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Math Help - Simple Calc Question - don't know how to start

  1. #1
    Junior Member eddie2042's Avatar
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    Simple Calc Question - don't know how to start

    Find y as a function of x if

    \frac{dy}{dx} = y(2x+1)

    and y = -3 when x = 0.

    Any idea on where to get started? I have no clue...
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by eddie2042 View Post
    Find y as a function of x if

    \frac{dy}{dx} = y(2x+1)

    and y = -3 when x = 3.

    Any idea on where to get started? I have no clue...
    This is a first order linear differential equation.

    Since it is separable, integrate on both sides to find y, that is,

    \int \frac{dy}{y} = \int (2x+1) dx
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  3. #3
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by harish21 View Post
    This is a first order linear differential equation.

    Since it is separable, integrate on both sides to find y, that is,

    \int \frac{dy}{y} = \int (2x+1) dx
    thanks though i must apologize. I made an edit on the first post. I said that y = -3 when x  = 3. It should be when x = 0.

    Thanks; i'll see if it works.
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by eddie2042 View Post
    thanks though i must apologize. I made an edit on the first post. I said that y = -3 when x  = 3. It should be when x = 0.

    Thanks; i'll see if it works.
    y=-3 when x=0 will give you the value of the constant C that is in your function. Integrate first to find an equation of y as a function of x. then plug in the given values of y and x, which will give you the value of the constant.
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  5. #5
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by harish21 View Post
    y=-3 when x=0 will give you the value of the constant C that is in your function. Integrate first to find an equation of y as a function of x. then plug in the given values of y and x, which will give you the value of the constant.
    Ok here's what i did so far. Tell me if i went wrong somewhere

    \int\frac{dy}{d} = \int(2x +1)dx

    ln |y| +C = x^2 + x + C

    ... then what..? The C's cancel out.. if i'm not mistaken..
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by eddie2042 View Post
    Ok here's what i did so far. Tell me if i went wrong somewhere

    \int\frac{dy}{d} = \int(2x +1)dx

    ln |y| +C = x^2 + x + C

    ... then what..? The C's cancel out.. if i'm not mistaken..
    The Cs are arbitrary constants, and when you do the integral of both the left/right side it is sufficient to have only one constant variable (i.e. only one C) either on the left side or right side of your equation.

    To obtain y, put everything up to the e. This will bring down y due to the rule of logs and that is your answer (assuming you've done the rest correct)
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by eddie2042 View Post
    Ok here's what i did so far. Tell me if i went wrong somewhere

    \int\frac{dy}{d} = \int(2x +1)dx

    ln |y| +C = x^2 + x + C

    ... then what..? The C's cancel out.. if i'm not mistaken..
    NO. Thats a bit too impulsive.

    The value of constant is not same on both the sides. if you have written C on the left side, then express the constant on the right hand side with another letter. (say C'). After that, bring all the constants on the same side. this gives you

    ln |y| = x^2 + x + (C'-C)

    note that C'-C, the difference of two constants, is also a constant. For convenience, denote it with another letter.

    then find y
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  8. #8
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by harish21 View Post
    NO. Thats a bit too impulsive.

    The value of constant is not same on both the sides. if you have written C on the left side, then express the constant on the right hand side with another letter. (say C'). The value of constant is NOT the same. After that, bring all the constants on the same side. this gives you

    ln |y| = x^2 + x + (C'-C)

    note that C'-C, the difference of two constants, is also a constant. For convenience, denote it with another letter.

    then find y

    ok great; thanks you two!
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  9. #9
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by eddie2042 View Post
    ok great; thanks you two!
    Hope you can finish up the problem now. Reply back if you have any problem(s).
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