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Thread: Solve the initial value problem:

  1. #1
    Newbie andvaka's Avatar
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    Question Solve the initial value problem:

    (dy/dx) = (y+1(x-1)) ; y(1)=3 variable separable

    We've only recently began the unit on Integration applications and Differentials, so I have absolutely no idea how to approach this. If any of you can show me how to solve this, it would be most appreciated.
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    Re: Solve the initial value problem:

    Quote Originally Posted by andvaka View Post
    (dy/dx) = (y+1(x-1)) ; y(1)=3 variable separable

    We've only recently began the unit on Integration applications and Differentials, so I have absolutely no idea how to approach this. If any of you can show me how to solve this, it would be most appreciated.
    Separating the variables we get dy/(y+1)=(x+1)dx Integrate both sides log (y+1)= x^2/2+x+C
    When x=1 y=3 so log4=(1/2)+1+C So C=log4-3/2 etc.
    Thanks from andvaka
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    Newbie andvaka's Avatar
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    Re: Solve the initial value problem:

    Quote Originally Posted by biffboy View Post
    Separating the variables we get dy/(y+1)=(x+1)dx Integrate both sides log (y+1)= x^2/2+x+C
    When x=1 y=3 so log4=(1/2)+1+C So C=log4-3/2 etc.
    Shouldn't the (x+1)dx in "dy/(y+1)=(x+1)dx" be (x-1)dx . or is there a reason for the sign change?
    Thanks.
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    Re: Solve the initial value problem:

    Quote Originally Posted by andvaka View Post
    (dy/dx) = (y+1(x-1)) ; y(1)=3 variable separable

    We've only recently began the unit on Integration applications and Differentials, so I have absolutely no idea how to approach this. If any of you can show me how to solve this, it would be most appreciated.
    if the DE is

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

    then

    \int \frac{dy}{y+1} = \int (x-1) \, dx

    ... most probably a typo by biff to put (x+1)
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  5. #5
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    Re: Solve the initial value problem:

    The typo was not by biffboy. Andvaka originally wrote "(dy/dx) = (y+1(x-1))" which is NOT "(dy/dx)= (y+1)(x-1)".
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  6. #6
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    Re: Solve the initial value problem:

    If the original function actually is as written, then it is \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}, which would need to be solved using an Integrating Factor...
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  7. #7
    Newbie andvaka's Avatar
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    Re: Solve the initial value problem:

    Quote Originally Posted by Prove It View Post
    If the original function actually is as written, then it is \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}, which would need to be solved using an Integrating Factor...
    Would you mind elaborating?
    Thanks!
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  8. #8
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    Re: Solve the initial value problem:

    Quote Originally Posted by andvaka View Post
    Would you mind elaborating?
    Thanks!
    I would, considering I have no idea if this is the actual problem. As the other posters have stated, it appears that you had tried to write \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} with a typo, thereby making an entirely different problem. The reason it looks wrong is because you would not usually write \displaystyle \begin{align*} 1(x - 1) \end{align*}.

    Which is the correct problem? \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} or \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}?
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  9. #9
    Newbie andvaka's Avatar
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    Re: Solve the initial value problem:

    Quote Originally Posted by Prove It View Post
    I would, considering I have no idea if this is the actual problem. As the other posters have stated, it appears that you had tried to write \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} with a typo, thereby making an entirely different problem. The reason it looks wrong is because you would not usually write \displaystyle \begin{align*} 1(x - 1) \end{align*}.

    Which is the correct problem? \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} or \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}?
    Ah. It was a typo. The first is correct.
    \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*}
    Last edited by andvaka; Jun 24th 2012 at 09:12 PM.
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  10. #10
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    Re: Solve the initial value problem:

    Sorry. Yes it should be (x-1)
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