dy}{dt} = g(t) h(y)$

Step 2: Rearrange

• Express as:
  • $dy = g(t) h(y) dt$

Step 3: Isolate Variables

• Change to:
• $\frac{dy}{h(y)} = g(t) dt$

Step 4: Integrate Both Sides

• Integrate to get:
  • $\int \frac{1}{h(y)} dy = \int g(t) dt$
• The result will yield:
  • $H(y) = G(t) + C$
  • Where C is a constant.

Step 5: Solve for y

• Rearrange to solve for y in terms of t.

Specific Examples

Example 3: Solving a Specific Separable Differential Equation

1. Start with:
   $\frac{dy}{dt} = (t-1)(y-2)$
2. Rearranging gives:
   $\frac{dy}{y-2} = (t-1) dt$
3. Integrate both sides:
   • $\int \frac{1}{y-2} dy = \int (t-1) dt$
4. Solutions yield:
   • Left Side yields: $\ln|y-2|$
   • Right Side yields: $\frac{1}{2}t^2 - t + C$
5. Thus combining gives:
   • $y - 2 = e^{(\frac{1}{2}t^2 - t + C)}$
6. Rearranging solves for y:
   • $y = 2 + Ce^{(\frac{1}{2}t^2 - t)}$

Particular Solutions

• Examine when $y = 2$:
  • It satisfies the differential equation because if dy = 0 it generates:
  • $(t-1)(0) = 0$

Example 4: Y-Cubed Separable Equation

1. Start with:
   • $\frac{dy}{dx} = y^2 - 4$
2. Factor yields:
   • $(y-2)(y+2)$
3. Set:
   • $g(x) = 1$
   • $h(y) = (y-2)(y+2)$
4. Rearranging yields:
   • $\frac{dy}{(y-2)(y+2)} = dx$
5. Using partial fractions to integrate:
   $\frac{A}{y-2} + \frac{B}{y+2} = 1$
6. Determines A, B by solving:

• Collect and integrate terms to get:
  • $\ln|y-2| - \ln|y+2| = 4x + C$

1. Exponential forms then lead to:
   • $y = 2 + Ce^{4x}$
   • This describes the one-parameter family of solutions.

Implicit Solution Example

• Result defined in implicit form:
  • General form:
    $G(x,y) = 0$
  • Note behavior for continuity and differentiability.
• End effects on bounds of solutions:
  • Can verify by applying initial conditions to yield specific solutions such as those given.

Conclusion

• This lecture provides tools for working through separable differential equations, showcasing how to factor, integrate, and apply initial conditions to derive explicit and implicit forms of solutions. All methods presented are crucial for solving related ordinary differential equations in various applications.