simple question about autonomous first-order DEs

sbankica

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
Have you solved it yet? What did you get?

chisigma

MHF Hall of Honor
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

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

Last edited:
zzzoak