68.1 First Order Linear Differential Equations Concept Overview

FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS | CONCEPT OVERVIEW The topic of DIFFERENTIAL EQUATIONS can be referenced on page 30 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

CONCEPT INTRO: A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is typically arranged in the following standard form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 Where: • P x and Q x are functions of "x", and in some cases may be constants • “𝑥” is the INDEPENDENT VARIABLE • “𝑦” is the DEPENDENT VARIABLE The STANDARD FORM of a FIRST ORDER LINEAR DIFFERENTIAL EQUATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.



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In order to solve a FIRST ORDER LINEAR DIFFERENTIAL EQUATION, we must start with the differential equation in this standard form. If the DIFFERENTIAL EQUATION is not originally presented in this form, then an effort should be made to rearrange it so that it fits this mold. The GENERAL FORMULA of the INTEGRATING FACTOR is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Often times we are presented FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS that can not easily be INTEGRATED or SOLVED. In these cases, we deploy what is called an INTEGRATING FACTOR, which is a function most commonly expressed as 𝜇 𝑥 . In practice, the INTEGRATING FACTOR is multiplied through the DIFFERENTIAL EQUATION resulting in new equivalent equation that can easily be INTEGRATED or SOLVED. Most FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS that we will encounter will be solved using the INTEGRATING FACTOR method.



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The PROCESS of solving a FIRST ORDER LINEAR DIFFERENTIAL EQUATION is fairly standard and can be outlined as such: 1. Confirm that the classification of the DIFFERENTIAL EQUATION is indeed FIRST ORDER and LINEAR. It is important to realize that the method we choose to evaluate the differential equation, is dependent on how we classify the differential equation with respect to order, linearity, homogeneous, separable, etc. 2. Compare the DIFFERENTIAL EQUATION with the standard form for a FIRST ORDER LINEAR DIFFERENTIAL EQUATION. If the leading coefficient is not 1, divide the equation through by the coefficient of !"

the !" term first and aim to get it in to the form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 After writing the equation in standard form, the 𝑃(𝑥) term can be identified. 3. Once we have identified the 𝑃 𝑥 term, we can determine the integrating factor by plugging the 𝑃(𝑥) term into the formula for the integrating factor: 𝜇 𝑥 = 𝑒∫ !



! !"

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4. Multiply the equation in standard form by the integrating factor, 𝜇 𝑥 and verify that the left side becomes the PRODUCT RULE (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. The integrating factor is defined so that equation becomes equivalent to: 𝑑 𝜇 𝑥 𝑦 = 𝜇 𝑥 𝑄(𝑥) 𝑑𝑥 5. We then integrate both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. 𝜇 𝑥 𝑦 = ∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥 The TOPICS of INTEGRAL CALCULUS, INTEGRATION BY PARTS, and INTEGRATION BY SUBSTITUTION can be referenced under the topic of INTEGRAL CALCULUS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When solving FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS using the process outlined above, it is important to realize that we may need to use ADVANCED INTEGRATION TECHNIQUES such as INTEGRATION BY SUBSTITUTION and INTEGRATION BY PARTS. We must also remember to include the CONSTANT OF INTEGRATION when evaluating integrals.



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Finally, division by the integrating factor, 𝜇(𝑥), gives “𝑦” EXPLICITLY in terms of “𝑥”, and thus gives the general solution to the differential equation. With the GENERAL SOLUTION of the DIFFERENTIAL EQUATION defined, we can use the INITIAL CONDITION to solve for the CONSTANT OF INTEGRATION, and obtain the PARTICULAR SOLUTION of the differential equation. The general solution of a FIRST ORDER LINEAR DIFFERENTIAL EQUATION can be represented by the equation:

𝑦 𝑥 =

∫ 𝜇 𝑥 𝑄 𝑥 𝑑𝑥 (+𝐶) 𝜇(𝑡)

Again, it is important to remember to include the CONSTANT OF INTEGRATION, which is included as “(+𝐶)” in the formula above for the general solution of a FIRST ORDER DIFFERENTIAL EQUATION. Like any INDEFINITE INTEGRAL, the general solution of a differential equation is a set of infinitely many functions containing one or more arbitrary constant(s).



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FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS | 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 for the given differential equation is best represented as: 𝑑𝑦 + 𝑦 = 𝑥; 𝑦 0 = 2 𝑑𝑥 A. 𝑦(𝑥) = 𝑥 − 1 + 3𝑒 ! B. 𝑦(𝑥) = 𝑥 − 1

!!

C. 𝑦(𝑥) = 𝑥 − 1

!

+ 3𝑒 !!

+3

D. 𝑦(𝑥) = 𝑥 − 1 + 3𝑒 !!

SOLUTION: The first thing we need to do here is classify the differential equation to verify which method we should use to solve for the general and particular solutions. In this problem, we are given the function: 𝑑𝑦 +𝑦 =𝑥 𝑑𝑥 This DIFFERENTIAL EQUATION is both FIRST ORDER and LINEAR. Made with by Prepineer | Prepineer.com

The STANDARD FORM of a FIRST ORDER LINEAR DIFFERENTIAL EQUATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. A FIRST ORDER LINEAR DIFFERENTIAL EQUATION is typically arranged in the standard form: 𝑑𝑦 + 𝑃 𝑥 𝑦 = 𝑄(𝑥) 𝑑𝑥 Where: • P x and Q x are functions of "x", and in some cases may be constants • “𝑥” is the INDEPENDENT VARIABLE • “𝑦” is the DEPENDENT VARIABLE So at this point, let’s confirm that we have it all set up in this form so we can move forward without hitting any obstacles. We see that this equation is already expressed in standard form with the standard terms represented as: 𝑃 𝑥 =1 𝑄 𝑥 =𝑥



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The FORMULA for the INTEGRATING FACTOR is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. With the 𝑃 𝑥 term now identified, we can determine the integrating factor by plugging this term into the standard form of the INTEGRATING FACTOR, which is: 𝜇 𝑥 = 𝑒∫ !

! !"

Plugging in 𝑃 𝑥 we get: 𝜇 𝑥 =𝑒

! !"

Carrying out the INTEGRATION, our INTEGRATING FACTOR becomes: 𝜇 𝑥 = 𝑒! We now multiply the equation in standard form by the integrating factor, 𝜇 𝑥 and verify that the left side becomes the PRODUCT RULE (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. Multiplying both sides of the differential equation by the integrating factor:

𝜇 𝑥



𝑑𝑦 + 𝑦 = 𝜇(𝑥)𝑥 𝑑𝑥

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Plugging in the function for the integrating factor:

𝑒!

𝑑𝑦 + 𝑒!𝑦 = 𝑒!𝑥 𝑑𝑥

We need to verify that the left side becomes the product rule (𝜇 𝑥 𝑦 𝑥 )′ and write it as such. Viewing our formula at this point, we can confirm that indeed the left side follows along with the PRODUCT RULE, which again reads as: The First, times the Derivative of the Second, plus the second times the derivative of the First Now adjusting the formula so that the left side is written in the form (𝜇 𝑥 𝑦 𝑥 )′ we get: 𝑑 ! 𝑒 𝑦 = 𝑒!𝑥 𝑑𝑥 We now need to INTEGRATE both sides of the equation with respect to the independent variable, which is “𝑥” in this differential equation. Rearranging a bit first, we get: 𝑑 𝑒 ! 𝑦 = 𝑒 ! 𝑥 𝑑𝑥



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And INTEGRATING: ∫ 𝑑 𝑒 ! 𝑦 = ∫ 𝑒 ! 𝑥 𝑑𝑥 The TOPIC of INTEGRATION BY PARTS can be referenced under the topic of INTEGRAL CALCULUS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The left side gives us no heart ache, but the right side of this INTEGRATION may present some difficulty if we didn’t know what we were doing. However, we will use INTEGRATION BY PARTS, to evaluate the right side: ∫ 𝑒 ! 𝑥 𝑑𝑥 INTEGRATION BY PARTS lays out as:

∫𝑢

𝑑𝑣 𝑑𝑢 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥

In our case, we will define: 𝑢=𝑥 And: 𝑑𝑣 = 𝑒! 𝑑𝑥

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We plug in these values: 𝑥𝑒 ! − ∫ 𝑒 ! 𝑑𝑥 And evaluate the integral: 𝑥𝑒 ! − 𝑒 ! = 𝑒 ! (𝑥 − 1) Carrying out the INTEGRATION on both sides, we now have: 𝑒!𝑦 = 𝑒! 𝑥 − 1 + 𝐶 This is the IMPLICIT SOLUTION to the FIRST ORDER LINEAR DIFFERENTIAL EQUATION. We can now determine the EXPLICIT SOLUTION. Recall that the EXPLICIT SOLUTION organizes the expression so that the INDEPENDENT and DEPENDENT VARIABLES are isolated on either side of the equation and takes the strict form of 𝑦 = 𝑓(𝑥), where 𝑦 is explicitly defined by a function 𝑓 𝑥 . We are close here, but do not have an expression that is EXPLICIT due to the exponential function being on the left side of the equation…we have to tackle that.



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Doing so, we can divide by the integrating factor, 𝜇(𝑥), getting “𝑦” explicitly in terms of “𝑥", and thus, giving us the general solution to the differential equation as: 𝑦(𝑥) = 𝑥 − 1 + 𝐶𝑒 !! As we have the general solution of the differential equation, we can now use the INITIAL CONDITION to solve for the CONSTANT OF INTEGRATION, and the PARTICULAR SOLUTION of the differential equation. Again, we were given the INITIAL CONDITION: 𝑦(0) = 2 Plugging this INITIAL CONDITION in to our GENERAL SOLUTION, we have: 2 = 0 − 1 + 𝐶𝑒 ! Which simplifies to: 2 = −1 + 𝐶 Rearranging and solving for the CONSTANT OF INTEGRATION we get: 𝐶=3



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We can now plug the CONSTANT OF INTEGRATION into our general solution and find the PARTICULAR SOLUTION, which is: 𝑦(𝑥) = 𝑥 − 1 + 3𝑒 !! The correct answer choice is D. 𝐲(𝐱) = 𝐱 − 𝟏 + 𝟑𝐞!𝐱



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