Math Problem Statement

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:

Solution

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

ad2ydt2+bdydt+cy=0,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:

ar2+br+c=0.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 Δ=b24ac\Delta = b^2 - 4ac:

1. Distinct Real Roots (Δ>0\Delta > 0)

  • When b24ac>0b^2 - 4ac > 0, the characteristic equation has two distinct real roots, say r1r_1 and r2r_2.
  • The general solution is given by: y(t)=C1er1t+C2er2t,y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}, where C1C_1 and C2C_2 are constants determined by initial conditions.

2. Repeated Real Roots (Δ=0\Delta = 0)

  • When b24ac=0b^2 - 4ac = 0, the characteristic equation has a repeated real root, say rr.
  • The general solution is: y(t)=(C1+C2t)ert.y(t) = (C_1 + C_2 t) e^{r t}.

3. Complex Conjugate Roots (Δ<0\Delta < 0)

  • When b24ac<0b^2 - 4ac < 0, the characteristic equation has complex conjugate roots, say r=α±βir = \alpha \pm \beta i.
  • The general solution is: y(t)=eαt(C1cos(βt)+C2sin(βt)).y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right). Here, α\alpha represents the real part, and β\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 C1C_1 and C2C_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 Δ=b24ac\Delta = b^2 - 4ac to quickly determine the nature of the solutions before solving further.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Homogeneous Equations
Characteristic Equation
Quadratic Equations

Formulas

a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0
a r^2 + b r + c = 0
\Delta = b^2 - 4ac

Theorems

Quadratic Roots Theorem
Discriminant of a Quadratic Equation

Suitable Grade Level

Undergraduate Mathematics or Advanced High School

Related Recommendation