Math Problem Statement

y''+4y'+5y=0

Solution

The given equation is a second-order linear homogeneous differential equation:

y+4y+5y=0y'' + 4y' + 5y = 0

To solve this, we'll look for a general solution of the form y=erty = e^{rt}, where rr is a constant to be determined. Substituting y=erty = e^{rt} into the differential equation:

r2ert+4rert+5ert=0r^2 e^{rt} + 4r e^{rt} + 5 e^{rt} = 0

Factoring out erte^{rt} (which is never zero):

ert(r2+4r+5)=0e^{rt} (r^2 + 4r + 5) = 0

Thus, we solve the characteristic equation:

r2+4r+5=0r^2 + 4r + 5 = 0

Using the quadratic formula:

r=4±424(1)(5)2(1)=4±16202=4±42r = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}

r=4±2i2=2±ir = \frac{-4 \pm 2i}{2} = -2 \pm i

So, the roots are r=2+ir = -2 + i and r=2ir = -2 - i, which are complex conjugates.

The general solution for the differential equation when the roots are complex is:

y(t)=e2t(C1cos(t)+C2sin(t))y(t) = e^{-2t} (C_1 \cos(t) + C_2 \sin(t))

Where C1C_1 and C2C_2 are constants determined by initial conditions (if given).

Would you like to go over the solution in more detail, or clarify anything?

Related Questions:

  1. How can we solve the equation with initial conditions y(0)=1y(0) = 1 and y(0)=0y'(0) = 0?
  2. What is the interpretation of the complex roots in the context of this differential equation?
  3. How can the solution be visualized as a damped oscillation?
  4. What would change if the equation had real roots instead of complex ones?
  5. 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)