Math Problem Statement
y''+4y'+5y=0
Solution
The given equation is a second-order linear homogeneous differential equation:
To solve this, we'll look for a general solution of the form , where is a constant to be determined. Substituting into the differential equation:
Factoring out (which is never zero):
Thus, we solve the characteristic equation:
Using the quadratic formula:
So, the roots are and , which are complex conjugates.
The general solution for the differential equation when the roots are complex is:
Where and are constants determined by initial conditions (if given).
Would you like to go over the solution in more detail, or clarify anything?
Related Questions:
- How can we solve the equation with initial conditions and ?
- What is the interpretation of the complex roots in the context of this differential equation?
- How can the solution be visualized as a damped oscillation?
- What would change if the equation had real roots instead of complex ones?
- How does the behavior of solutions to second-order linear differential equations depend on the discriminant of the characteristic equation?
Tip:
Complex roots indicate oscillatory behavior, often seen in systems involving damping or wave-like motion.
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Second-Order Linear Homogeneous Equations
Complex Numbers
Oscillatory Behavior
Formulas
y'' + 4y' + 5y = 0
General solution: y(t) = e^{rt}
Characteristic equation: r^2 + 4r + 5 = 0
Quadratic formula: r = (-b ± sqrt(b^2 - 4ac)) / 2a
Complex solution: y(t) = e^{-2t} (C_1 cos(t) + C_2 sin(t))
Theorems
Quadratic Formula
General Solution for Complex Roots in Differential Equations
Suitable Grade Level
Undergraduate level (Calculus II or Differential Equations)
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