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  • Ii. eliminate the arbitrary of con1. y = cx+c²+1​...

Ii. eliminate the arbitrary of con
1. y = cx+c²+1​


  • Réponse publiée par: jasminsexy


    Given c urve is, y=cx−c

    ANSWER +y=0

    Step-by-step explanation:


  • Réponse publiée par: snow01

    The equation is 3y' - 4xy' + 4xy - 2y - y" - xy" = 0

    Step-by-step explanation:

    To eliminate the arbitrary constants, derive the equation 2 times.

    Deriving the equation

    y=C_{1} x^2 +C_{2} e^{2x}          equation 1

    y'=2C_{1} x +2C_{2} e^{2x}       equation 2

    y"=2C_{1}  +4C_{2} e^{2x}        equation 3

    Solving the equation

    1. Use equation 1, then substitute to equation 2

    y=C_{1} x^2 +C_{2} e^{2x}

    C_{2} e^{2x}=y-C_{1} x^2

    y'=2C_{1} x +2C_{2} e^{2x}

    y'=2C_{1} x +2(y-C_{1}x^2)

    y'=2C_{1} x +2y-2C_{1}x^2

    y'=2C_{1} x(1-x) +2y          equation 4

    2. Use equation 2, then substitute to equation 3

    y'=2C_{1} x +2C_{2} e^{2x}

    2C_{2} e^{2x}=y'-2C_{1} x

    y"=2C_{1}  +4C_{2} e^{2x}

    y"=2C_{1}  +2(y'-2C_{1} x)

    y"=2C_{1}  +2y'-4C_{1} x

    y"=2C_{1}(1-2x)  +2y'          equation 5

    3. Solve for C_{1} using equation 5, then substitute to equation 4

    y"=2C_{1}(1-2x)  +2y'

    y"-2y'=2C_{1}(1-2x)             divide both side by 2(1-2x)

    C_{1} =\frac{y"-2y'}{2(1-2x)}                            substitute to equation 4

    y'=2C_{1} x(1-x) +2y

    y'=2\frac{y"-2y'}{2(1-2x)}  x(1-x) +2y

    y'-2y=\frac{y"-2y'}{1-2x}  x(1-x)                            cross-multiply


    Get the product of both sides of the equation


    Simplify by combining like terms

    3y' - 4xy' + 4xy - 2y - y" - xy" = 0

    Therefore the final answer is 3y' - 4xy' + 4xy - 2y - y" - xy" = 0

    To learn more about arbitrary constants, just click on the following links:

    Definition of an arbitrary constant


    Additional example




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Ii. eliminate the arbitrary of con1. y = cx+c²+1​...