# simple question about autonomous first-order DEs

• May 12th 2010, 02:01 PM
sbankica
simple question about autonomous first-order DEs
consider the autonomous first-order equation $dy/dx = y - y^3$. and the initial condition $y(0)=y_0$. How do you go about drawing the graph of a typical solution $y(x)$ when
a) $y_0>1$
b) $0
• May 12th 2010, 03:00 PM
pickslides
Have you solved it yet? What did you get?
• May 13th 2010, 12:39 AM
chisigma
The DE is of 'separable variables type' and can be written as...

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

Because is...

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

... is also...

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

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

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

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

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

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

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

... and repeat the procedure obtaining...

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

Kind regards

$\chi$ $\sigma$