Chapter 31: Problem 12
Solve the given differential equations. $$D^{2} y+4 y=2 D y$$
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 Type of Differential Equation
The given equation is a linear homogeneous differential equation with constant coefficients. It can be rewritten as \(D^2 y - 2D y + 4y = 0\).
02
Form the Characteristic Equation
The characteristic equation for a differential equation of the form \(aD^2 y + bD y + c y = 0\) is \(a r^2 + b r + c = 0\). For this equation, it becomes \(r^2 - 2r + 4 = 0\).
03
Solve the Characteristic Equation
To solve the characteristic equation \(r^2 - 2r + 4 = 0\), use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -2\), and \(c = 4\). Calculate the discriminant: \((-2)^2 - 4 \times 1 \times 4 = 4 - 16 = -12\).
04
Determine the Roots
Since the discriminant is negative, the roots are complex. Using the quadratic formula, they are \(r = \frac{2 \pm \sqrt{-12}}{2}\), which simplifies to \(r = 1 \pm i\sqrt{3}\).
05
Write the General Solution
For complex roots \(r = \alpha \pm i\beta\), the general solution is \(y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))\). Here, \(\alpha = 1\) and \(\beta = \sqrt{3}\). Therefore, the solution is \(y(t) = e^{t} (C_1 \cos(\sqrt{3} t) + C_2 \sin(\sqrt{3} t))\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Homogeneous Differential Equations
A linear homogeneous differential equation with constant coefficients is a structured type of equation essential in understanding how certain physical systems behave. Identifying such an equation starts with looking for a form that might resemble:
- \(aD^2 y + bD y + cy = 0\).
Characteristic Equation
The characteristic equation is a transformational tool used to solve linear homogeneous differential equations. Converting the differential equation into an algebraic equation makes it more manageable. To find it, replace the differentiation operator \(D\) with a dummy variable \(r\):
- For example, from \(D^2 y - 2D y + 4y = 0\), derive \(r^2 - 2r + 4 = 0\).
Complex Roots
Complex roots occur when the characteristic equation's solutions involve the square root of a negative number, implying oscillatory components in solutions. Given a quadratic equation like \(r^2 - 2r + 4 = 0\), calculate the discriminant \(b^2 - 4ac\):
- If it's negative, such as \(-12\) in this scenario, the roots are complex.
- Using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), determine the roots, expressed in terms of a real part \(\alpha\) and an imaginary part \(\beta\).
General Solution
The general solution of a homogeneous linear differential equation with complex roots expresses the anticipated behavior of a dynamic system. Through insights gained from the roots of the characteristic equation, the general solution aligns with expected system responses:
- For complex roots \(\alpha \pm i\beta\), the solution takes the form: \(y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))\).