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PAGE 1

Exam 1 Review

Slope Field

\( y' = -2 + t - y \)

draw slopes for \( (t, y) \) pairs

nullcline: curve on which \( y' = 0 \)

\( -2 + t - y = 0 \implies y = t - 2 \)

Coordinate graph showing the nullcline y = t - 2 and slope field behavior
Figure: Coordinate graph showing the nullcline y = t - 2 and slope field behavior
Figure: A slope field diagram on the \( (t, y) \) plane. The dashed line represents the nullcline \( y = t - 2 \) where the slope segments are horizontal (\( y' = 0 \)). A curved solution trajectory is sketched crossing the \( t \)-axis and asymptotic to the nullcline. Slope segments above \( y = t - 2 \) point downward (\( y' < 0 \)), while slope segments below \( y = t - 2 \) point upward (\( y' > 0 \)).

\( y' = -2 + t - y \)

above \( y = t - 2 \)

  • \( \rightarrow y > t - 2 \)
  • \( y' = (t - 2) - y < 0 \)

below \( y = t - 2 \)

  • \( \rightarrow y < t - 2 \)
  • \( y' = (t - 2) - y > 0 \)
Key Takeaway: Along the nullcline \( y = t - 2 \), the slope is zero (\( y' = 0 \)). Above this line, trajectories have negative slopes (\( y' < 0 \)), and below it, trajectories have positive slopes (\( y' > 0 \)), causing solutions to tend toward the line.
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1st-order Eqs

linear:

\( y' + p(t)y = q(t) \)

Solved by multiplying by integrating factor \( \mu = e^{\int p(t)\,dt} \)

then integrate to solve

\[ \frac{d}{dt}\bigl[ \mu y \bigr] = \mu q \]

Example

\( t y' + 2y = 4t^2 \)

↑ coefficient must be 1 to read \( p(t) \) correctly

\( y' + \frac{2}{t}y = 4t \)

\( p(t) = \frac{2}{t} \quad \implies \quad \mu = e^{\int \frac{2}{t}\,dt} = e^{2\ln t} = t^2 \)
\[ \frac{d}{dt}\bigl[ t^2 \cdot y \bigr] = t^2 \cdot 4t = 4t^3 \]

Integrating both sides:

\( t^2 y = t^4 + c \)

\( y = t^2 + \frac{c}{t^2} \)

Separable:

can be written as \( f(y)\,dy = g(x)\,dx \)

Solved by integrating both sides

PAGE 3

Separable and Homogeneous Differential Equations

Separable Equation Example

\[ \frac{dy}{dx} = \frac{4x - x^3}{4 + y^3} \] \[ (4 + y^3) dy = (4x - x^3) dx \]

integrate

\[ 4y + \frac{1}{4} y^4 = 2x^2 - \frac{1}{4} x^4 + C \]
This is a separable first-order differential equation. Separating the variables allows direct integration of each side with respect to \( y \) and \( x \) respectively.

Homogeneous:

  • Can be written as \( y' = f\left(\frac{y}{x}\right) \)
  • Solved by making the subs \( v = \frac{y}{x} \)
  • Transform into linear/separable in \( v \) and \( x \)
  • Solve, go back to \( y \).

Example:

\[ \frac{dy}{dx} = \frac{y - 4x}{x - y} \]

let \( v = \frac{y}{x} \implies y = vx \)

\[ \frac{dy}{dx} = v + v'x \quad \text{(left)} \] \[ \frac{\frac{y - 4x}{x}}{\frac{x - y}{x}} = \frac{\frac{y}{x} - 4}{1 - \frac{y}{x}} = \frac{v - 4}{1 - v} \quad \text{(right)} \]
By substituting \( y = vx \) and applying the product rule to differentiate, \( \frac{dy}{dx} \) becomes \( v + x\frac{dv}{dx} \). Dividing both numerator and denominator of the right-hand side by \( x \) expresses it purely in terms of \( v \).
PAGE 4

Homogeneous Equation (Continued)

new eq:

\[ v + v'x = \frac{v - 4}{1 - v} \] \[ \begin{aligned} v'x &= \frac{v - 4}{1 - v} - v = \frac{v - 4}{1 - v} - \frac{v - v^2}{1 - v} \\[8pt] v'x &= \frac{-4 + v^2}{1 - v} \quad \text{separable} \end{aligned} \] \[ \frac{1 - v}{v^2 - 4} \, dv = \frac{1}{x} \, dx \]
Subtracting \( v \) from both sides and finding a common denominator isolates \( x\frac{dv}{dx} \), resulting in a separable ODE that can be integrated using partial fractions for the \( v \) term.

Exact:

Can be written as \( M(x, y)\,dx + N(x, y)\,dy = 0 \)

Such that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \)

Solution: \( f(x, y) = C \) where \( \frac{\partial f}{\partial x} = M \), \( \frac{\partial f}{\partial y} = N \)

Example:

\[ 2x + y^2 + 2xy\,y' = 0 \] \[ (2x + y^2)\,dx + (2xy)\,dy = 0 \]

where:

  • \( M = 2x + y^2 \)
  • \( N = 2xy \)
\[ \frac{\partial M}{\partial y} = 2y, \quad \frac{\partial N}{\partial x} = 2y \quad \text{exact} \]
Because the mixed partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) are equal, the differential form is exact, guaranteeing the existence of a potential function \( f(x, y) \) whose level curves \( f(x,y)=C \) define the general solution.
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Solving Exact Equations and Integrating Factors

Find \( f(x, y) \)

Given:

\[\begin{aligned}\frac{\partial f}{\partial x} &= M = 2x + y^2 \implies f(x, y) = x^2 + xy^2 + h(y) \\[1em]\frac{\partial f}{\partial y} &= N = 2xy\end{aligned}\]

Differentiating \( f(x, y) \) with respect to \( y \):

\[\frac{\partial f}{\partial y} = 2xy + \frac{dh}{dy}\]

Matching \( \frac{\partial f}{\partial y} \) with \( N = 2xy \):

\[2xy = 2xy + \frac{dh}{dy} \implies \frac{dh}{dy} = 0 \implies h = C\]

Solution is \( f(x, y) = C \):

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

Integrating Factor for Exact Equations

non-exact \( \rightarrow \) exact

Given the differential equation:

\[y + (2xy - e^{-2y})y' = 0\]

In differential form:

\[y\,dx + (2xy - e^{-2y})\,dy = 0\]

Checking exactness:

\[\frac{\partial}{\partial y}(y) \neq \frac{\partial}{\partial x}(2xy - e^{-2y})\]

where:

  • \( M = y \implies \frac{\partial M}{\partial y} = 1 \)
  • \( N = 2xy - e^{-2y} \implies \frac{\partial N}{\partial x} = 2y \)

Multiply by \( \mu \) to make exact

PAGE 6

Finding the Integrating Factor \( \mu(y) \)

Multiplying the equation by \( \mu \):

\[(\mu y)\,dx + \bigl[\mu(2xy - e^{-2y})\bigr]\,dy = 0\]

Choose \( \mu = \mu(x) \) or \( \mu = \mu(y) \).

Let's go with \( \mu = \mu(y) \):

We want exactness:

\[\frac{\partial}{\partial y}\bigl[\mu(y)y\bigr] = \frac{\partial}{\partial x}\bigl[\mu(y)(2xy - e^{-2y})\bigr]\]
\[\mu + \mu' y = \mu (2y)\]\[\mu' y = \mu (2y - 1)\]

(separable)

\[\frac{1}{\mu}\,d\mu = \frac{2y - 1}{y}\,dy\]\[\frac{1}{\mu}\,d\mu = \left(2 - \frac{1}{y}\right)dy\]

Integrating both sides: \( \ln|\mu| = 2y - \ln|y| \implies \mu = e^{2y - \ln y} = \frac{1}{y}e^{2y} \)

\[\mu = \frac{1}{y} e^{2y}\]

PAGE 7

Existence & Uniqueness of Solutions (1st-order)

Linear Equations

\[ y' + p(t)y = g(t), \quad y(t_0) = y_0 \]

Unique solution on interval containing \( t_0 \) on which BOTH \( p(t) \) and \( g(t) \) are continuous.

Theorem (Linear 1st-Order): If \( p \) and \( g \) are continuous on an open interval \( I = (\alpha, \beta) \) containing \( t_0 \), then there exists a unique solution \( y = \phi(t) \) on \( I \).

Nonlinear Equations

\[ y' = f(t, y), \quad y(t_0) = y_0 \]

Common interval containing \( t_0 \) where \( f(t, y) \) and \( \frac{\partial f}{\partial y} \) are continuous.

Example

\[ y' = (1 - t^2 - y^2)^{1/2} \]

where \( f(t, y) = (1 - t^2 - y^2)^{1/2} \).

\( f \) is continuous on \( t^2 + y^2 \le 1 \).

\[ \begin{aligned} \frac{\partial f}{\partial y} &= \frac{1}{2} (1 - t^2 - y^2)^{-1/2}(-2y) \\[8pt] &= \frac{-y}{(1 - t^2 - y^2)^{1/2}} \end{aligned} \]

Continuous on \( t^2 + y^2 < 1 \).

PAGE 8

Region of Existence and Uniqueness

Coordinate diagram of dashed open disk in the (t, y) plane
Figure: Coordinate diagram of dashed open disk in the (t, y) plane
Figure: A Cartesian coordinate system with horizontal axis \( t \) and vertical axis \( y \). A dashed circular boundary represents \( t^2 + y^2 = 1 \), indicating an open disk \( t^2 + y^2 < 1 \) where the condition for existence and uniqueness holds for initial points \( (t_0, y_0) \).

If \( (t_0, y_0) \) inside here,

then there is a unique solution on some interval of \( t \) inside.

Autonomous Eqs. and Stability

\[ y' = f(y) \text{ only} \]

\( y' = 0 \rightarrow \text{equilibrium solutions} \)

  • Unstable if nearby solutions diverge from it
  • Asymptotically stable if nearby solutions converge onto it
  • Semi-stable if unstable on one side and stable the other
PAGE 9

Stability Analysis of an Autonomous Differential Equation

\[ y' = (y + 1)(y - 1)^3 (y - 4)^2 \]

Equilibrium: \( y = -1 \), \( y = 1 \), \( y = 4 \)

Equilibrium solutions (or critical points) occur where \( y' = 0 \). Here, solving \( (y+1)(y-1)^3(y-4)^2 = 0 \) yields \( y = -1 \), \( y = 1 \), and \( y = 4 \).

Phase Line Analysis

Phase line showing signs of y' and arrows indicating stability of critical points at y = -1 , y = 1 , and y = 4 .
Figure: Phase line showing signs of \( y' \) and arrows indicating stability of critical points at \( y = -1 \), \( y = 1 \), and \( y = 4 \).
Figure: Phase line diagram depicting the sign of \( y' \) across intervals: positive for \( y < -1 \) (rightward arrow), zero at \( y = -1 \) (labeled as asymptotically stable), negative for \( -1 < y < 1 \) (leftward arrow), zero at \( y = 1 \) (labeled as unstable), positive for \( 1 < y < 4 \) (rightward arrow), zero at \( y = 4 \) (labeled as semi-stable), and positive for \( y > 4 \) (rightward arrow).

Solution Trajectories in the \((t, y)\)-Plane

Graph of solution curves y(t) versus t showing convergence to y = -1 , divergence from y = 1 , and semi-stable behavior at y = 4 .
Figure: Graph of solution curves \( y(t) \) versus \( t \) showing convergence to \( y = -1 \), divergence from \( y = 1 \), and semi-stable behavior at \( y = 4 \).
Figure: Plot in the \((t, y)\)-plane displaying horizontal equilibrium lines at \( y = 4 \), \( y = 1 \), and \( y = -1 \) (dashed lines). Solution curves with arrows illustrate behavior over time: trajectories above \( y = 4 \) and between \( y = 1 \) and \( y = 4 \) both increase toward or away from \( y = 4 \), curves between \( y = 1 \) and \( y = -1 \) decrease asymptotically toward \( y = -1 \), and curves below \( y = -1 \) increase toward \( y = -1 \).