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    Exact DE

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    Mohamad Alrifai
    This document provides an overview of exact first-order ordinary differential equations (ODEs) with the following key points: 1. It defines an exact first-order ODE and the condition for an ODE to be exact. 2. It outlines the steps to solve an exact first-order ODE, which involve finding an integral function whose partial derivatives equal the ODE. 3. Examples are provided to demonstrate solving exact first-order ODEs, finding integrating factors to make ODEs exact, and solving initial value problems with exact ODEs.

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    Exact DE

    Uploaded by

    Mohamad Alrifai
    59 views10 pages
    This document provides an overview of exact first-order ordinary differential equations (ODEs) with the following key points: 1. It defines an exact first-order ODE and the condition for an ODE to be exact. 2. It outlines the steps to solve an exact first-order ODE, which involve finding an integral function whose partial derivatives equal the ODE. 3. Examples are provided to demonstrate solving exact first-order ODEs, finding integrating factors to make ODEs exact, and solving initial value problems with exact ODEs.

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    This document provides an overview of exact first-order ordinary differential equations (ODEs) with the following key points: 1. It defines an exact first-order ODE and the condition for an ODE to be exact. 2. It outlines the steps to solve an exact first-order ODE, which involve finding an integral function whose partial derivatives equal the ODE. 3. Examples are provided to demonstrate solving exact first-order ODEs, finding integrating factors to make ODEs exact, and solving initial value problems with exact ODEs.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    59 views10 pages

    Exact DE

    Uploaded by

    Mohamad Alrifai
    This document provides an overview of exact first-order ordinary differential equations (ODEs) with the following key points: 1. It defines an exact first-order ODE and the condition for an ODE to be exact. 2. It outlines the steps to solve an exact first-order ODE, which involve finding an integral function whose partial derivatives equal the ODE. 3. Examples are provided to demonstrate solving exact first-order ODEs, finding integrating factors to make ODEs exact, and solving initial value problems with exact ODEs.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 10

    MATH142

    Essentials of Engineering Mathematics

    First-Order ODEs

    I. Separable
    II. Homogeneous
    III. Linear
    IV. Exact
    V. Bernoulli

    IV. Exact Equations

    Prerequisite -First Partial Derivative of a function of two variables 𝑧 = 𝑓(𝑥, 𝑦)


    𝜕𝑓 𝜕𝑓
    The first partial derivatives of 𝑓 with respect to x and y are denoted by 𝑓𝑥 = 𝑎𝑛𝑑 𝑓𝑦 =
    𝜕𝑥 𝜕𝑦

    Example 1: Find the first partial derivatives of the following functions.

    a. 𝑓(𝑥, 𝑦) = 4𝑥 + 2𝑦 + 2𝑥 3 𝑦 2

    b. ℎ(𝑥, 𝑦) = 3𝑥 − 𝑥 2 𝑦 2 + 2𝑥 3 𝑦

    1
    2𝑦
    c. 𝑔(𝑥, 𝑦) = 𝑥𝑒 𝑥

    d. 𝑧 = 𝑥 4 𝑠𝑖𝑛(𝑥𝑦 3 )

    2
    Definition: A first-order ODE of the form 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0 is said to be exact if
    𝜕𝑀 𝜕𝑁
    and only if 𝜕𝑦
    = 𝜕𝑥
    (test for exactness).
    In this case, there exists a function 𝑓(𝑥, 𝑦) satisfying:

    𝜕𝑓 𝜕𝑓
    𝜕𝑥
    = 𝑀(𝑥, 𝑦), 𝜕𝑦
    = 𝑁(𝑥, 𝑦) and hence 𝑦 is implicitly defined by 𝑓(𝑥, 𝑦) = 𝑐.

    Steps for Solving Exact Equations

    1. Write the given DE in differential form 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0 (𝐼)

    𝜕𝑀 𝜕𝑁
    2. Test for exactness 𝜕𝑦
    = 𝜕𝑥
    . If the partial derivatives are equal, proceed to the
    following steps.

    𝜕𝑓
    3. Write 𝜕𝑥 = 𝑀(𝑥, 𝑦) and integrate w.r.t 𝑥 ⇒ 𝑓(𝑥, 𝑦) = ∫ 𝑀(𝑥, 𝑦)𝑑𝑥 + ℎ(𝑦) (𝐼𝐼)

    4. Take the partial derivative w.r.t 𝑦 of 𝑓(𝑥, 𝑦) in (𝐼𝐼) and equate it to 𝑁(𝑥, 𝑦)

    5. Find ℎ′ (𝑦) from step 4 and integrate to get ℎ(𝑦)

    6. Replace ℎ(𝑦) in (𝐼𝐼) to find 𝑓(𝑥, 𝑦)

    7. Finally, 𝑦 is implicitly defined by 𝑓(𝑥, 𝑦) = 𝑐 . Solve for 𝑦 if possible.

    3
    Example 2: Solve the following IVP.

    𝑐𝑜𝑠𝑥 − 2𝑥𝑦 + (𝑒 𝑦 − 𝑥 2 )𝑦 ′ = 0 𝑦(1) = 4

    4
    Example 3: Solve (𝑥 + 𝑠𝑖𝑛𝑦)𝑑𝑥 + (𝑥𝑐𝑜𝑠𝑦𝑦 − 2𝑦)𝑑𝑦 = 0

    5
    2+𝑦𝑒 𝑥𝑦
    Example 4: 𝑦 ′ =
    2𝑦−𝑥𝑒 𝑥𝑦

    6
    Special Integrating Factors

    𝜕𝑀 𝜕𝑁
    Given 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0 (𝐼) and suppose 𝜕𝑦
    ≠ 𝜕𝑥

    Find:
    𝜕𝑀 𝜕𝑁
    −
    𝜕𝑦 𝜕𝑥
    1. = 𝑓(𝑥) ⇒ 𝜇(𝑥) = 𝑒 ∫ 𝑓(𝑥)𝑑𝑥
    𝑁

    𝜕𝑁 𝜕𝑀
    −
    𝜕𝑥 𝜕𝑦
    2. 𝑀
    = 𝑔(𝑦) ⇒ 𝜇(𝑦) = 𝑒 ∫ 𝑔(𝑦)𝑑𝑦

    and multiply (𝐼) by one of the integrating factors to change it to an exact DE.

    𝑦 1 𝑑𝑦
    Example 5: Solve 𝑥 2 + 1 + 𝑥 𝑑𝑥 = 0

    7
    8
    Example 6: Solve 2𝑥𝑦𝑑𝑥 + (𝑦 2 − 3𝑥 2 )𝑑𝑦 = 0

    9
    10

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