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Chapter 6: Problem 14

Find the general solution to the linear differential equation. $$ 4 y^{\prime \prime}-12 y^{\prime}+9 y=0 $$

Short Answer

Expert verified
The general solution is \( y(x) = c_1 e^{\frac{3}{2}x} + c_2 xe^{\frac{3}{2}x} \).

Step by step solution

01

Identify the differential equation

The given differential equation is a linear homogeneous second-order differential equation with constant coefficients:\[ 4y'' - 12y' + 9y = 0 \]
02

Write the characteristic equation

To solve the differential equation, we first find the characteristic equation by replacing \(y''\) with \(m^2\), \(y'\) with \(m\), and \(y\) with 1:\[ 4m^2 - 12m + 9 = 0 \]
03

Solve the characteristic equation

The characteristic equation is a quadratic equation in \(m\). Use the quadratic formula:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a=4\), \(b=-12\), \(c=9\):\[ m = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 4 \times 9}}{2 \times 4} \]\[ m = \frac{12 \pm \sqrt{144 - 144}}{8} \]\[ m = \frac{12}{8} = \frac{3}{2} \]There is one repeated root, \(m = \frac{3}{2}\).
04

Write the general solution from repeated roots

Since the characteristic equation has a repeated root \(m = \frac{3}{2}\), the general solution of the differential equation is:\[ y(x) = c_1 e^{\frac{3}{2}x} + c_2 xe^{\frac{3}{2}x} \]where \(c_1\) and \(c_2\) are arbitrary constants.

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Key Concepts

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

Second-Order Differential Equations
Second-order differential equations involve the second derivative of a function. These equations are essential in various fields, such as physics, engineering, and mathematics, as they can model phenomena like motion, waves, and heat transfer. In this exercise, we have a second-order differential equation:
  • The function is with respect to a variable, commonly time or position.
  • The equation involves the first and second derivatives of this function.
  • Solutions to these equations describe the behavior of the system being modeled.
For the specific example, the equation is linear, indicating that the function and its derivatives are each raised only to the first power, making it easier to solve using standard techniques. Each type of second-order differential equation has specific methods for finding solutions, and knowing which type you're dealing with is crucial.
Homogeneous Equations
A homogeneous differential equation is one where the function equals zero, as seen in our example: 4y'' - 12y' + 9y = 0. In homogeneous equations:
  • All terms involve the dependent variable (like y) or its derivatives.
  • There are no external inputs or additional functions.
  • Solutions typically take a form involving exponential functions due to the nature of the equations.
The zero on the right-hand side signifies homogeneity, meaning there's a uniform distribution or influence within the system without external forces acting on it. Homogeneous equations are fundamental in understanding systems in equilibrium, where only internal factors define the system's behavior.
Characteristic Equation
The characteristic equation bridges the differential equation to algebra, allowing us to find solutions using well-known algebraic techniques. For a second-order linear homogeneous differential equation with constant coefficients:
  • Replace each derivative with powers of a variable, often 'm'.
  • Transform the differential equation into a quadratic equation.
  • This approach is especially useful since quadratic equations have predictable solution methods.
In the exercise, the characteristic equation derived was 4m^2 - 12m + 9 = 0. From this equation, solutions in terms of 'm' give insights into the behavior of the original differential equation. Thus,
  • Each root of the characteristic equation corresponds to an exponential function of the form e^(mx) in the differential equation's solution.
Using the characteristic equation is a standard procedure for finding solutions quickly and effectively.
Repeated Roots
When solving the characteristic equation, if you find that roots are repeated, it shapes the form of the solution significantly. In this case:
  • Typically, roots provide exponential solutions like e^(mx).
  • If a root is repeated, the solution includes not just e^(mx) but also introduces a factor of x, resulting in forms like xe^(mx).
In our exercise, we found a repeated root, m = 3/2. Because the root is repeated, the general solution for the differential equation does not just include a single term:
  • It includes terms involving both e^(m * x) and x*e^(m *x), as seen in y(x) = c_1 e^(3/2 x) + c_2 x e^(3/2 x).
This form ensures the solution space is fully covered, accounting for all nuances of how the dependent variable behaves over time or space, especially when confronted with such repeated behavior.

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Most popular questions from this chapter

The following problems consider the "beats" that occur when the forcing term of a differential equation causes "slow" and "fast" amplitudes. Consider the general differential equation \(a y^{\prime \prime}+b y=\cos (\omega t)\) that governs undamped motion. Assume that \(\sqrt{\frac{b}{a}} \neq \omega\).Assuming the system starts from rest, show that the particular solution can be written as \(y=\frac{2}{a\left(\omega_{b}^{2}-\omega^{2}\right)} \sin \left(\frac{\omega_{b}-\omega t}{2}\right) \sin \left(\frac{\omega_{0}+\omega t}{2}\right) .\)

Solve the following equations using the method of undetermined coefficients. $$ y^{\prime \prime}-4 y^{\prime}+4 y=8 x^{2}+4 x $$

A \(32-\mathrm{lb}\) weight ( 1 slug) stretches a vertical spring 4 in. The resistance in the spring-mass system is equal to four times the instantaneous velocity of the mass. a. Find the equation of motion if it is released from its equilibrium position with a downward velocity of \(12 \mathrm{ft} / \mathrm{sec}\). b. Determine whether the motion is overdamped, critically damped, or underdamped.

For the following problem, set up and solve the differential equation.The motion of a swinging pendulum for small angles \(\theta\) can be approximated by \(\frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \theta=0\), where \(\theta\) is the angle the pendulum makes with respect to a vertical line, \(g\) is the acceleration resulting from gravity, and \(L\) is the length of the pendulum. Find the equation describing the angle of the pendulum at time \(t\), assuming an initial displacement of \(\theta_{0}\) and an initial velocity of zero.

For the following problem, set up and solve the differential equations.An opera singer is attempting to shatter a glass by singing a particular note. The vibrations of the glass can be modeled by \(y^{\prime \prime}+a y=\cos (b t)\), where \(y^{\prime \prime}+a y=0\) represents the natural frequency of the glass and the singer is forcing the vibrations at \(\cos (b t) .\) For what value \(b\) would the singer be able to break that glass? (Note: in order for the glass to break, the oscillations would need to get higher and higher.)
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