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Chapter 8: Problem 20

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 is \(y(t) = C_1 + C_2t + \frac{5}{3} e^{3t} - \frac{1}{16} \sin(4t)\).

Step by step solution

01

Solve the homogeneous equation for Complementary Function (CF)

We start by solving the homogeneous equation: $$y''(t) = 0$$ The general solution of this equation is: $$y_{CF}(t) = C_1 + C_2t$$
02

Guess the form of Particular Solution (PS)

Now, let's guess the form of a particular solution for the differential equation, considering the forcing terms (right-hand side of the equation): $$y_{PS}(t) = A e^{3t} + B \cos(4t) + C \sin(4t)$$ Differentiate \(y_{PS}(t)\) with respect to \(t\) to obtain \(y_{PS}'(t)\): $$y_{PS}'(t) = 3A e^{3t} - 4B \sin(4t) + 4C \cos(4t)$$ Differentiate \(y_{PS}'(t)\) to get \(y_{PS}''(t)\): $$y_{PS}''(t) = 9A e^{3t} - 16B \cos(4t) - 16C \sin(4t)$$
03

Use given differential equation to find coefficients A, B, and C

We substitute \(y_{PS}''(t)\) into the given differential equation and equate like terms: $$9A e^{3t} - 16B \cos(4t) - 16C \sin(4t) = 15 e^{3t} + \sin{4t}$$ Comparing coefficients of \(e^{3t}\), \(\cos(4t)\), and \(\sin(4t)\), we get: $$9A = 15 \Rightarrow A = \frac{5}{3}$$ $$-16B = 0 \Rightarrow B = 0$$ $$-16C = 1 \Rightarrow C = -\frac{1}{16}$$ Thus, the Particular Solution (PS) becomes: $$y_{PS}(t) = \frac{5}{3} e^{3t} - \frac{1}{16} \sin(4t)$$
04

Combine Complementary Function (CF) and Particular Solution (PS) to get the general solution

Finally, we combine the Complementary Function (CF) and Particular Solution (PS) to obtain the general solution of the given differential equation: $$y(t) = y_{CF}(t) + y_{PS}(t)$$ $$y(t) = (C_1 + C_2t) + \left(\frac{5}{3} e^{3t} - \frac{1}{16} \sin(4t)\right)$$ Thus, the general solution 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.

Complementary Function
When tackling differential equations, the first step is often finding the Complementary Function (CF). This function represents the solution to the related homogeneous equation, which is formed by setting the right-hand side of the given differential equation to zero.
For the equation provided, \( y''(t) = 0 \), solving this gives us the Complementary Function. Such equations, lacking any external forces (right-side terms), represent a system's natural behavior:
  • The solution is of the form \( y_{CF}(t) = C_1 + C_2t \).
  • \( C_1 \) and \( C_2 \) are arbitrary constants determined by boundary conditions.
  • These constants allow the solution to adapt to various initial conditions.
This linear structure of the CF is a hallmark of second-order differential equations. By accounting for different scenarios with arbitrary constants, the CF forms the backbone of the general solution.
Particular Solution
The Particular Solution (PS) focuses on the unique aspects of the differential equation, specifically the non-homogeneous part, also known as the forcing term. This requires assuming a form based on the nature of these terms, which are typically functions like exponential or trigonometric expressions.
For this specific exercise, we have the non-homogeneous equation \( y''(t) = 15e^{3t} + \sin 4t \), which relies on external inputs. We must choose a trial function reflecting these influences:
  • The assumed form was \( y_{PS}(t) = A e^{3t} + B \cos(4t) + C \sin(4t) \).
  • The derivative calculations follow, aligning terms with original equation counterparts.
  • Sequentially, the coefficients \( A \), \( B \), and \( C \) are found by equating terms based on a match with the forcing function:
    • Results are: \( A = \frac{5}{3} \), \( B = 0 \), \( C = -\frac{1}{16} \).
Through plugging back into the differential equation and simplifying, the PS emerges, indicating the system's response to specific inputs.
General Solution
The General Solution is the cornerstone of solving differential equations, as it incorporates both the natural tendencies of the system and the specific responses to external forces. It combines:
  • The Complementary Function, addressing the homogeneous equation.
  • The Particular Solution, dealing with non-homogeneous elements.
For this problem, once both components are derived:
  • CF: \( C_1 + C_2t \)
  • PS: \( \frac{5}{3} e^{3t} - \frac{1}{16} \sin(4t) \)
They join to provide the General Solution:
  • \( y(t) = C_1 + C_2t + \frac{5}{3} e^{3t} - \frac{1}{16} \sin(4t) \)
The General Solution captures all potential behaviors of the differential equation under different initial conditions, making it versatile and comprehensive.

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Most popular questions from this chapter

An object in free fall may be modeled by assuming that the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass \(\times\) acceleration \(=\) the sum of the external forces), the velocity of the object satisfies the differential equation $$\underbrace {m}_{\text {mass}}\quad \cdot \underbrace{v^{\prime}(t)}_{\text {acceleration }}=\underbrace {m g+f(v)}_{\text {external forces}}$$ where \(f\) is a function that models the air resistance (assuming the positive direction is downward). One common assumption (often used for motion in air) is that \(f(v)=-k v^{2},\) where \(k>0\) is a drag coefficient. a. Show that the equation can be written in the form \(v^{\prime}(t)=g-a v^{2},\) where \(a=k / m\) b. For what (positive) value of \(v\) is \(v^{\prime}(t)=0 ?\) (This equilibrium solution is called the terminal velocity.) c. Find the solution of this separable equation assuming \(v(0)=0\) and \(0 < v^{2} < g / a\) d. Graph the solution found in part (c) with \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) \(m=1,\) and \(k=0.1,\) and verify that the terminal velocity agrees with the value found in part (b).

The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. Let \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\). a. Draw the direction field, for \(0 \leq t \leq 100,0 \leq y \leq 600\). b. What is the equilibrium solution? c. For what initial values \(m(0)=A\) are solutions increasing? Decreasing?

Explain how the growth rate function determines the solution of a population model.

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Use the method outlined in Exercise 43 to solve the following Bernoulli equations. a. \(y^{\prime}(t)+y=2 y^{2}\) b. \(y^{\prime}(t)-2 y=3 y^{-1}\) c. \(y^{\prime}(t)+y=\sqrt{y}\)
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