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Math Help - Diffrential equations...can't seem to get answer =(.

  1. #1
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    Diffrential equations...can't seem to get answer =(.

    There are two questions, maybe im doing the same mistake or not doing something in both? Idk...

    Answer = <br />
y = 3 - \frac{4}{x^2}<br />
    Question = Solve for y in terms of x, the differential equation given that x = 2, when y = 2
    x \frac{dy}{dx} = 2(3 - y)
    I dunno what im doing wrong .
    x \frac{dy}{dx} = 2(3 - y)
    So, take over stuff and integrate both sides.
    \int\frac{1}{3-y} = \int\frac{2}{x}
    Gives:
    \ln {3-y} = 2\ln {x} + c
    Rearrange gives y = 3 - x^2 + c
    Subbing in x = 2 and y = 2 i get c = -3
    So my answer is y = 3 - x^2 - 3
    or y = x^2

    Which is wrrrrooooonnng lol ..

    Question 2:
    Given that y = 1 when x = 0, express y in terms of x if
    \frac{dy}{dx} = \frac{e^x}{e^y}
    Now when i take stuff to side and integrate i get
    e^y = e^ x + c
    And therefore y = x + c
    Subbing in values gives c = 1
    And i get the answer y = x + 1...which fits the co-ordinates but its WRONG. .

    The answer is : y = \ln(e^x + e - 1)
    I have no idea how they get that.

    Any help would be really appreciated!
    Thank-you very much in advance for anything you can give!
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  2. #2
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    For #1) when you integrate 1/(3-y) you forgot to add in a negative sign. When you take the derivative of ln(3-y) by using the chain rule you'll multiply the whole thing by -1 because of the -y, thus when integrating you must compensate for this. That negative applied to 2ln(x) makes the exponent x^(-2) instead of x^2.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by AshleyT View Post
    Question 2:
    Given that y = 1 when x = 0, express y in terms of x if
    \frac{dy}{dx} = \frac{e^x}{e^y}
    Now when i take stuff to side and integrate i get
    e^y = e^ x + c
    And therefore y = x + c
    Subbing in values gives c = 1
    And i get the answer y = x + 1...which fits the co-ordinates but its WRONG. .

    The answer is : y = \ln(e^x + e - 1)
    I have no idea how they get that.

    Any help would be really appreciated!
    Thank-you very much in advance for anything you can give!
    \begin{aligned}\frac{dy}{dx}&=\frac{e^x}{e^y}\\<br />
&\implies{e^ydy=e^xdx}\\<br />
&\implies{e^y=e^x+c}\\<br />
&\implies\ln(e^y)=y=\ln\left(e^x+c\right)<br />
\end{aligned}

    So know we know that

    1=\ln\left(e^0+c\right)=\ln(1+c)

    We know \ln(x)=1 when x=e. So 1+c=e\Rightarrow{c=e-1}

    \therefore{y=\ln\left(e^x+e-1\right)}
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  4. #4
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    Ooo.
    Thankyou very much for both answers and speedy replies! This is greatly appreciated!!
    Thank-you so much!
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  5. #5
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    Smile

    Diffrential equations...can't seem to get answer =(.-untitled.jpg
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  6. #6
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    Quote Originally Posted by Math_Helper View Post
    Click image for larger version. 

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    Cheers . Ima see if i can have another go though before looking at your solutions .
    p.s I might be going to Calgary next year!
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