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In Problems \(1-4\), the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. $$ \begin{aligned} &y=c_{1} e^{4 x}+c_{2} e^{-x},(-\infty, \infty) ; y^{\prime \prime}-3 y^{\prime}-4 y=0, y(0)=1 \\ &y^{\prime}(0)=2 \end{aligned} $$

Short Answer

Expert verified
The solution is \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \).

Step by step solution

01

Understand the General Solution

The given family of functions is the general solution: \( y = c_{1} e^{4x} + c_{2} e^{-x} \). This is derived from the differential equation \( y'' - 3y' - 4y = 0 \). Our task is to find specific constants \( c_1 \) and \( c_2 \) such that this function satisfies the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \).
02

Find the First Derivative

Compute the first derivative of the general solution. Given: \( y = c_{1} e^{4x} + c_{2} e^{-x} \). Using differentiation: \( y' = \frac{d}{dx}(c_{1} e^{4x}) + \frac{d}{dx}(c_{2} e^{-x}) = 4c_{1} e^{4x} - c_{2} e^{-x} \).
03

Apply Initial Condition y(0)=1

Substitute \( x = 0 \) into the general solution \( y(0) = c_{1} e^{4(0)} + c_{2} e^{-0} = c_{1} + c_{2} = 1 \). This gives us the first equation: \( c_1 + c_2 = 1 \).
04

Apply Initial Condition y'(0)=2

Substitute \( x = 0 \) into the derivative: \( y'(0) = 4c_{1} e^{4(0)} - c_{2} e^{-0} = 4c_{1} - c_{2} = 2 \). This gives us the second equation: \( 4c_1 - c_2 = 2 \).
05

Solve the System of Equations

We have the system: 1. \( c_1 + c_2 = 1 \).2. \( 4c_1 - c_2 = 2 \).Add the two equations to eliminate \( c_2 \): \( (c_1 + c_2) + (4c_1 - c_2) = 1 + 2 \), which simplifies to \( 5c_1 = 3 \), leading to \( c_1 = \frac{3}{5} \). Insert \( c_1 \) back into \( c_1 + c_2 = 1 \), giving \( \frac{3}{5} + c_2 = 1 \), thus \( c_2 = \frac{2}{5} \).
06

Write the Particular Solution

With calculated values of \( c_1 \) and \( c_2 \), the particular solution of the initial-value problem is: \( y = \frac{3}{5} e^{4x} + \frac{2}{5} e^{-x} \).

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Key Concepts

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

Initial Value Problem
An initial value problem is a type of differential equation problem where you are given additional information called initial conditions. This information helps define a unique solution from a family of possible solutions. The initial conditions typically specify the value of the function and its derivatives at a particular point.
In our example, the initial value problem involves the differential equation \(y'' - 3y' - 4y = 0\) with two initial conditions:
  • \(y(0) = 1\)
  • \(y'(0) = 2\)
These conditions are crucial because they allow us to determine the specific constants in the general solution, thus finding the unique solution that fits them.
General Solution
The general solution of a differential equation contains all possible solutions and is expressed in terms of arbitrary constants. For a second-order linear differential equation like the one given \(y'' - 3y' - 4y = 0\), the general solution is:
  • \(y = c_{1} e^{4x} + c_{2} e^{-x}\)
Here, \(c_{1}\) and \(c_{2}\) are arbitrary constants that represent a family of solutions. To find a particular solution from this family, we use initial conditions to determine specific values for these constants. This converts the general solution into a specific one that satisfies the given initial conditions.
System of Equations
A system of equations comes into play when we have multiple conditions or equations that we need to solve together. In this case, applying the initial conditions to our functions and derivatives creates a system of linear equations:
  • \(c_1 + c_2 = 1\)
  • \(4c_1 - c_2 = 2\)
These equations allow us to solve for \(c_1\) and \(c_2\) by eliminating one variable in favor of the other. Solving these equations gives us the precise constants \(c_1 = \frac{3}{5}\) and \(c_2 = \frac{2}{5}\). By resolving these constants, we arrive at the particular solution to the initial value problem.
Exponential Functions
Exponential functions are a type of mathematical function that involve an exponent and are expressed in the form \(e^{x}\), where \(e\) is the base of the natural logarithm. In the context of solving differential equations, exponential functions often represent solutions because they naturally describe rates of change and behavior of systems over time.
The general solution in the exercise involves exponential terms \(e^{4x}\) and \(e^{-x}\). Different constants \(c_1\) and \(c_2\) modify the influence of these exponential terms on the solution. This shows how exponential functions contribute to the shape and behavior of the solution over the interval \((-\infty, \infty)\). Such functions are crucial as they can model growth, decay, oscillations, and other dynamic behaviors observed in various fields like physics, biology, and economics.

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Most popular questions from this chapter

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

Motion in a Force Field A mathematical model for the position \(x(t)\) of a body moving rectilinearly on the \(x\) -axis in an inverse-square force field is given by $$ \frac{d^{2} x}{d t^{2}}=-\frac{k^{2}}{x^{2}} $$ Suppose that at \(t \quad 0\) the body starts from rest from the position \(x \quad x_{0}, x_{0}>0\). Show that the velocity of the body at time \(t\) is given by \(v^{2} \quad 2 k^{2}\left(1 / x-1 / x_{0}\right)\). Usethelastexpression anda CAS to carry out the integration to express time \(t\) in terms of \(x\).

Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1}+c_{2} x+c_{3} e^{7 x} $$

Consider a pendulum that is released from rest from an initial displacement of \(\theta_{0}\) radians. Solving the linear model (7) subject to the initial conditions \(\theta(0)=\theta_{0}, \theta^{\prime}(0)=0\) gives \(\theta(t)=\theta_{0} \cos \sqrt{g} \| t .\) Theperiod of oscillations predicted by this modelisgivenbythefamiliarformula \(T=2 \pi / \sqrt{g l l}=2 \pi \sqrt{U g}\). The interesting thing about this formula for \(T\) is that it does not depend on the magnitude of the initial displacement \(\theta_{0}\). In other words, the linear model predicts that the time that it would take the pendulum to swing from an initial displacement of, say, \(\theta_{0}=\pi / 2\left(=90^{\circ}\right)\) to \(-\pi / 2\) and back again would be exactly the same time to cycle from, say, \(\theta_{0}=\pi / 360\left(=0.5^{\circ}\right)\) to \(-\pi / 360\). This is intuitively unreasonable; the actual period must depend on \(\theta_{0}\). If we assume that \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) and \(l=32 \mathrm{ft}\), then the period of oscillation of the linear model is \(T=2 \pi \mathrm{s}\). Let us compare this last number with the period predicted by the nonlinear model when \(\theta_{0}=\pi / 4\). Using a numerical solver that is capable of generating hard data, approximate the solution of $$\frac{d^{2} \theta}{d t^{2}}+\sin \theta=0, \quad \theta(0)=\frac{\pi}{4}, \quad \theta^{\prime}(0)=0$$ for \(0 \leq t \leq 2\). As in Problem 24 , if \(t_{1}\) denotes the first time the pendulum reaches the position \(O P\) in Figure 3.11.3, then the period of the nonlinear pendulum is \(4 t_{1} .\) Here is another way of solving the equation \(\theta(t)=0\). Expeniment with small step sizes and advance the time staning at \(t=0\) and ending at \(t=2\). From your hard data, observe the time \(t_{1}\) when \(\theta(t)\) changes, for the first time, from positive to negative. Use the value \(t_{1}\) to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period estimated by \(T=2 \pi\).

Consider the boundary-value problem \(y^{\prime \prime}+\lambda y=0, y(0)=0\), \(y(\pi / 2)=0 .\) Discuss: Is it possible to determine values of \(\lambda\) so that the problem possesses (a) trivial solutions? (b) nontrivial solutions?

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