Solving Homogeneous Equations
For equations of the form:
a (d^2 y)/(dt^2 )  +dy/dt+cy=0
the characteristic equation is ar^2+br+c=0. Solutions depend on the nature of the roots:


In solving second-order linear homogeneous differential equations of the form:

\[
a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0,
\]

we use the **characteristic equation** associated with the differential equation, which is a quadratic equation:

\[
a r^2 + b r + c = 0.
\]

The nature of the solutions depends on the roots of this quadratic equation. There are three main cases based on the discriminant \( \Delta = b^2 - 4ac \):

### 1. **Distinct Real Roots** (\( \Delta > 0 \))
- When \( b^2 - 4ac > 0 \), the characteristic equation has two distinct real roots, say \( r_1 \) and \( r_2 \).
- The general solution is given by:
  \[
  y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t},
  \]
  where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

### 2. **Repeated Real Roots** (\( \Delta = 0 \))
- When \( b^2 - 4ac = 0 \), the characteristic equation has a repeated real root, say \( r \).
- The general solution is:
  \[
  y(t) = (C_1 + C_2 t) e^{r t}.
  \]

### 3. **Complex Conjugate Roots** (\( \Delta < 0 \))
- When \( b^2 - 4ac < 0 \), the characteristic equation has complex conjugate roots, say \( r = \alpha \pm \beta i \).
- The general solution is:
  \[
  y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right).
  \]
  Here, \( \alpha \) represents the real part, and \( \beta \) is the imaginary part.

These are the basic solutions depending on the discriminant of the characteristic equation.

Would you like details on how to apply these solutions in specific cases or explore further examples?

### Relative Questions:
1. How do you solve a second-order homogeneous equation with complex roots?
2. What are the key steps in deriving the characteristic equation from the original differential equation?
3. How can initial conditions be used to find specific values of \( C_1 \) and \( C_2 \)?
4. What happens when the differential equation has variable coefficients instead of constant ones?
5. How can you apply this method to solve physical problems like oscillations?

### Tip:
Always check the discriminant \( \Delta = b^2 - 4ac \) to quickly determine the nature of the solutions before solving further.

Solving Homogeneous Equations. For equations of the form: a (d^2 y)/(dt^2 )  +dy/dt+cy=0 the characteristic equation is ar^2+br+c=0. Solutions depend on the nature of the roots.

Learn how to solve second-order linear homogeneous differential equations, including distinct, repeated, and complex roots. Explore the characteristic equation, its discriminant, and solutions based on the nature of the roots. Suitable for advanced high school and undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation