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Consider the differential equation \(y^{\prime}(t)+9 y(t)=10\) a. How many arbitrary constants appear in the general solution of the differential equation? b. Is the differential equation linear or nonlinear?

Short Answer

Expert verified
#Answer# a) The general solution of the given differential equation will have 1 arbitrary constant. b) The given differential equation is a linear differential equation.

Step by step solution

01

Analyze the given differential equation

The given differential equation is $$y'(t) + 9y(t) = 10.$$ Here, we notice that it's a first-order differential equation, as the highest order derivative is \(y'(t)\). So, the number of arbitrary constants will be equal to the order of the equation, which in this case, is 1.
02

Answering Question (a)

As the order of the given differential equation is 1, the general solution will have 1 arbitrary constant.
03

Determine the linearity of the given differential equation

The equation we have is $$y'(t) + 9y(t) = 10.$$ To determine if this is linear or nonlinear, let's look at if it satisfies the properties of a linear differential equation. We can rewrite it as $$L[y(t)] = y'(t) + 9y(t),$$ where \(L\) is a linear operator. A linear differential equation has the property that \(L[Ay_1(t) + By_2(t)] = AL[y_1(t)] + BL[y_2(t)]\) for all constants \(A\) and \(B\). The given equation is linear since it satisfies the linearity properties, as all the dependent variable terms (\(y(t)\) and \(y'(t)\)) appear either multiplied by a constant or by themselves and not in any other non-linear terms like \(y^2(t)\) or \(\sin(y(t))\).
04

Answering Question (b)

As established in step 3, the given differential equation is linear. So, the final answers are: a) The general solution of the given differential equation will have 1 arbitrary constant. b) The given differential equation is a linear differential equation.

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Key Concepts

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

Arbitrary constants in differential equations
In differential equations, arbitrary constants play a crucial role in determining the general solution. They reflect the family of solutions, indicating that there are infinite solutions fitting the equation, each defined by different constant values.

For first-order differential equations, as in the example provided, the equation's order indicates the number of arbitrary constants. The **order** of a differential equation refers to the highest derivative present in the equation. In this case, it's the first derivative, represented as \( y'(t) \).

The general solution to a first-order equation contains one arbitrary constant. This is because you need this constant to account for the initial conditions or specific scenarios not immediately visible in the equation. Essentially, the arbitrary constant provides flexibility in solving practical problems with unique starting conditions.
Linear differential equations
Differential equations can be classified as linear or nonlinear. This classification depends on how the dependent variable and its derivatives appear in the equation.

A linear differential equation is one where the dependent variable, \( y(t) \), and its derivatives (like \( y'(t) \)) appear to the power of one. Importantly, they are not multiplied by each other or any function of \( y(t) \) that is not a constant. The structure of a linear differential equation is kept simple and straightforward.

For instance, the equation \( y'(t) + 9y(t) = 10 \) fits this definition well. Here, both \( y(t) \) and \( y'(t) \) are linear terms because they are not squared, cubed, or parts of any nonlinear mathematical functions such as sine, cosine, or exponentials (where \( y(t) \) is present). This makes solving linear differential equations often simpler, allowing predictable methods, such as integrating factors or characteristic equations, to be used effectively.
Linearity properties of differential equations
The linearity properties of differential equations are fundamental in understanding how solutions can be manipulated and combined to satisfy new conditions or create new solutions that also solve the original differential equation.

Linearity in this context means that if you have multiple solutions to a linear differential equation, you can create new solutions by taking linear combinations of these solutions. Specifically, if \( y_1(t) \) and \( y_2(t) \) are solutions to a linear differential equation, then any linear combination \( Ay_1(t) + By_2(t) \) (where \( A \) and \( B \) are constants) will also be a solution.

In assessing the linearity of \( y'(t) + 9y(t) = 10 \), the use of linear operators becomes handy. A **linear operator** acts on functions and maintains the additive and scalar properties. So, applying this to our equation confirms that any combination of solutions, if one existed, will adhere to the linear operators' properties, thereby reaffirming the equation's linear nature. This property is essential because it simplifies solving and analyzing differential equations in different fields of science and engineering.

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Most popular questions from this chapter

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.

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.004 B-800, B(0)=40,000$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.

Equilibrium solutions \(A\) differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on y. The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=6-2 y$$

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.0075 B-1500, B(0)=100,000$$

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