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Thread: differential equation with y^2

  1. #1
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    differential equation with y^2

    Hello.

    I'm stuck at this differential equation:
    $\displaystyle x\cdot y'+y^{2}=1,$ $\displaystyle y(1)=0$

    Not even sure where to start, because everything I tried failed.

    Appreciate any help.
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  2. #2
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    Did you try this?

    $\displaystyle x y'+y^{2}=1$

    $\displaystyle x y'= 1-y^{2}$

    $\displaystyle \frac{dy}{dx}= \frac{1-y^{2}}{x}$

    $\displaystyle \frac{dy}{1-y^{2}}= \frac{dx}{x}$

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

    continue..
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Did you try this?

    $\displaystyle x y'+y^{2}=1$

    $\displaystyle x y'= 1-y^{2}$

    $\displaystyle \frac{dy}{dx}= \frac{1-y^{2}}{x}$

    $\displaystyle \frac{dy}{1-y^{2}}= \frac{dx}{x}$

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

    continue..
    Ahh thank you I'll try to continue from there. I started in a similar way but I I'm so tired I accidentally swapped the $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{dx}{dy}$
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  4. #4
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    I got stuck again.
    I got to:
    $\displaystyle \frac{1}{2}ln(y+1)-\frac{1}{2}ln(y-1)=lnx+C$
    How can I continue from here?

    I know $\displaystyle y(1)=0$ but inserting 1 will not work here..
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  5. #5
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    $\displaystyle \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1)$

    factor out $\displaystyle \frac{1}{2}$ then combine logs, let me know how you go.
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  6. #6
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    Quote Originally Posted by pickslides View Post
    $\displaystyle \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1)$

    factor out $\displaystyle \frac{1}{2}$ then combine logs, let me know how you go.
    Alright. I got here now, not sure if I made any mistakes though there are so many steps.

    $\displaystyle \frac{y+1}{y-1}=x^{2}+e^{2C}$

    $\displaystyle y=\frac{x^{2}+e^{2C}+1}{x^{2}+e^{2C}-1}$

    Is this correct? How can I use the $\displaystyle y(1)=0$ here?
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  7. #7
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    Quote Originally Posted by pumbster View Post
    Alright. I got here now, not sure if I made any mistakes though there are so many steps.

    $\displaystyle \frac{y+1}{y-1}=x^{2}+e^{2C}$

    $\displaystyle y=\frac{x^{2}+e^{2C}+1}{x^{2}+e^{2C}-1}$

    Is this correct? How can I use the $\displaystyle y(1)=0$ here?
    I solved it! I realized a made an error, it should of course be:
    $\displaystyle y=\frac{x^{2}e^{2C}+1}{x^{2}e^{2C}-1}$

    And I ended up with the answer:
    $\displaystyle y = \frac{x^2-1}{x^2+1}$

    Thank you so much for all the help! Really appreciated!
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  8. #8
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    Do you know if your answer is correct?

    I was thinking

    $\displaystyle \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1) = \ln x +C $

    $\displaystyle \frac{1}{2}\ln\frac{(y+1)}{(y-1)} = \ln x +\ln C $

    $\displaystyle \ln\sqrt{\frac{(y+1)}{(y-1)}} = \ln x +\ln C $

    $\displaystyle \ln\sqrt{\frac{(y+1)}{(y-1)}} = \ln (Cx) $

    $\displaystyle \sqrt{\frac{(y+1)}{(y-1)}} = Cx $

    $\displaystyle \frac{(y+1)}{(y-1)} = Cx^2 $

    $\displaystyle \frac{2}{(y-1)}+1 = Cx^2 $

    .
    .
    .
    .

    $\displaystyle y = \frac{2}{Cx^2-1}+1$

    Then use the initial condition $\displaystyle y(1)=0$ to find $\displaystyle C$
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