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Chapter 9: Problem 26

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$y^{\prime \prime}(t)=15 e^{3 t}+\sin 4 t$$

Short Answer

Expert verified
Answer: The general solution of the given differential equation is \(y(t) = C_1 + C_2t + \frac{5}{3}e^{3t} - \frac{1}{16}\sin(4t)\).

Step by step solution

01

Identify terms in the differential equation

The non-homogeneous part of the equation is composed of two terms: an exponential term (15e^(3t)) and a sinusoidal term (sin(4t)). The given differential equation is: $$y^{\prime \prime}(t) = 15e^{3t} + \sin(4t)$$
02

Find the particular solution for each term

To find the particular solutions for each term, we will first guess a function form for the exponential term and then the sinusoidal term. For the exponential term, we guess a particular solution of the form: $$y_p(t) = Ae^{3t}$$ For the sinusoidal term, we guess a particular solution of the form: $$y_p(t) = B\cos(4t) + C\sin(4t)$$
03

Solve for the particular solutions

For the exponential term, differentiate the assumed function form twice and substituting into the equation to solve for A. $$y_p(t) = Ae^{3t} \Rightarrow y_p'(t) = 3Ae^{3t} \Rightarrow y_p''(t) = 9Ae^{3t}$$ Now, we substitute \(y_p''(t)\) into the equation: $$9Ae^{3t} = 15e^{3t}$$ Solving for A, we get: $$A = \frac{15}{9} = \frac{5}{3}$$ So the particular solution for the exponential term is: $$y_p(t) = \frac{5}{3}e^{3t}$$ For the sinusoidal term, differentiate the assumed function form twice and substituting into the equation to solve for B and C. $$y_p(t) = B\cos(4t) + C\sin(4t) \Rightarrow y_p'(t) = -4B\sin(4t) + 4C\cos(4t)$$ $$\Rightarrow y_p''(t) = -16B\cos(4t) - 16C\sin(4t)$$ Now, we substitute \(y_p''(t)\) into the equation, equating the coefficients of sin and cos: $$-16B\cos(4t) - 16C\sin(4t) = \sin(4t)$$ We have: $$-16B = 0 \Rightarrow B = 0$$ $$-16C = 1 \Rightarrow C = -\frac{1}{16}$$ So the particular solution for the sinusoidal term is: $$y_p(t) = -\frac{1}{16}\sin(4t)$$
04

Find the general solution of the homogeneous equation

To find the general solution of the homogeneous equation, we first rewrite the given equation without the non-homogeneous part: $$y^{\prime\prime}(t) = 0$$ The corresponding homogeneous equation is a second-order equation, so we guess a solution of the form: $$y_h(t) = C_1e^{rt}$$ Plugging this into the homogeneous equation: $$(C_1r^2e^{rt}) = 0$$ Here we have \(r^2 = 0\), so \(r=0\). The general solution of the homogeneous equation is: $$y_h(t) = C_1 + C_2t$$
05

Add the particular solutions and the homogeneous solution

Finally, we add the particular solutions we found from Step 3 and the homogeneous solution found in Step 4: $$y(t) = y_h(t) + y_{p\_exp}(t) + y_{p\_sin}(t)$$ Thus, the general solution of the given differential equation is: $$y(t) = C_1 + C_2t + \frac{5}{3}e^{3t} - \frac{1}{16}\sin(4t)$$

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Key Concepts

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

Second-order Differential Equation
A second-order differential equation involves a function and its derivatives up to the second order. The most general form is \(y'' = f(t, y, y')\) where \(y''\) represents the second derivative of the function \(y\) with respect to \(t\), \(y'\) is the first derivative, and \(f\) is some function that relates these derivatives to each other.

When dealing with second-order differential equations, our main objective is to find the function \(y(t)\) that satisfies this relationship for given initial or boundary conditions. The solution process incorporates finding a homogeneous solution, which solves the equation when \(f\) is zero, and a particular solution, which solves the equation when \(f\) is not zero. The sum of both gives us the general solution to the equation. In the context of the provided exercise, the given second-order differential equation is non-homogeneous due to the presence of terms independent of \(y\) and \(y'\).
Particular Solution
In differential equations, the particular solution \(y_p(t)\) is a specific function that satisfies the non-homogeneous differential equation and includes the response of the system to an external force or input. To obtain the particular solution, we can use methods such as the method of undetermined coefficients or variation of parameters, which often involve 'guessing' a form of the solution that relates to the non-homogeneous part.

In our exercise, we see two guesses, one for an exponential term and another for a sinusoidal term. After assuming a form for the solution, we differentiate it accordingly, then substitute back into the equation, and solve for the constants. This procedure gives us the particular solution corresponding to each term that when added together, along with the homogeneous solution, will give the full description of the system's behavior under the influence of external inputs.
Homogeneous Equation
A homogeneous differential equation is one in which the function \(f(t, y, y')\) is equal to zero. This implies there are no terms in the equation that are only functions of \(t\); instead, every term is either \(y\), \(y'\), or \(y''\). The general solution of a homogeneous second-order differential equation takes the form \(y_h(t)\) and often relies on characteristic equations which can give exponential, sinusoidal, or polynomial functions.

In the exercise, to find the homogeneous solution, the problem is simplified to \(y'' = 0\), a second-order homogeneous equation with a straightforward solution. The corresponding characteristic equation indicates the nature of the solution to be linear in nature and provides a foundation over which the effects of external inputs can be added (in this case, the particular solution). The technique exemplified here is essential in many fields of science and engineering.
Non-homogeneous Differential Equation
A non-homogeneous differential equation includes a non-zero function on the right-hand side, which acts as an external force or input affecting the system described by the equation. The complete solution of such an equation is the sum of the general solution to the associated homogeneous equation \(y_h(t)\) and a particular solution \(y_p(t)\) to the non-homogeneous equation. This principle is crucial because most real-world phenomena have external influences that must be accounted for in the analysis.

The exercise provided demonstrates the process of finding a particular solution influenced by exponential and sinusoidal inputs and subsequently shows how to integrate these with the homogeneous solution to reach the general solution. The combined solutions outline the behavior of the system under both its inherent properties (homogeneous solution) and external forces (particular solutions).

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Most popular questions from this chapter

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$v^{\prime}(t)=\frac{4}{t^{2}-4}$$

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem. A one-million-liter pond is contaminated by a chemical pollutant with a concentration of \(20 \mathrm{g} / \mathrm{L} .\) The source of the pollutant is removed, and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr}\). Assuming the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form \(y^{\prime}(t)=-k y^{n}(t),\) where \(y(t)\) is the concentration of the compound, for \(t \geq 0, k>0\) is a constant that determines the speed of the reaction, and \(n\) is a positive integer called the order of the reaction. Assume the initial concentration of the compound is \(y(0)=y_{0}>0\) a. Consider a first-order reaction \((n=1)\) and show that the solution of the initial value problem is \(y(t)=y_{0} e^{-k t}\) b. Consider a second-order reaction \((n=2)\) and show that the solution of the initial value problem is \(y(t)=\frac{y_{0}}{y_{0} k t+1}\) c. Let \(y_{0}=1\) and \(k=0.1 .\) Graph the first-order and secondorder solutions found in parts (a) and (b). Compare the two reactions.

Consider the differential equation \(y^{\prime \prime}(t)-k^{2} y(t)=0,\) where \(k>0\) is a real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} e^{t}+C_{2} e^{-t} .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} e^{2 t}+C_{2} e^{-2 t}\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture. d. For a positive real number \(k\), verify that the general solution of the equation may also be expressed in the form \(y(t)=C_{1} \cosh k t+C_{2} \sinh k t,\) where cosh and sinh are the hyperbolic cosine and hyperbolic sine, respectively (Section \(7.3)\)

Properties of stirred tank solutions a. Show that for general positive values of \(R, V, C_{i},\) and \(m_{0},\) the solution of the initial value problem $$m^{\prime}(t)=-\frac{R}{V} m(t)+C_{i} R, \quad m(0)=m_{0}$$is \(m(t)=\left(m_{0}-C_{i} V\right) e^{-R t / V}+C_{i} V\) b. Verify that \(m(0)=m_{0}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t)\) and give a physical interpretation of the result. d. Suppose \(m_{0}\) and \(V\) are fixed. Describe the effect of increasing \(R\) on the graph of the solution.
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