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Chapter 3: Problem 9

In each of Problems 1 through 10 find the general solution of the given differential equation. \(25 y^{\prime \prime}-20 y^{\prime}+4 y=0\)

Short Answer

Expert verified
Answer: The general solution to the given differential equation is \(y(x) = C_1e^{\frac{2}{5}x} + C_2xe^{\frac{2}{5}x}\), where \(C_1\) and \(C_2\) are constants determined by any given initial conditions.

Step by step solution

01

Determine the Characteristic Equation

Rewrite the given differential equation in terms of algebraic operators, replacing the derivative terms with polynomial terms, and then finding the characteristic equation. Here, we replace \(y^{\prime\prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with 1. Doing so, we get: \(25r^2 - 20r + 4 = 0\)
02

Solve the Characteristic Equation

Now, we need to solve the quadratic equation for \(r\): \(25r^2 - 20r + 4 = 0\) We can factor the equation as: \((5r - 2)(5r - 2) = 0\) Thus, we have a repeated root: \(r = \frac{2}{5}\)
03

Write the General Solution

Since we have a repeated root, we'll write the general solution using the form: \(y(x) = C_1e^{r_1x} + C_2xe^{r_1x}\) Where \(r_1\) is the repeated root, and \(C_1\) and \(C_2\) are constants determined by any given initial conditions. In our case, \(r_1 = \frac{2}{5}\). So the general solution is: \(y(x) = C_1e^{\frac{2}{5}x} + C_2xe^{\frac{2}{5}x}\) And that's the general solution to the given differential equation.

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Key Concepts

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

Characteristic Equation
In the realm of differential equations, the characteristic equation is a key concept for solving linear homogeneous differential equations with constant coefficients. When faced with such an equation, the first step is to translate it into an algebraic form using a characteristic equation. For example, consider the differential equation \(25 y^{\prime \prime}-20 y^{\prime}+4 y=0\). To find its characteristic equation, we replace the derivatives with polynomials:
  • Replace \(y^{\prime \prime}\) with \(r^2\)
  • Replace \(y^{\prime}\) with \(r\)
  • Replace \(y\) with \(1\)
This transforms the differential equation into a quadratic form: \(25r^2 - 20r + 4 = 0\). Solving this polynomial equation gives us potential solutions for \(r\), which are crucial for constructing the general solution.
Repeated Roots
Once you solve the characteristic equation, you may encounter repeated roots, which are specific solutions where the roots are equal. In the example equation, \(25r^2 - 20r + 4 = 0\) factors into \((5r - 2)(5r - 2) = 0\), leading directly to a repeated root of \(r = \frac{2}{5}\).
If you have repeated roots in a characteristic equation:
  • The root appears more than once due to repeated factors in the equation.
  • This influences the form of the general solution significantly.
Understanding how to handle repeated roots is important as it affects the structure of the general solution, ensuring that the solution captures all possible behaviors of the differential equation.
General Solution
Once the roots from the characteristic equation are found, it is time to write down the general solution. In cases of repeated roots, the general solution takes a special form to cover all solutions of the differential equation.
For instance, with a repeated root \(r_1 = \frac{2}{5}\), the general solution is:
  • \(y(x) = C_1e^{\frac{2}{5}x} + C_2xe^{\frac{2}{5}x}\)
  • \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions.
This structure ensures that the solution generalizes for all initial conditions and scenarios. Each component plays a role:
  • \(C_1e^{\frac{2}{5}x}\) accounts for the simple exponential growth or decay.
  • \(C_2xe^{\frac{2}{5}x}\) introduces the linear term, addressing the distinctiveness of having repeated roots.
By constructing this solution correctly, we ensure that every potential outcome of the differential equation is captured.

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Most popular questions from this chapter

determine \(\omega_{0}, R,\) and \(\delta\) so as to write the given expression in the form \(u=R \cos \left(\omega_{0} t-\delta\right)\) $$ u=-2 \cos \pi t-3 \sin \pi t $$

Consider the forced but undamped system described by the initial value problem $$ u^{\prime \prime}+u=3 \cos \omega t, \quad u(0)=0, \quad u^{\prime}(0)=0 $$ (a) Find the solution \(u(t)\) for \(\omega \neq 1\). (b) Plot the solution \(u(t)\) versus \(t\) for \(\omega=0.7, \omega=0.8,\) and \(\omega=0.9\). Describe how the response \(u(t)\) changes as \(\omega\) varies in this interval. What happens as \(\omega\) takes on values closer and closer to \(1 ?\) Note that the natural frequency of the unforced system is \(\omega_{0}=1\)

determine \(\omega_{0}, R,\) and \(\delta\) so as to write the given expression in the form \(u=R \cos \left(\omega_{0} t-\delta\right)\) $$ u=4 \cos 3 t-2 \sin 3 t $$

The differential equation $$ x y^{\prime \prime}-(x+N) y^{\prime}+N y=0 $$ where \(N\) is a nonnegative integer, has been discussed by several authors. 6 One reason it is interesting is that it has an exponential solution and a polynomial solution. (a) Verify that one solution is \(y_{1}(x)=e^{x}\). (b) Show that a second solution has the form \(y_{2}(x)=c e^{x} \int x^{N} e^{-x} d x\). Calculate \(y_{2 (x)\) for \(N=1\) and \(N=2 ;\) convince yourself that, with \(c=-1 / N !\) $$ y_{2}(x)=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\cdots+\frac{x^{N}}{N !} $$ Note that \(y_{2}(x)\) is exactly the first \(N+1\) terms in the Taylor series about \(x=0\) for \(e^{x},\) that is, for \(y_{1}(x) .\)

Follow the instructions in Problem 28 to solve the differential equation $$ y^{\prime \prime}+2 y^{\prime}+5 y=\left\\{\begin{array}{ll}{1,} & {0 \leq t \leq \pi / 2} \\ {0,} & {t>\pi / 2}\end{array}\right. $$ $$ \text { with the initial conditions } y(0)=0 \text { and } y^{\prime}(0)=0 $$ $$ \begin{array}{l}{\text { Behavior of Solutions as } t \rightarrow \infty \text { , In Problems } 30 \text { and } 31 \text { we continue the discussion started }} \\ {\text { with Problems } 38 \text { through } 40 \text { of Section } 3.5 \text { . Consider the differential equation }}\end{array} $$ $$ a y^{\prime \prime}+b y^{\prime}+c y=g(t) $$ $$ \text { where } a, b, \text { and } c \text { are positive. } $$
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