65.1 Initial Value Problems Problem Set
INITIAL VALUE PROBLEMS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The particular solution of the differential equation noted is best represented as: ππ¦ = 9π₯ ! β 4π₯ + 5; π¦ β1 = 0 ππ₯ A. π¦(π₯) = π₯ ! β 2π₯ ! + 5π₯ + 12 B. π¦(π₯) = 2π₯ ! β 3π₯ ! + 6π₯ + 4 C. π¦(π₯) = 3π₯ ! β 2π₯ ! + 5π₯ + 10 D. π¦(π₯) = 4π₯ ! β 3π₯ ! + 5π₯ + 10
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PROBLEM 2: The particular solution of the differential equation noted is best represented as: ππ¦ + 5π¦ = 0; π¦ 0 = 1 ππ₯ A. 1 B. 10π₯ C. 9 D. 2π₯ !
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PROBLEM 3: Solve the initial value problem: ππ¦ = π₯ + 4, π¦ 1 = 2 ππ₯ A. π¦(π₯) = π₯ ! + 5π₯ B. π¦(π₯) =
!! !
+ 4π₯ β 2.5
C. π¦(π₯) = 2π₯ ! + π₯ D. π¦(π₯) = β10 + πΆ
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PROBLEM 4: Solve the initial value problem: ππ¦ = π₯ ! + 6π₯, π¦ 0 = 1 ππ₯ A. π¦(π₯) = 6π₯ ! B. π¦(π₯) = C. π¦(π₯) = D. π¦(π₯) =
!! ! ! !! ! !! !
+ π₯! + πΆ
+ π₯! + 3π₯ ! + 1
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PROBLEM 5: Solve the initial value problem: ππ¦ = π₯ ! + 3π₯ ! + 2π₯ + 5, π¦ 1 = 2 ππ₯ A. π¦(π₯) =
!! !
+ π₯ ! + π₯ ! + 5π₯ β 5.25
B. π¦(π₯) = π₯ ! + π₯ ! + π₯ + πΆ C. π¦(π₯) = 2π₯ + π₯ ! + π₯ ! + 3π₯ !
D. π¦(π₯) = π₯ ! + π₯ β 4 !
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INITIAL VALUE PROBLEMS | SOLUTIONS SOLUTION 1: 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. 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.
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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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ = 9π₯ ! β 4π₯ + 5 ππ₯ We are also given the INITIAL CONDITION: π¦ β1 = 0 The goal is to find a function, π¦(π₯), 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 upfront grunt work to a more efficient solution.
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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 letβs start with step 1. 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: ππ¦ = 9π₯ ! β 4π₯ + 5 ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (9π₯ ! β 4π₯ + 5)ππ₯ We find that the GENERAL SOLUTION of the differential equation is: 9π₯ ! 4π₯ ! π¦(π₯) = β + 5π₯ + πΆ 3 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.
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We are given the INITIAL CONDITION: π¦ β1 = 0 Plugging this in to our GENERAL SOLUTION, we have: 0 = 3 β1
!
β 2 β1
!
+ 5 β1 + πΆ
Which simplifies to: 0 = β3 β 2 β 5 + πΆ Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ = 10 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π¦(π₯) = 3π₯ ! β 2π₯ ! + 5π₯ + 10 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, βπΆβ. This is our PARTICULAR SOLUTION.
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The correct answer choice is C. π(π) = πππ β πππ + ππ + ππ
SOLUTION 2: 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. 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.
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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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ + 5π¦ = 0 ππ₯ Which can be rewritten in to the more preferable form: ππ¦ = β5π¦ ππ₯ We are also given the INITIAL CONDITION: π¦ 0 =1 The goal is to find a function, π¦(π₯), 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.
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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 upfront 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 letβs start with step 1. 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: ππ¦ = β5π¦ ππ₯ Grouping the variables where we want them in our equation, we can rearrange such that:
β
1 ππ¦ = ππ₯ 5π¦
We then integrate each side of the equation, such that:
β
1 5
1 ππ¦ = π¦
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We find that the GENERAL SOLUTION of the differential equation is: 1 β ππ π¦ = π₯ + πΆ 5 We want to get y on the left side alone, so letβs multiply each side by -5: ππ π¦ = β5π₯ β 5πΆ And get rid of the natural log on the left side, such that: π !" ! = π !!!!!! Which gives us: π¦(π₯) = π !!!!!! 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: π¦ 0 =1
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Plugging this in to our GENERAL SOLUTION, we have: 1 = π !!(!)!!! Which can be rewritten as: ππ 1 = ππ π !!! And we know that: ππ 1 = 0 So: 0 = β5πΆππ π Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ=0 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π¦(π₯) = π !!(!)
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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, βπΆβ. But something interested has happened here. Our PARTICULAR SOLUTION has evolved to be: π¦(π₯) = π ! Which is: π¦(π₯) = 1 The correct answer choice is A. π
SOLUTION 3: 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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by Prepineer | Prepineer.com
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, and its what we are asked for when asked to βSolve the Initial Value Problemβ. 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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ =π₯+4 ππ₯
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We are also given the INITIAL CONDITION: π¦ 1 =2 The goal is to find a function, π¦(π₯), 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 letβs 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: ππ¦ = (π₯ + 4)ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (π₯ + 4)ππ₯ We find that the GENERAL SOLUTION of the differential equation is:
π¦(π₯) =
π₯! + 4π₯ + πΆ 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: π¦ 1 =2 Plugging this in to our GENERAL SOLUTION, we have: 1! 2= + 4(1) + πΆ 2
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Which can be simplified as: !
2=!+4+πΆ Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ = β2.5 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π₯! π¦(π₯) = + 4π₯ β 2.5 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, βπΆβ. This is our PARTICULAR SOLUTION and the ultimately the solution to our INITIAL VALUE PROBLEM.
The correct answer choice is B. π(π) =
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ππ π
+ ππ β π. π
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SOLUTION 4: 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. 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, and its what we are asked for when asked to βSolve the Initial Value Problemβ. There are two mandatory parts to defining the PARTICULAR SOLUTION: 1. A DIFFERENTIAL EQUATION
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2. A set of INITIAL or BOUNDARY CONDITIONS Without both of these pieces present, a conclusive solution within a given range canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ = π₯ ! + 6π₯ ππ₯ We are also given the INITIAL CONDITION: π¦ 0 =1 The goal is to find a function, π¦(π₯), 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.
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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 letβs start with step 1. 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: ππ¦ = (π₯ ! + 6π₯)ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (π₯ ! + 6π₯)ππ₯ We find that the GENERAL SOLUTION of the differential equation is: π₯! π¦(π₯) = + 3π₯ ! + πΆ 3 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.
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We are given the INITIAL CONDITION: π¦ 0 =1 Plugging this in to our GENERAL SOLUTION, we have: 0! 1= +3 0 3
!
+πΆ
Which can be simplified as: 1=0+0+πΆ Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ=1 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π₯! π¦(π₯) = + 3π₯ ! + 1 3 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, βπΆβ.
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This is our PARTICULAR SOLUTION and the ultimately the solution to our INITIAL VALUE PROBLEM.
The correct answer choice is D. π(π) =
ππ π
+ πππ + π
SOLUTION 5: 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. 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.
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This is referred to as our PARTICULAR SOLUTION, and its what we are asked for when asked to βSolve the Initial Value Problemβ. 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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ = π₯ ! + 3π₯ ! + 2π₯ + 5 ππ₯ We are also given the INITIAL CONDITION: π¦ 1 =2 The goal is to find a function, π¦(π₯), 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.
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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 letβs start with step 1. 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: ππ¦ = (π₯ ! + 3π₯ ! + 2π₯ + 5)ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (π₯ ! + 3π₯ ! + 2π₯ + 5)ππ₯ We find that the GENERAL SOLUTION of the differential equation is: π₯! π¦(π₯) = + π₯ ! + π₯ ! + 5π₯ + πΆ 4
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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: π¦ 1 =2 Plugging this in to our GENERAL SOLUTION, we have: 1! 2= + 1! + 1! + 5(1) + πΆ 4 Which can be simplified as:
2=
1 +1+1+5+πΆ 4
Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ = β5.25 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π₯! π¦(π₯) = + π₯ ! + π₯ ! + 5π₯ β 5.25 4
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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, βπΆβ. This is our PARTICULAR SOLUTION and the ultimately the solution to our INITIAL VALUE PROBLEM.
The correct answer choice is A. π(π) =
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ππ π
+ ππ + ππ + ππ β π. ππ
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PROBLEM 1: The particular solution of the differential equation noted is best represented as: ππ¦ = 9π₯ ! β 4π₯ + 5; π¦ β1 = 0 ππ₯ A. π¦(π₯) = π₯ ! β 2π₯ ! + 5π₯ + 12 B. π¦(π₯) = 2π₯ ! β 3π₯ ! + 6π₯ + 4 C. π¦(π₯) = 3π₯ ! β 2π₯ ! + 5π₯ + 10 D. π¦(π₯) = 4π₯ ! β 3π₯ ! + 5π₯ + 10
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PROBLEM 2: The particular solution of the differential equation noted is best represented as: ππ¦ + 5π¦ = 0; π¦ 0 = 1 ππ₯ A. 1 B. 10π₯ C. 9 D. 2π₯ !
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PROBLEM 3: Solve the initial value problem: ππ¦ = π₯ + 4, π¦ 1 = 2 ππ₯ A. π¦(π₯) = π₯ ! + 5π₯ B. π¦(π₯) =
!! !
+ 4π₯ β 2.5
C. π¦(π₯) = 2π₯ ! + π₯ D. π¦(π₯) = β10 + πΆ
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PROBLEM 4: Solve the initial value problem: ππ¦ = π₯ ! + 6π₯, π¦ 0 = 1 ππ₯ A. π¦(π₯) = 6π₯ ! B. π¦(π₯) = C. π¦(π₯) = D. π¦(π₯) =
!! ! ! !! ! !! !
+ π₯! + πΆ
+ π₯! + 3π₯ ! + 1
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PROBLEM 5: Solve the initial value problem: ππ¦ = π₯ ! + 3π₯ ! + 2π₯ + 5, π¦ 1 = 2 ππ₯ A. π¦(π₯) =
!! !
+ π₯ ! + π₯ ! + 5π₯ β 5.25
B. π¦(π₯) = π₯ ! + π₯ ! + π₯ + πΆ C. π¦(π₯) = 2π₯ + π₯ ! + π₯ ! + 3π₯ !
D. π¦(π₯) = π₯ ! + π₯ β 4 !
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INITIAL VALUE PROBLEMS | SOLUTIONS SOLUTION 1: 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. 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.
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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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ = 9π₯ ! β 4π₯ + 5 ππ₯ We are also given the INITIAL CONDITION: π¦ β1 = 0 The goal is to find a function, π¦(π₯), 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 upfront grunt work to a more efficient solution.
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by Prepineer | Prepineer.com
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 letβs start with step 1. 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: ππ¦ = 9π₯ ! β 4π₯ + 5 ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (9π₯ ! β 4π₯ + 5)ππ₯ We find that the GENERAL SOLUTION of the differential equation is: 9π₯ ! 4π₯ ! π¦(π₯) = β + 5π₯ + πΆ 3 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.
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We are given the INITIAL CONDITION: π¦ β1 = 0 Plugging this in to our GENERAL SOLUTION, we have: 0 = 3 β1
!
β 2 β1
!
+ 5 β1 + πΆ
Which simplifies to: 0 = β3 β 2 β 5 + πΆ Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ = 10 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π¦(π₯) = 3π₯ ! β 2π₯ ! + 5π₯ + 10 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, βπΆβ. This is our PARTICULAR SOLUTION.
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The correct answer choice is C. π(π) = πππ β πππ + ππ + ππ
SOLUTION 2: 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. 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.
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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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ + 5π¦ = 0 ππ₯ Which can be rewritten in to the more preferable form: ππ¦ = β5π¦ ππ₯ We are also given the INITIAL CONDITION: π¦ 0 =1 The goal is to find a function, π¦(π₯), 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.
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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 upfront 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 letβs start with step 1. 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: ππ¦ = β5π¦ ππ₯ Grouping the variables where we want them in our equation, we can rearrange such that:
β
1 ππ¦ = ππ₯ 5π¦
We then integrate each side of the equation, such that:
β
1 5
1 ππ¦ = π¦
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We find that the GENERAL SOLUTION of the differential equation is: 1 β ππ π¦ = π₯ + πΆ 5 We want to get y on the left side alone, so letβs multiply each side by -5: ππ π¦ = β5π₯ β 5πΆ And get rid of the natural log on the left side, such that: π !" ! = π !!!!!! Which gives us: π¦(π₯) = π !!!!!! 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: π¦ 0 =1
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Plugging this in to our GENERAL SOLUTION, we have: 1 = π !!(!)!!! Which can be rewritten as: ππ 1 = ππ π !!! And we know that: ππ 1 = 0 So: 0 = β5πΆππ π Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ=0 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π¦(π₯) = π !!(!)
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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, βπΆβ. But something interested has happened here. Our PARTICULAR SOLUTION has evolved to be: π¦(π₯) = π ! Which is: π¦(π₯) = 1 The correct answer choice is A. π
SOLUTION 3: 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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by Prepineer | Prepineer.com
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, and its what we are asked for when asked to βSolve the Initial Value Problemβ. 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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ =π₯+4 ππ₯
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We are also given the INITIAL CONDITION: π¦ 1 =2 The goal is to find a function, π¦(π₯), 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 letβs 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: ππ¦ = (π₯ + 4)ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (π₯ + 4)ππ₯ We find that the GENERAL SOLUTION of the differential equation is:
π¦(π₯) =
π₯! + 4π₯ + πΆ 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: π¦ 1 =2 Plugging this in to our GENERAL SOLUTION, we have: 1! 2= + 4(1) + πΆ 2
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Which can be simplified as: !
2=!+4+πΆ Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ = β2.5 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π₯! π¦(π₯) = + 4π₯ β 2.5 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, βπΆβ. This is our PARTICULAR SOLUTION and the ultimately the solution to our INITIAL VALUE PROBLEM.
The correct answer choice is B. π(π) =
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ππ π
+ ππ β π. π
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SOLUTION 4: 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. 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, and its what we are asked for when asked to βSolve the Initial Value Problemβ. There are two mandatory parts to defining the PARTICULAR SOLUTION: 1. A DIFFERENTIAL EQUATION
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2. A set of INITIAL or BOUNDARY CONDITIONS Without both of these pieces present, a conclusive solution within a given range canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ = π₯ ! + 6π₯ ππ₯ We are also given the INITIAL CONDITION: π¦ 0 =1 The goal is to find a function, π¦(π₯), 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.
Made with
by Prepineer | Prepineer.com
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 letβs start with step 1. 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: ππ¦ = (π₯ ! + 6π₯)ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (π₯ ! + 6π₯)ππ₯ We find that the GENERAL SOLUTION of the differential equation is: π₯! π¦(π₯) = + 3π₯ ! + πΆ 3 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.
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We are given the INITIAL CONDITION: π¦ 0 =1 Plugging this in to our GENERAL SOLUTION, we have: 0! 1= +3 0 3
!
+πΆ
Which can be simplified as: 1=0+0+πΆ Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ=1 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π₯! π¦(π₯) = + 3π₯ ! + 1 3 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, βπΆβ.
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This is our PARTICULAR SOLUTION and the ultimately the solution to our INITIAL VALUE PROBLEM.
The correct answer choice is D. π(π) =
ππ π
+ πππ + π
SOLUTION 5: 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. 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.
Made with
by Prepineer | Prepineer.com
This is referred to as our PARTICULAR SOLUTION, and its what we are asked for when asked to βSolve the Initial Value Problemβ. 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 canβt be determined. In this problem we are given the DIFFERENTIAL EQUATION: ππ¦ = π₯ ! + 3π₯ ! + 2π₯ + 5 ππ₯ We are also given the INITIAL CONDITION: π¦ 1 =2 The goal is to find a function, π¦(π₯), 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.
Made with
by Prepineer | Prepineer.com
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 letβs start with step 1. 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: ππ¦ = (π₯ ! + 3π₯ ! + 2π₯ + 5)ππ₯ We then integrate each side of the equation, such that: β« ππ¦ = β« (π₯ ! + 3π₯ ! + 2π₯ + 5)ππ₯ We find that the GENERAL SOLUTION of the differential equation is: π₯! π¦(π₯) = + π₯ ! + π₯ ! + 5π₯ + πΆ 4
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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: π¦ 1 =2 Plugging this in to our GENERAL SOLUTION, we have: 1! 2= + 1! + 1! + 5(1) + πΆ 4 Which can be simplified as:
2=
1 +1+1+5+πΆ 4
Rearranging and solving for the CONSTANT OF INTEGRATION, βπΆβ, we find: πΆ = β5.25 We can now plug in the value for the constant of integration into the general solution, giving us the PARTICULAR SOLUTION: π₯! π¦(π₯) = + π₯ ! + π₯ ! + 5π₯ β 5.25 4
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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, βπΆβ. This is our PARTICULAR SOLUTION and the ultimately the solution to our INITIAL VALUE PROBLEM.
The correct answer choice is A. π(π) =
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ππ π
+ ππ + ππ + ππ β π. ππ
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