Chapter 17: Problem 17
Find the general solution of the given equation. $$y^{\prime \prime}+2 y^{\prime}+4 y=0$$
Short Answer
Expert verified
The general solution is \(y(t) = e^{-t}(C_1 \cos(\sqrt{3} t) + C_2 \sin(\sqrt{3} t))\).
Step by step solution
01
Identify the Characteristic Equation
The differential equation is given by \(y'' + 2y' + 4y = 0\). Start by finding the characteristic equation by assuming a solution of the form \(y = e^{rt}\). This leads to the characteristic equation: \(r^2 + 2r + 4 = 0\).
02
Solve the Characteristic Equation
Solve the quadratic equation \(r^2 + 2r + 4 = 0\) using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), \(c = 4\). Calculate the discriminant \(b^2 - 4ac = 2^2 - 4 \times 1 \times 4 = 4 - 16 = -12\). Since the discriminant is negative, the roots are complex.
03
Express Complex Roots
The roots are \(r = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2}\). Simplifying, we get \(r = -1 \pm i\sqrt{3}\).
04
Form the General Solution
With complex roots \(r = \alpha \pm i\beta\), where \(\alpha = -1\) and \(\beta = \sqrt{3}\), the general solution to the differential equation is \(y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t))\). Substitute \(\alpha = -1\) and \(\beta = \sqrt{3}\) to get \(y(t) = e^{-t}(C_1 \cos(\sqrt{3} t) + C_2 \sin(\sqrt{3} t))\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
In the study of linear differential equations, a crucial first step is the formation of the characteristic equation. For the given second order differential equation, \( y'' + 2y' + 4y = 0 \), we assume a solution of the form \( y = e^{rt} \), where \( r \) is a constant. This assumption transforms the differential equation into a quadratic polynomial expression called the characteristic equation: \( r^2 + 2r + 4 = 0 \). This equation forms the backbone of finding the roots that will guide us to the general solution of the differential equation.
- The characteristic equation applies only to linear differential equations with constant coefficients.
- The roots of this polynomial - real or complex - determine the structure of the solution.
- The form \( y = e^{rt} \) simplifies the differentiation process to just multiplication, which is why it's generally assumed for such analysis.
Complex Roots
When solving the characteristic equation, \( r^2 + 2r + 4 = 0 \), by employing the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), students may encounter complex roots. Here, \( a = 1 \), \( b = 2 \), and \( c = 4 \). First, we compute the discriminant: \( b^2 - 4ac = 4 - 16 = -12 \). Because the discriminant is negative, we acknowledge that our roots are complex. The roots are determined to be \( r = -1 \pm i\sqrt{3} \). These roots take the form \( \alpha \pm i\beta \), where \( \alpha = -1 \) and \( \beta = \sqrt{3} \). Such complex roots indicate that the solution not only decays or grows exponentially because of \( \alpha \), but also oscillates due to the imaginary part \( \beta \).
- Complex roots arise in differential equations when the system described has oscillatory behavior.
- With complex roots, solutions to differential equations often involve trigonometric functions: cosine and sine, representing oscillations.
- Negative \( \alpha \) suggests that the solution’s oscillations will dampen over time.
General Solution
Once the characteristic roots are known, the general solution of a differential equation can be formulated. For complex roots \( \alpha \pm i\beta \), the general solution takes on a special trigonometric form: \(y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))\)Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions. Substitute \( \alpha = -1 \) and \( \beta = \sqrt{3} \), we present approximately: \(y(t) = e^{-t} (C_1 \cos(\sqrt{3}t) + C_2 \sin(\sqrt{3}t))\)The term \( e^{-t} \) illustrates exponential decay, which implies that over time, the solution's steady oscillations diminish.
- The general solution captures all possible behaviors of the system described by the differential equation.
- Initial conditions allow us to solve for \( C_1 \) and \( C_2 \), resulting in a specific solution.
- Trigonometric terms \( \cos \) and \( \sin \) signify the periodic oscillations present due to complex roots.