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PAGE 1

2.5 (Continued)

asymptotically stable: neighboring solutions converge onto this equilibrium

unstable: solutions run away from this

Graph of y versus t showing solution curves converging to a stable equilibrium and diverging from an unstable one.

another type

\[ \frac{dy}{dt} = y(y-1)^2 = f(y) \]

equilibrium / critical point: \( f(y) = 0 \)

here, \( y = 0, y = 1 \)

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phase line:

Phase line diagram for y showing signs of f(y) around critical points y=0 and y=1.
Direction field and solution curves on a y-t graph corresponding to the phase line.

semi-stable: stable from one side only

example

\[ y' = y^2(y-1)(y+2)(y^2-9) \]

critical pts: \( 0, 1, -2, -3, 3 \)

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Sign of \( f(y) = y^2(y-1)(y+2)(y^2-9) \)

A phase line for the variable y with critical points marked at -3, -2, 0, 1, and 3. Arrows indicate the direction of flow between these points.
  • -3 : asymp stable
  • -2 : unstable
  • 0 : semi-stable
  • 1 : stable
  • 3 : unstable
A coordinate graph with y on the vertical axis and t on the horizontal axis. Horizontal dashed lines represent equilibrium solutions at y = 3, 1, 0, -2, and -3, with green solution curves showing stability behavior.
PAGE 4

2.6 Exact Diff. Eqs.

from calculus, we saw level curves \( \psi(x, y) = C \)

implicit function of x
A small coordinate graph showing a curve labeled psi(x,y)=C on x and y axes.

differentiate \( \psi(x, y) \) with respect to x

\[ \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \]

let \( \frac{\partial \psi}{\partial x}(x, y) = M(x, y) \) and \( \frac{\partial \psi}{\partial y}(x, y) = N(x, y) \)

then we have

\[ M(x, y) + N(x, y) y' = 0 \]

if \( \psi \) has at least up to 2nd-order derivatives being continuous, then \( \psi_{xy} = \psi_{yx} \)

\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad \text{or} \quad M_y = N_x \]
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Exact Differential Equations

A differential equation of the form

\[ M(x, y) + N(x, y) y' = 0 \text{ such that } M_y = N_x \]

is called an exact diff. eq.

Sometimes written in differential form as \( M dx + N dy = 0 \).

So, there must exist a function \( \psi(x, y) = c \) which we can use as an implicit solution.

Example

\[ (3x^2 + 2y^2) + (4xy + 6y^2) y' = 0 \]

Term M

\( M = 3x^2 + 2y^2 \)

Attached to \( dx \) or nothing in the "usual" form.

Term N

\( N = 4xy + 6y^2 \)

Attached to \( y' \) or \( dy \) in differential form.

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It looks exact, but is it?

Check: \( M_y = N_x \)

  • If so, it is exact.
  • If not, it might be or not.
\[ M = 3x^2 + 2y^2 \]\[ N = 4xy + 6y^2 \]
\[ M_y = 4y \]\[ N_x = 4y \]

Equal, so is exact.

So there is a \( \psi(x, y) \) such that:

\[ \psi_x = M \]\[ \psi_y = N \]
\[ \psi_x = 3x^2 + 2y^2 \]\[ \psi_y = 4xy + 6y^2 \]
Recover \( \psi \) by integration.
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Integration with Respect to x

Let's integrate \(\Psi_x\) with respect to \(x\):

\[\Psi(x,y) = \int \Psi_x \, dx = \int M \, dx = \int (3x^2 + 2y^2) \, dx\]

Variable is \(x\), so \(y\) is "constant".

\[\Psi(x,y) = x^3 + 2y^2x + h(y)\]

\(h(y)\) disappears when \(\frac{\partial}{\partial x}\) and no trace of it shows up in \(M\), so it could be a constant but more generally a function of \(y\).

Whatever \(h(y)\) is, the partial of \(\Psi(x,y) = x^3 + 2y^2x + h(y)\) must be equal to \(N\) because \(N = \Psi_y\).

\[\frac{\partial}{\partial y} \Psi = \frac{\partial}{\partial y} (x^3 + 2y^2x + h(y)) = 4xy + 6y^2\]
\[4yx + h'(y) = 4xy + 6y^2 \quad \text{(given } N\text{)}\]

So, \(h'(y) = 6y^2\). Integrating gives \(h(y) = 2y^3 + C_1\).

Choose \(C_1 = 0\) usually.

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So, \(\Psi(x,y) = x^3 + 2xy^2 + 2y^3 + C_1\).

The solution is implicitly defined as \(\Psi(x,y) = C\).

So we get:

\[x^3 + 2xy^2 + 2y^3 = C\]

Separable Equations and Exactness

Many separable equations are exact:

\[\frac{dy}{dx} = \frac{x}{y} \rightarrow y \frac{dy}{dx} = x \rightarrow x - y y' = 0\]
  • \(M = x\)
  • \(N = -y\)
  • \(M_y = 0\), \(N_x = 0\)

But not all separable equations are exact:

\[\frac{dy}{dx} = \frac{y}{x} \rightarrow y - xy' = 0\]
  • \(M = y\)
  • \(N = -x\)
  • \(M_y = 1\), \(N_x = -1\)

NOT exact