simple question about autonomous first-order DEs

Nov 2009
37
0
consider the autonomous first-order equation \(\displaystyle dy/dx = y - y^3\). and the initial condition \(\displaystyle y(0)=y_0\). How do you go about drawing the graph of a typical solution \(\displaystyle y(x)\) when
a)\(\displaystyle y_0>1\)
b)\(\displaystyle 0<y_0<1\)
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
Have you solved it yet? What did you get?
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
The DE is of 'separable variables type' and can be written as...

\(\displaystyle \frac{dy}{y\cdot (1-y^{2})} = dx\) (1)

Because is...

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

... is also...

\(\displaystyle \int \frac{dy}{y\cdot (1-y^{2})} = \ln y - \frac{1}{2} \cdot \ln (1-y^{2}) + \ln c\) (3)

... so that the solution of (1) is...

\(\displaystyle \frac{y}{\sqrt{1-y^{2}}} = c \cdot e^{x} \) (4)

... that can be 'explicitate' as function of x as...

\(\displaystyle y= \pm \frac{c\cdot e^{x}}{\sqrt{1+c^{2}\cdot e^{2x}}}\) (5)

If now we intend to impose the initial condition \(\displaystyle y(0)= y_{0}\) we 'discover' that we can do it only if is \(\displaystyle |y_{0}|<1\). To overcome the handycap in case \(\displaystyle |y_{0}|>1\) we write (2) as...

\(\displaystyle \frac{1}{y\cdot (1-y^{2})} = \frac{1}{y} - \frac{y}{y^{2}-1}\) (6)

... and repeat the procedure obtaining...

\(\displaystyle y= \pm \frac{c\cdot e^{x}}{\sqrt{c^{2}\cdot e^{2x}-1}}\) (7)

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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