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Chapter 4: Problem 16

In Problems 1-36 find the general solution of the given differential equation. \(2 y^{\prime \prime}-3 y^{\prime}+4 y=0\)

Short Answer

Expert verified
The general solution is \(y(t) = e^{\frac{3}{4}t}(C_1\cos(\frac{\sqrt{23}}{4}t) + C_2\sin(\frac{\sqrt{23}}{4}t))\).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is a linear homogeneous second-order differential equation with constant coefficients: \(2y'' - 3y' + 4y = 0\).
02

Formulate the Characteristic Equation

To solve the differential equation, we first need to find the characteristic equation. Assuming solutions of the form \(y = e^{rt}\), differentiate to get \(y' = re^{rt}\) and \(y'' = r^2e^{rt}\). Substituting into the differential equation gives: \(2r^2 - 3r + 4 = 0\).
03

Solve the Characteristic Equation

The characteristic equation is a quadratic equation \(2r^2 - 3r + 4 = 0\). We find the roots using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = 4\). This yields: \(r = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 4}}{4}\).
04

Calculate the Roots of the Characteristic Equation

Compute the discriminant: \(b^2 - 4ac = 9 - 32 = -23\). Since the discriminant is negative, the roots are complex: \(r = \frac{3 \pm i\sqrt{23}}{4}\). This means the roots are \(r_1 = \frac{3}{4} + \frac{i\sqrt{23}}{4}\), \(r_2 = \frac{3}{4} - \frac{i\sqrt{23}}{4}\).
05

Write the General Solution

With complex roots \(r = \alpha \pm i\beta\), the general solution is given by \(y(t) = e^{\alpha t}(C_1\cos(\beta t) + C_2\sin(\beta t))\). Here, \(\alpha = \frac{3}{4}\) and \(\beta = \frac{\sqrt{23}}{4}\), thus the general solution is: \(y(t) = e^{\frac{3}{4}t}(C_1\cos(\frac{\sqrt{23}}{4}t) + C_2\sin(\frac{\sqrt{23}}{4}t))\).

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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 is a specific type of differential equation that has great importance in mathematical modeling. It is called 'homogeneous' because it equals zero on the right-hand side and does not contain any terms independent of the function being solved for. In the given problem, we have a second-order linear homogeneous differential equation with constant coefficients:
  • The general form of such an equation is: \[an y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y^{\prime} + a_0 y = 0\]
  • Where each term is a derivative of the function \(y\), multiplied by a constant coefficient.
  • In our problem, the equation is \(2y'' - 3y' + 4y = 0\).
To solve this kind of equation, we rely on finding the characteristic equation, which is derived from these constant coefficients.
Characteristic Equation
The characteristic equation is a crucial step in solving linear homogeneous differential equations. It transforms the differential equation into a more manageable algebraic equation. This is done by assuming solutions of the form \(y = e^{rt}\), where \(r\) is a constant. Substituting this form into the differential equation allows us to derive the characteristic equation:
  • The process involves differentiating: - \(y' = re^{rt}\) - \(y'' = r^2e^{rt}\)
  • Substituting these into the given equation: \(2r^2 - 3r + 4 = 0\)
This quadratic equation will help us find the roots, which are essential for determining the form of the general solution, especially when dealing with the behavior of complex roots.
Complex Roots
Complex roots arise when the discriminant in the characteristic equation is negative. In the given problem, the characteristic equation is quadratic: \(2r^2 - 3r + 4 = 0\). The discriminant for this equation is calculated as \(b^2 - 4ac = 9 - 32 = -23\), which is negative. Here鈥檚 how complex roots are approached:
  • When the discriminant is negative, it indicates that the roots are complex and occur in conjugate pairs.
  • For our equation, the roots are: \[r_1 = \frac{3}{4} + \frac{i\sqrt{23}}{4},\ r_2 = \frac{3}{4} - \frac{i\sqrt{23}}{4}\]
  • The format is \(r = \alpha \pm i\beta\), where \(\alpha\) and \(i\beta\) are the real and imaginary parts, respectively.
These complex roots influence the form of the general solution, leading us to involve both sine and cosine functions in representing each part of the complex root.
General Solution of Differential Equations
The general solution of linear homogeneous differential equations provides a comprehensive expression that includes all possible specific solutions. For a differential equation with complex roots, such as the one we derived:
  • The general solution is represented as: \( y(t) = e^{\alpha t}(C_1\cos(\beta t) + C_2\sin(\beta t)) \)
  • Using the previously found values: - \(\alpha = \frac{3}{4}\) - \(\beta = \frac{\sqrt{23}}{4}\)
  • The equation becomes: \( y(t) = e^{\frac{3}{4}t}(C_1\cos(\frac{\sqrt{23}}{4}t) + C_2\sin(\frac{\sqrt{23}}{4}t)) \)
This general solution confirms the behavior of the system over time, characterized by exponential damping and oscillatory components due to the imaginary part, and can be tailored to initial conditions through constants \(C_1\) and \(C_2\).

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Most popular questions from this chapter

verify that \(y_{1}\) and \(y_{2}\) are solutions of the given differential equation but that \(y=c_{1} y_{1}+c_{2} y_{2}\) is, in general, not a solution. $$ y y^{\prime \prime}=\frac{1}{2}\left(y^{\prime}\right)^{2} ; \quad y_{1}=1, y_{2}=x^{2} $$

Solve each differential equation by variation of parameters. State an interval on which the general solution is defined. $$3 y^{\prime \prime}-6 y^{\prime}+30 y=e^{x} \tan 3 x$$

Solve each differential equation by variation of parameters. State an interval on which the general solution is defined. $$y^{\prime \prime}+y=\sec x \tan x$$

Determine whether the given functions are linearly independent or dependent on \((-\infty, \infty)\). $$ f_{1}(x)=\cos 2 x, \quad f_{2}(x)=1, \quad f_{2}(x)=\cos ^{2} x $$

Solve the given differential equation by undetermined coefficients. $$ y^{\prime \prime}+4 y=3 \sin 2 x $$
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