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Math Help - separable differential equation xy'+y=y^2 when y(1)=-1

  1. #1
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    separable differential equation xy'+y=y^2 when y(1)=-1

    The question is to find the solution for
    xy'+y=Y^2 for when y(1)=-1

    here is my working

    <br />
\begin{array}{l}<br />
 xy' + y = y^2  \\ <br />
 y(1) =  - 1 \\ <br />
 x\frac{{dy}}{{dx}} = y^2  - y \\ <br />
 \int {\frac{{dy}}{{y^2  - y}} = \int {\frac{{dx}}{x}} }  \\ <br />
 \int {\frac{1}{{y^2  - y}}dy = \int {\frac{1}{x}dx} }  \\ <br />
 \end{array}<br />

    then by partial fractions the left hand side is
    <br />
\begin{array}{l}<br />
 LHS = \int {\frac{1}{{y^2  - y}}dy}  \\ <br />
 \frac{1}{{y^2  - y}} = \frac{A}{y} + \frac{B}{{(y - 1)}} \\ <br />
 1 = (A + B)y - A \\ <br />
 A =  - 1 \\ <br />
 A + B = 0 \\ <br />
 B = 1 \\ <br />
 \int {\frac{1}{{y^2  - y}}dy}  = \int {\frac{{ - 1}}{y}dy + \int {\frac{1}{{y - 1}}dy} }  \\ <br />
  =  - \ln y + \ln |y - 1| \\ <br />
 \end{array}<br />

    and right hand side is
    <br />
\begin{array}{l}<br />
 RHS = \int {\frac{1}{x}dx}  \\ <br />
  = \ln x + c \\ <br />
 \end{array}<br />

    so
    <br />
 - \ln y + \ln |y - 1| = \ln x + c<br />

    raising by power of e

    <br />
\begin{array}{l}<br />
  - y + y - 1 = x + e^c  \\ <br />
  - 1 = x + e^c  \\ <br />
 \end{array}<br />

    But this does not appear to me to be right, as the y term has fallen off all together, and I don't have the answer for this problem to check, plus am confused where the y(1)=-1 comes in.
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  2. #2
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    Quote Originally Posted by Craka View Post
    The question is to find the solution for
    xy'+y=Y^2 for when y(1)=-1

    here is my working

    <br />
\begin{array}{l}<br />
xy' + y = y^2 \\ <br />
y(1) = - 1 \\ <br />
x\frac{{dy}}{{dx}} = y^2 - y \\ <br />
\int {\frac{{dy}}{{y^2 - y}} = \int {\frac{{dx}}{x}} } \\ <br />
\int {\frac{1}{{y^2 - y}}dy = \int {\frac{1}{x}dx} } \\ <br />
\end{array}<br />

    then by partial fractions the left hand side is
    <br />
\begin{array}{l}<br />
LHS = \int {\frac{1}{{y^2 - y}}dy} \\ <br />
\frac{1}{{y^2 - y}} = \frac{A}{y} + \frac{B}{{(y - 1)}} \\ <br />
1 = (A + B)y - A \\ <br />
A = - 1 \\ <br />
A + B = 0 \\ <br />
B = 1 \\ <br />
\int {\frac{1}{{y^2 - y}}dy} = \int {\frac{{ - 1}}{y}dy + \int {\frac{1}{{y - 1}}dy} } \\ <br />
= - \ln y + \ln |y - 1| \\ <br />
\end{array}<br />

    and right hand side is
    <br />
\begin{array}{l}<br />
RHS = \int {\frac{1}{x}dx} \\ <br />
= \ln x + c \\ <br />
\end{array}<br />

    so
    <br />
- \ln y + \ln |y - 1| = \ln x + c<br />

    raising by power of e

    Mr F says: Below is where your error lies (assuming previous working is correct - I have not checked thoroughly).

    <br />
\begin{array}{l}<br />
- y + y - 1 = x + e^c \\ <br />
- 1 = x + e^c \\ <br />
\end{array}<br />

    But this does not appear to me to be right, as the y term has fallen off all together, and I don't have the answer for this problem to check, plus am confused where the y(1)=-1 comes in.
    - \ln |y| + \ln |y - 1| = \ln |x| + c

    \Rightarrow \ln \left| \frac{y - 1}{y} \right| = \ln |x| + c

    Exponentiate both sides to base e:

    \frac{y - 1}{y} = x e^c = Ax.

    Now your job is to make y the subject.
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  3. #3
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    Doh, thanks. Why can I never remember my log rules!!!
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  4. #4
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    an easier way would to use Bernoulli differentiation
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  5. #5
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    Even easier is to notice that

    xy' + y = y^2 becomes \frac{d}{dx} xy = y^2 so \frac{d (xy)}{(xy)^2} = \frac{dx}{x^2} so  \frac{1}{xy} = \frac{1}{x} + c
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  6. #6
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    idont think that is correct??
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  7. #7
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    Quote Originally Posted by mpl06c View Post
    idont think that is correct??
    1. You should quote the post you're referring to so that what you say can be understood.

    2. If you're going to say something like this, you should back it up with a reason.

    3. All forms of the solution given in this thread are correct, as can be easily confirmed by substitution into the given DE.
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