# Thread: simple question about autonomous first-order DEs

1. ## simple question about autonomous first-order DEs

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)$\displaystyle 0<y_0<1$

2. Have you solved it yet? What did you get?

3. 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

$\displaystyle \chi$ $\displaystyle \sigma$