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Chapter 31: Problem 9

Solve the given differential equations. $$2 D^{2} y-3 y=D y$$

Short Answer

Expert verified
The general solution is \(y(t) = C_1 e^{1.5t} + C_2 e^{-t}\).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is \(2 D^{2} y - 3y = D y\). This is a linear homogeneous differential equation with constant coefficients. We can rewrite it in differential operator form as \(2 D^2 y - D y - 3y = 0\).
02

Form the Characteristic Equation

To solve this equation, first write the associated characteristic equation, which is found by replacing \(D\) with \(r\) (where \(r\) is a root of the characteristic equation). This yields: \[2r^2 - r - 3 = 0\]
03

Solve the Characteristic Equation

Solve the quadratic characteristic equation \(2r^2 - r - 3 = 0\) using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 2\), \(b = -1\), and \(c = -3\). Substituting these values: \[r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}\]This gives the roots \(r_1 = 1.5\) and \(r_2 = -1\).
04

Write the General Solution

The general solution of the differential equation, given the roots \(r_1 = 1.5\) and \(r_2 = -1\), is: \[y(t) = C_1 e^{1.5t} + C_2 e^{-t}\]where \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions, if any are provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
When dealing with linear homogeneous differential equations with constant coefficients, a key step is to form the characteristic equation. This is done by transforming the differential equation into an algebraic equation using a placeholder variable, often denoted as \(r\). In the original exercise, we transformed the differential equation into: \[ 2 D^2 y - D y - 3y = 0 \]By replacing the differential operator \(D\) with \(r\), we derive the characteristic equation:\[ 2r^2 - r - 3 = 0 \]The characteristic equation is crucial because it allows us to find the roots, \(r_1\) and \(r_2\), which help build the general solution to the differential equation.
  • Each root corresponds to a fundamental solution of the differential equation.
  • Understanding characteristic equations is vital for solving these types of differential equations efficiently.
Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations, which are polynomial equations of the form \( ax^2 + bx + c = 0 \). The formula is:\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the context of our differential equation, we used the quadratic formula to solve the characteristic equation: \[ 2r^2 - r - 3 = 0 \]Identifying \(a = 2\), \(b = -1\), and \(c = -3\), we substitute these into the quadratic formula:\[r = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}\]This calculation provides two roots:
  • \(r_1 = 1.5\)
  • \(r_2 = -1\)
Finding these roots using the quadratic formula is significant because it directly affects the final form of the solution to the differential equation.
General Solution
With the roots of the characteristic equation \(r_1 = 1.5\) and \(r_2 = -1\), we can write the general solution of the differential equation.For linear homogeneous differential equations, the general solution combines these root-based components:\[y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\]Substituting values, we get:\[y(t) = C_1 e^{1.5t} + C_2 e^{-t}\]
  • Here, \(C_1\) and \(C_2\) are constants. These are typically determined by initial conditions specific to the problem at hand.
  • Each term in the solution corresponds to one of the roots found earlier, showcasing how they form the core of the equation's solution.
The general solution provides a comprehensive representation of all possible solutions that satisfy the differential equation.

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