Linear Equations

Lecture I.5: Linear Equations

Introduction

• Focus on the study of the linear equation denoted as:
  • $a(t) \frac{dy}{dt} + P(t) y = g(t) \quad (Egn \, 5.1)$
• Dependent variable is denoted as $y$.

Definitions

• Homogeneous Linear Equation:
  • Defined when $g(t) = 0$ in $(Egn \, 5.1)$, hence it becomes:
  • $a(t) \frac{dy}{dt} + P(t) y = 0$
• Standard Form of the equation:
  • $\frac{dy}{dt} + P(t) y = f(t)$

Example of Linear Equation

• Given:
  • $a(t) = t^2, \ P(t) = 3$
• The equation simplifies to:
  • $\frac{dy}{dt} + 3y = t^2$

Steps to simplify:

1. Divide by $a(t)$ to normalize:
   • $\frac{1}{a(t)} \frac{dy}{dt} + \frac{P(t)}{a(t)}y = \frac{g(t)}{a(t)}$
2. Resulting form:
   • $\frac{dy}{dt} + P(t) = \frac{g(t)}{a(t)}$
   • Therefore, $P(t) = \frac{3}{t^2}$

Solutions of Linear Equations

• General Solution $y(t)$:
  • Composed of:
  • Homogeneous solution, $y_h(t)$
  • Particular solution, $y_p(t)$
• Generally stated as:
  • $y(t) = y<em>h(t) + y</em>p(t)$

Solving the Linear Equation:

Step 1: Rewrite the Equation

• Rearrange the equation into standard form:
  • $\frac{dy}{dt} + P(t)y = g(t)$

Step 2: Solve the Homogeneous ODE

• The homogeneous equation becomes:
  • $a(t) \frac{dy}{dt} + P(t)y = 0$

Step 3: Find the Particular Solution

• Use the formula:
  • $\frac{dy}{dt} + P(t)y = g(t)$

Step 4: General Formulation

• Write the general solution as:
  • $y(t) = y<em>h(t) + y</em>p(t)$

Homogeneous Solutions

• The homogeneous version, assuming associate behavior with known solutions.
  • If the homogeneous part of a problem is:
  • $a(t) \frac{dy}{dt} + P(t)y = 0$

Example 1

• Given:
  • $\frac{dy}{dt} + ext{sin}(t)y = 0$
• This is a separable equation:
  • $\frac{dy}{y} = -P(t) dt$

Solving the Homogeneous Problem:

1. Start with equation:
   • $a(t) \frac{dy}{dt} + P(t)y = 0$
2. Compute:
   • $\int \frac{dy}{y} = -\int P(t) dt$
3. If $C$ is the constant, the solution becomes:
   • $y_h = C e^{-P(t)}$

Step-by-Step: Solving a Homogeneous Equation

Step 1: Setup

• Homogeneous equation is:
  • $2y + P(t)y = 0$
  • Assume:
  • $P(t) = ext{some function of } t$

Example of a Simple Homogeneous Solution

• Let $P(t) = -4$:
  • $\frac{dy}{dx} - 4y = 0$
  • Solutions tend to form as:
  • $y(x) = Ce^{4x}$

Conclusion:

• The equation explores various methods of finding solutions for linear equations, including integrating factors and variation of parameters, to determine both homogeneous and particular solutions effectively.