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    Solving Separable Differential Equation Using Antidifferential

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    Jiejhay Jiipie
    A chemical reaction is known to follow the rate law: Rate = k[A]^2[B] Where [A] and [B] are the concentrations of reactants A and B and k is the rate constant. If initially there are 2 moles of A and 1 mole of B, find an expression for the concentration of A as a function of time. Step-by-Step process Explanation Rate = k[A]2[B] Given rate law d[A] Rate is equal to the rate of change of [A] dt d[A] = -k[A]2[B]dt Rate law expressed as derivative

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    Solving Separable Differential Equation Using Antidifferential

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    Jiejhay Jiipie
    269 views9 pages
    A chemical reaction is known to follow the rate law: Rate = k[A]^2[B] Where [A] and [B] are the concentrations of reactants A and B and k is the rate constant. If initially there are 2 moles of A and 1 mole of B, find an expression for the concentration of A as a function of time. Step-by-Step process Explanation Rate = k[A]2[B] Given rate law d[A] Rate is equal to the rate of change of [A] dt d[A] = -k[A]2[B]dt Rate law expressed as derivative

    Original Description:

    Sample Report in Basic Calculus STEM

    Original Title

    Basic CalculusReport

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    A chemical reaction is known to follow the rate law: Rate = k[A]^2[B] Where [A] and [B] are the concentrations of reactants A and B and k is the rate constant. If initially there are 2 moles of A and 1 mole of B, find an expression for the concentration of A as a function of time. Step-by-Step process Explanation Rate = k[A]2[B] Given rate law d[A] Rate is equal to the rate of change of [A] dt d[A] = -k[A]2[B]dt Rate law expressed as derivative

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    269 views9 pages

    Solving Separable Differential Equation Using Antidifferential

    Uploaded by

    Jiejhay Jiipie
    A chemical reaction is known to follow the rate law: Rate = k[A]^2[B] Where [A] and [B] are the concentrations of reactants A and B and k is the rate constant. If initially there are 2 moles of A and 1 mole of B, find an expression for the concentration of A as a function of time. Step-by-Step process Explanation Rate = k[A]2[B] Given rate law d[A] Rate is equal to the rate of change of [A] dt d[A] = -k[A]2[B]dt Rate law expressed as derivative

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 9

    Basic Calculus | Separable Equation

    Topic: Solving Separable Differential Equation using


    Antidifferentiation

    Subject Teacher: Mrs. Ma. Cristina Javier Barbacena

    Solving Separable
    Differential Equation
    Using Antidifferential

    John Rommel Octavo


    11- STEM B
    1
    TOPIC OBJECTIVES:

    SOLVE SEPARABLE DIFFERENTIATION USING ANTIDIFFERENTIATION

    o Determining separable derivative function


    o Solve the general solution of separable differential equation
    o Solve the particular solution of separable differential equation
    Determining Separable 2
    Derivative Equations

     A differential equation is said to be separable if the variables can be separated.


     A differential equation is separable if and only if it can be written as
    𝒅𝒚
    = 𝒇(𝒙)𝒈(𝒚) or 𝑮(𝒚)𝒅(𝒚) = 𝑭(𝒙)𝒅(𝒙)
    𝒅𝒙

    Table 1: Shows first-order differential equations and determines if it is separable or not. It also shows the
    resulting equation if the differential equation if separable.
    Note: Resulting equations can be obtain through the manipulation of variables using laws of algebra so that
    variables are separated from both sides of equation. Resulting equations are important in solving for the
    antiderivative.

    Differential Equations Separable or Not Resulting Equations

    𝒅𝒚 𝒙𝟐
    = Separable (𝟐 − 𝒚𝟐 )𝒅𝒚 = 𝒙𝟐 𝒅𝒙
    𝒅𝒙 𝟐 − 𝒚𝟐

    𝒅𝟒 𝒚 𝒅𝟐 𝒚
    + = 𝒇(𝒙) Not Separable
    𝒅𝒙𝟒 𝒅𝒙𝟐

    𝒅𝒚 𝒙𝟐 + 𝟏
    = Separable 𝟐𝒚𝒅𝒚 = (𝒙𝟐 + 𝟏)𝒅𝒙
    𝒅𝒙 𝟐𝒚

    𝒅𝒚 𝒙 + 𝒚
    = Not Separable
    𝒅𝒙 𝒙 − 𝒚

    𝒅𝒚 𝒙
    = Separable 𝒚𝒅𝒚 = 𝒙𝒅𝒙
    𝒅𝒙 𝒚

    𝒙 𝒅𝒙 + 𝒔𝒆𝒄 𝒙 𝒔𝒊𝒏 𝒚 𝒅𝒚 = 𝟎 Separable 𝒔𝒊𝒏 𝒚 𝒅𝒚 = −𝒙 𝒄𝒐𝒔 𝒙 𝒅𝒙


    3
    Solving General Solution

     The general solution of the differential equation can be obtain only when the
    separable equation is integrated from both side of the equation. Note that not
    all equations can be evaluated in terms of y.

    Example: Step by step process of solving the general equation of the differential equation.

    Step-by-Step process Explanation

    𝒅𝒚 𝒙𝟐 + 𝟏
    = Given
    𝒅𝒙 𝟐𝒚

    The variables were separated from both side of the


    𝟐𝒚𝒅𝒚 = (𝒙𝟐 + 𝟏)𝒅𝒙
    equation.

    ∫ 𝟐𝒚𝒅𝒚 = ∫(𝒙𝟐 + 𝟏)𝒅𝒙 Solve for integral of both sides of the equations

    𝟏 𝟑
    𝒚𝟐 = 𝒙 +𝒙+𝑪 Theorems (no. 11) on antidifferentiation
    𝟑
    𝟏
    𝒚 = √ 𝒙𝟑 + 𝒙 + 𝑪 Evaluate in terms of y
    𝟑

    Change the variable of the constant function into A.


    𝟏 𝟑
    𝒚 = √𝑨 + 𝟑
    𝒙 + 𝒙 ; where A = C This is important so that constant may be derive
    when a point is given.

    𝟏
    𝒚 = √𝑨 + 𝒙𝟑 + 𝒙 ; where A = C General solution of the differential equation
    𝟑
    4
    Example

    Step-by-Step process Explanation

    𝟐𝐲𝐝𝐱 − 𝟑𝐱𝐝𝐲 = 𝟎 Given

    𝟐 𝟑 The variables were separated from both side of the


    𝒅𝒙 = 𝒅𝒚
    𝒙 𝒚 equation.

    𝟐 𝟑
    ∫ 𝒅𝒙 = ∫ 𝒅𝒚 Solve for integral of both sides of the equations
    𝒙 𝒚

    𝟐 𝒍𝒏 |𝒙| = 𝟑 𝒍𝒏 |𝒚| + 𝑪 Theorem

    𝒍𝒏 |𝒚|𝟑 = 𝟐 𝒍𝒏 |𝒙| − 𝑪 Evaluate in terms of y; Laws of Logarithm

    𝟐
    |𝒚|𝟑 = 𝒆𝒍𝒏 |𝒙| 𝒆−𝑪 Evaluate in terms of y; Laws of Logarithm

    |𝒚|𝟑 = 𝒆−𝑪 𝒙𝟐 Evaluate in terms of y; Laws of Logarithm

    𝟑
    𝒚 = √ 𝒆−𝑪 𝒙𝟐 Evaluate in terms of y

    Change the variable of the constant function into A.


    𝟑
    𝒚= √𝑨𝒙𝟐 ; where A = e-3 This is important so that constant may be derive
    when a point is given.

    𝟑 Simplify; General solution to the separable


    𝒚 = √𝑨𝒙𝟐/𝟑 ; where A = e-3 differential equation
    5
    Solving Particular Solution

    The particular solution of the differential equation can be obtain only when there are
    initial conditions, or if the solution passes through a point. The unknown value ‘A’ will
    be solved during substitution.

    Example: Continuity of the example on page number 3.

    Step-by-Step process Explanation

    𝟐𝐲𝐝𝐱 − 𝟑𝐱𝐝𝐲 = 𝟎 Given

    𝟐 𝟑 The variables were separated from both side of the


    𝒅𝒙 = 𝒅𝒚
    𝒙 𝒚 equation.

    𝟑
    𝒚 = √𝑨𝒙𝟐/𝟑 ; where A = e-3 General solution solved on page 4

    Point (1,1) Example point where the solution pass through

    𝟑
    𝟏 = √𝑨𝟏𝟐/𝟑 Substitution

    Substitution; Derivation of the value of the


    𝑨=𝟏 constant A.

    𝟑 Substitution of A; Particular Separable Differential


    𝒚 = √𝒙𝟐 Equation
    6
    Application

    Problem:
    In a class experiment, you have observed that the population of bacteria in a petri
    dish is reproduced one-fourth of its population at any point of time. Let y denote the
    population so that the change is one-fourth y. Solve for the particular solution of the
    equation to recover the population y using integration given that the initial population is
    100.

    Step-by-Step process Explanation


    𝟏
    𝒚 Given
    𝟒
    𝒅𝒚 𝟏
    = 𝒚 Application of derivative
    𝒅𝒙 𝟒
    𝒅𝒚 𝟏 The variables were separated from both side of the
    = 𝒅𝒙
    𝒚 𝟒 equation.
    𝒅𝒚 𝟏 Integration of both side to solve the general
    ∫ = ∫ 𝒅𝒙
    𝒚 𝟒 solution
    𝟏
    𝒍𝒏 |𝒚| = 𝒕+𝑪 Theorem (no. 11) on antidifferentiation
    𝟒
    𝟏
    |𝒚| = 𝒆𝟒𝒕 𝒆𝒄 Evaluate in terms of y; Laws of logarithm
    𝟏
    𝒚 = 𝑨𝒆𝟒𝒕 ; where A = eC General Solution

    Point (0,100) Initial time: 0; Initial population: 100


    𝟏
    𝟏𝟎𝟎 = 𝑨𝒆𝟒(𝟎) Substitution

    A = 100 Evaluate; Derivation of the constant A value

    𝟏 Substitute; Particular solution to the separable


    𝒚 = 𝟏𝟎𝟎𝒆𝟒𝒕 differential equation

    𝟏
    Answer: The particular solution or the equation to be used in the problem is 𝒚 = 𝟏𝟎𝟎𝒆𝟒𝒕 .
    7
    Application

    Problem:
    𝒅𝒓
    Find the curve 𝒓 = 𝒓(𝜽) in polar coordinate given the formula = −𝒓 𝒕𝒂𝒏𝜽 and
    𝒅𝜽
    the point (pi, 2) that solves the solution of the initial-value problem.

    Step-by-Step process Explanation


    𝐝𝐫
    = −𝒓 𝒕𝒂𝒏 𝜃 Given
    𝒅𝜃
    𝒅𝒓 The variables were separated from both side of the
    = −𝐭𝐚𝐧𝜃 𝒅𝜃
    𝒓 equation.
    𝒅𝒓 Integration of both side to solve the general
    ∫ = ∫ −𝐭𝐚𝐧𝜃 𝒅𝜃
    𝒓 solution
    𝟏 −𝐬𝐢𝐧𝜽
    ∫ 𝒅𝒓 = ∫ 𝒅𝜽 Simplify; Trigonometric Identities
    𝒓 𝒄𝒐𝒔𝜽
    𝟏 𝟏 𝟏
    ∫ 𝒓 𝒅𝒓 = ∫ 𝒖 𝒅𝒖 U-substitution; where u=cos𝜽 and dx =− 𝒔𝒊𝒏𝜽 𝒅𝒖

    𝒍𝒏 |𝒓| = 𝒍𝒏|𝒖| + 𝒄 Theorem

    𝒍𝒏 |𝒓| = 𝒍𝒏|𝒄𝒐𝒔(𝜽)| + 𝒄 Substitute

    𝒆𝒍𝒏 |𝒓| = 𝒆𝒍𝒏|𝒄𝒐𝒔(𝜽)| 𝒆𝒄 Evaluate in terms of r; Laws of logarithm

    General solution; Substitute variable of the


    𝒓 = 𝑨 𝒄𝒐𝒔(𝜽) ; where A=ec
    constant to A

    Point(pi, 2) Given

    𝟐 = 𝑨 𝒄𝒐𝒔(𝝅)  A = -2 Substitute; Simplify


    Particular solution to the separable differential
    𝒓 = 𝟐 𝒄𝒐𝒔(𝜽)
    equation

    Answer: The particular solution of the initial-value problem is


    𝒓 = 𝟐 𝒄𝒐𝒔(𝜽) .
    8
    References

    Balmaceda, Lemence, Ortega, Pilar-Arceo, & Vallejo, (2016). Basic Calculus. Commission
    on Higher Education Diliman Quezon City.

    Separable Equations. (2016). Retrieved on February 28, 2019, from


    https://www.cliffsnotes.com/study-guides/differential-equations/first-order-
    equations/separable-equations

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