64.1 Initial Value Problems Concept Overview

INITIAL VALUE PROBLEMS | CONCEPT OVERVIEW The TOPIC of INITIAL VALUE PROBLEMS is not directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Therefore, we must memorize these formulas and understand their applications independent of the NCEES Supplied Reference Handbook.

CONCEPT INTRO: As we can do with any algebraic expression, given a defined DIFFERENTIAL EQUATION, a solution can be derived. The goal in solving DIFFERENTIAL EQUATIONS is to find an expression for the differential function in terms of its independent variable. This expression does not need to be EXPLICIT, meaning given in the general form of 𝑦 𝑡 , but it will provide a functional relation between the variables involved, not containing derivatives or differentials, that satisfies the differential equation. The GENERAL SOLUTION of a DIFFERENTIAL EQUATION of the 𝑛!! order will contain “𝑛” constants, and is a solution which contains a number of arbitrary independent functions equal to the order of the original DIFFERENTIAL EQUATION. The PARTICULAR SOLUTION of a DIFFERENTIAL EQUATION is obtained from the general solution by assigning values to the constant.



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This is typically done by evaluating the BOUNDARY CONDITIONS, otherwise known as INITIAL CONDITIONS.

INITIAL VALUE PROBLEMS: When we solve a differential equation, we will derive a general solution that represents an infinite series of functions that are all solutions of the given equation. However, when working DIFFERENTIAL EQUATION problems, we are usually concerned with defining a more specific function that both satisfies a given equation and a set of some initial conditions or boundary conditions. When you are given a differential equation in the form: 𝑑𝑦 = 𝑓(𝑡, 𝑦) 𝑑𝑡 The goal is to find a function, 𝑓(𝑡), that solves the differential equation. Luckily, depending on the type of DIFFERENTIAL EQUATION we are working with, a standard GENERAL SOLUTION has been set and derived for us, even given to us in the NCEES Reference Handbook, which we will get to. We will take this GENERAL SOLUTION, apply it, and then set it against the INITIAL CONDITIONS to wholly quantify the values throughout the solution.



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An INITIAL-VALUE PROBLEM is one in which provides us a differential equation along with secondary conditions on the value of the independent variable with respect to the unknown function and its derivatives. A BOUNDARY-VALUE PROBLEM is a differential equation that is given with more than one value of the independent variable such that multiple conditions create a boundary, and are referred to as BOUNDARY CONDITIONS. The solution to an INITIAL-VALUE or BOUNDARY-VALUE PROBLEM is a function that solves both the differential equation and satisfies all given subsidiary conditions. There are two parts of these types of problems: 1. A DIFFERENTIAL EQUATION 2. A set of INITIAL or BOUNDARY CONDITIONS Without both of these pieces present, a conclusive solution within a given range can’t be determined. Generally speaking, the steps in solving these types of process will follow this flow: 1. First solve the equation to find the general solution (which contains one or more arbitrary constants or coefficients.) 2. Use the initial or boundary condition(s) to determine the exact value(s) of those constant(s).



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The result is what we would call the PARTICULAR SOLUTION of the DIFFERENTIAL EQUATION. Applying the initial condition allows you to select one solution among the infinite number that result from integration, as represented by the CONSTANT OF INTEGRATION, “𝐶”. Let’s take a look at the most basic differential equation that has the form: 𝑑𝑦 = 𝑓(𝑥) 𝑑𝑥 If we want to explicitly solve to find the function “𝑦”, we would want to cross multiply and integrate both sides, such that: ∫ 𝑑𝑦 = ∫ 𝑥 𝑑𝑥 It is important to remember that when evaluating an indefinite integral, we must always add the CONSTANT OF INTEGRATION to account for the infinite number of solutions that would be possible for differential equation absent any boundary conditions, generally giving us: 𝑦 =𝑥+𝐶 This is a GENERAL SOLUTION.



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If we wish to put context around the CONSTANT OF INTEGRATION, “𝐶”, and further solve for a PARTICULAR SOLUTION, we must be given an INITIAL CONDITION or BOUNDARY CONDITION that we can use to quantify the unknown value.



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INITIAL VALUE PROBLEMS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The particular solution of the noted differential equation is best represented as: 𝑑𝑦 = 10 − 𝑥; 𝑦 0 = −1 𝑑𝑥 A. 𝑦(𝑥) = 10𝑥 − B. 𝑦(𝑥) = 20𝑥 − C. 𝑦(𝑥) = 30𝑥 − D. 𝑦(𝑥) = 40𝑥 −

!! ! !! ! !! ! !! !

−1 +1 −1 +1

SOLUTION: The TOPIC of INITIAL VALUE PROBLEMS is not directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Therefore, we must memorize these formulas and understand their applications independent of the NCEES Supplied Reference Handbook. As we can do with any algebraic expression, given a defined DIFFERENTIAL EQUATION, a solution can be derived.



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The goal in solving DIFFERENTIAL EQUATIONS is to find an expression for the differential function in terms of its independent variable. When solving DIFFERENTIAL EQUATIONS, we always start with defining a GENERAL SOLUTION that represents an infinite series of functions that are all solutions of the given equation. However, as we are in this problem, we are usually concerned with defining a more specific function that both satisfies a given equation and a set of some initial conditions or boundary conditions. This is referred to as our PARTICULAR SOLUTION. There are two mandatory parts to defining the PARTICULAR SOLUTION: 1. A DIFFERENTIAL EQUATION 2. A set of INITIAL or BOUNDARY CONDITIONS Without both of these pieces present, a conclusive solution within a given range can’t be determined. In this problem we are given the DIFFERENTIAL EQUATION: 𝑑𝑦 = 10 − 𝑥 𝑑𝑥



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We are also given the INITIAL CONDITION: 𝑦 0 = −1 The goal is to find a function, 𝑦(𝑥), that solves this DIFFERENTIAL EQUATION given the INITIAL CONDITION. Luckily, depending on the type of DIFFERENTIAL EQUATION we are working with, a standard GENERAL SOLUTION has been set and derived for us, even given to us in the NCEES Reference Handbook, which we will get to. Although in this problem we will derive the solution from scratch, in future teachings, we will learn how to identify the type of DIFFERENTIAL EQUATION we are working with to help us circumvent all the up front grunt work to a more efficient solution. Generally speaking, the steps in solving these types of process will follow this flow: 1. First solve the equation to find the general solution (which contains one or more arbitrary constants or coefficients.) 2. Use the initial or boundary condition(s) to determine the exact value(s) of those constant(s). So let’s start with step 1.



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The first thing we can do is cross-multiply the differential equation to get all of the variables on the same side of the equation, and re-write it as: 𝑑𝑦 = 10 − 𝑥 𝑑𝑥 We then integrate each side of the equation, such that: ∫ 𝑑𝑦 = ∫ 10 − 𝑥 𝑑𝑥 We find that the GENERAL SOLUTION of the differential equation is:

𝑦(𝑥) = 10𝑥 −

𝑥! +𝐶 2

We will take this GENERAL SOLUTION and then set it against the INITIAL CONDITIONS to wholly quantify any unknown values, which in this case, is only the CONSTANT OF INTEGRATION. We are given the INITIAL CONDITION: 𝑦 0 = −1 Plugging this in to our GENERAL SOLUTION, we have:

−1 = 10 0 −



0 +𝐶 2

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Rearranging and solving for the CONSTANT OF INTEGRATION, “𝐶”, we find: 𝐶 = −1 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: 𝑥! 𝑦(𝑥) = 10𝑥 − − 1 2 Applying the initial condition allowed us to define one solution among the infinite number that result from the initial integration, as represented by the CONSTANT OF INTEGRATION, “𝐶”. This is our PARTICULAR SOLUTION.

The correct answer choice is A. 𝒚(𝒙) = 𝟏𝟎𝒙 −



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𝒙𝟐 𝟐

−𝟏

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