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Thread: need someone to solve this question

  1. #1
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    need someone to solve this question

    2yy' +3 = y^2 + 3x with initial value y(0) = 8
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    Re: need someone to solve this question

    Quote Originally Posted by mmiraj View Post
    2yy' +3 = y^2 + 3x with initial value y(0) = 8
    One way to start this is to note that \left ( y^2 \right ) ' = 2y~y', so your equation becomes:
    \left ( y^2 \right )' + 3 = y^2 + 3x

    Now, y^2 is just a function, let's call it z = y^2, turning your equation into
    z' + 3 = z + 3x

    Can you finish from here?

    -Dan
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  3. #3
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    Re: need someone to solve this question

    How is 2yy' = (y^2) ???
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  4. #4
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    Re: need someone to solve this question

    Quote Originally Posted by mmiraj View Post
    How is 2yy' = (y^2) ???
    Its not.

    topsquark posted $2yy = (y^2)$ ... note the prime symbol after the parentheses, i.e. the derivative of $y^2$ is $2y \cdot y$
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  5. #5
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    Re: need someone to solve this question

    Quote Originally Posted by topsquark View Post
    One way to start this is to note that \left ( y^2 \right ) ' = 2y~y', so your equation becomes:
    \left ( y^2 \right )' + 3 = y^2 + 3x

    Now, y^2 is just a function, let's call it z = y^2, turning your equation into
    z' + 3 = z + 3x

    Can you finish from here?

    -Dan
    Let $u = y^2+3x$. Now you have $u' = u$.
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