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

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March 18th 2009, 03:32 AM
Craka

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

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

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

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

so
$ - \ln y + \ln |y - 1| = \ln x + c $

raising by power of e

$ \begin{array}{l} - y + y - 1 = x + e^c \\ - 1 = x + e^c \\ \end{array} $

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.
March 18th 2009, 04:12 AM
mr fantastic

Quote:

Originally Posted by Craka

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

here is my working

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

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

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

so
$ - \ln y + \ln |y - 1| = \ln x + c $

raising by power of e

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

$ \begin{array}{l} - y + y - 1 = x + e^c \\ - 1 = x + e^c \\ \end{array} $

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.
March 18th 2009, 04:23 AM
Craka

Doh, thanks. Why can I never remember my log rules!!!
March 18th 2009, 07:49 AM
mpl06c

an easier way would to use Bernoulli differentiation
March 18th 2009, 09:39 AM
Danny

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 $\frac{1}{xy} = \frac{1}{x} + c$
March 18th 2009, 06:04 PM
mpl06c

idont think that is correct??
March 18th 2009, 06:34 PM
mr fantastic

Quote:

Originally Posted by mpl06c

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.