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PAGE 1

Exam 1 Review

Direction Field

General form: \( y' = f(t, y) \)

  • Pick \( (t, y) \), graph \( y' \)
  • If no \( t \) in \( f(t, y) \), same slope across
  • If it contains \( t \), best to find out where \( y' = 0 \), then above and below
\[ y' = t^2 - y \quad \rightarrow \quad y' = 0 \implies y = t^2 \]
Direction field graph with t and y axes, showing slope segments and a red parabola curve y = t^2 .

\( y > t^2 \) (above)

\( y' < 0 \)

\( y < t^2 \) (below)

\( y' > 0 \)

PAGE 2

Types of Diff. Eqs.

1st-order linear

\[ y' + p(t)y = g(t) \]

Where \( p(t) \) and \( g(t) \) are functions of \( t \) or constants.

Solution: integrating factor \( \mu = e^{\int p(t) dt} \)

Multiply both sides of diff. eq.

Left side becomes \( \frac{d}{dt}(\mu y) \)
\[ \frac{d}{dt}(\mu y) = \mu g \quad \text{then integrate and solve} \]

Example:

\[ ty' + 4y = 12t^2 \] \[ y' + \left( \frac{4}{t} \right) y = 12t \]

Here, \( p(t) = \frac{4}{t} \)

\[ \mu = e^{\int \frac{4}{t} dt} = e^{4 \ln t} = t^4 \] \[ \frac{d}{dt}(t^4 y) = 12t^5 \] \[ t^4 y = 2t^6 + C \] \[ y = 2t^2 + \frac{C}{t^4} \]
PAGE 3

Separable

\[ \frac{dy}{dx} = M(x)N(y) \]

Solution:

\[ \frac{1}{N(y)} dy = M(x) dx \]

integrate and solve

Example:

\[ \frac{dy}{dx} = xy + x = x(y + 1) \]\[ \frac{1}{y + 1} dy = x dx \]\[ \ln |y + 1| = \frac{1}{2}x^2 + C \]\[ |y + 1| = e^C \cdot e^{\frac{1}{2}x^2} \]\[ y + 1 = C e^{\frac{1}{2}x^2} \]\[ y = C e^{\frac{1}{2}x^2} - 1 \]
PAGE 4

Homogeneous

\[ \frac{dy}{dx} = f(x, y) = g\left(\frac{y}{x}\right) \]

Solution: make sub \( v = \frac{y}{x} \)

  • rewrite eq. as one involving \( v \) and \( x \)
  • (could be separable or linear or others)
  • solve for \( v \), then \( y \)

Example:

\[ \frac{dy}{dx} = \frac{3x - y}{x + y} \]\[ = \frac{3 - (\frac{y}{x})}{1 + (\frac{y}{x})} = g\left(\frac{y}{x}\right) \]
\[ v = \frac{y}{x} \]
\[ y = vx \]\[ \frac{dy}{dx} = v + v'x \]
\[ v + v'x = \frac{3 - v}{1 + v} \]no y!
PAGE 5

Solving Separable Differential Equations

\[ x v' = \frac{3-v}{1+v} - \frac{v+v^2}{1+v} \]
\[ x v' = \frac{3-2v-v^2}{1+v} \quad \text{Separable} \]
\[ \frac{1+v}{v^2+2v-3} dv = -\frac{1}{x} dx \]

integrate, find \( v \), then \( y \).

Let \( u = v^2+2v-3 \)

\( du = (2v+2) dv \)

\( = 2(v+1) dv \)

\[ \frac{1}{2} \int \frac{1}{u} du = -\int \frac{1}{x} dx \]

\( \vdots \)

PAGE 6

Exact Differential Equations

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

or

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

such that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) (or \( M_y = N_x \))

Solution: \( \psi(x,y) = C \)

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

Example

\[ (e^x \sin y - 2y \sin x - 1) dx + (e^x \cos y + 2 \cos x + 1) dy = 0 \]

Where \( M = e^x \sin y - 2y \sin x - 1 \) and \( N = e^x \cos y + 2 \cos x + 1 \)

\[ M_y = e^x \cos y - 2 \sin x \]
\[ N_x = e^x \cos y - 2 \sin x \]

so exact

\[ \psi_x = e^x \sin y - 2y \sin x - 1 \]
\[ \psi_y = e^x \cos y + 2 \cos x + 1 \]
PAGE 7

Solving for the Potential Function

Pick one to integrate:

\[ \psi_y = e^x \cos y + 2 \cos x + 1 \]
\[ \psi = \int (e^x \cos y + 2 \cos x + 1) dy \]

Where \( x \) is treated as a constant during integration.

\[ = e^x \sin y + 2y \cos x + y + h(x) \]

Now, differentiate with respect to \( x \) to find \( h(x) \):

\[ \psi_x = e^x \sin y - 2y \sin x + h'(x) = M = e^x \sin y - 2y \sin x - 1 \]

So, \( h'(x) = -1 \implies h(x) = -x \).

The potential function is:

\[ \psi(x, y) = e^x \sin y + 2y \cos x + y - x \]

Solution: \( \psi = C \)

No exact integrating factor on exam.

PAGE 8

Equilibrium and Stability

\[ y' = f(t, y) \]
  • Autonomous: \( y' = f(y) \)
  • Equilibrium/Critical pt: \( f(y) = 0 \) (where \( y' = 0 \))
  • Stable: solutions nearby converge onto it
  • Unstable: solutions nearby run away from it
  • Semi-stable: some converge onto it, others run away

Example

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

Setting \( y' = 0 \implies y = 0, y = 1, y = 2 \).

Testing signs of \( y' \):

\[ y' \quad + \quad 0 \quad + \quad 0 \quad - \quad 0 \quad + \]
A horizontal phase line with critical points at 0, 1, and 2. Arrows indicate flow direction.
  • \( y = 0 \): Semi-stable
  • \( y = 1 \): Stable
  • \( y = 2 \): Unstable
PAGE 9

Existence and Uniqueness

1st-order linear

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

\( p, g \) continuous and contain \( t_0 \)

1st-order nonlinear

\[ y' = f(t, y) \]

\( f \) and \( \frac{\partial f}{\partial y} \) are continuous containing \( (t_0, y_0) \)

Euler's method

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

decide \( h \) (step size)

\[ \begin{aligned} &t_0 \\ &y_0 \\ &t_1 = t_0 + h \\ &y_1 = y_0 + f(t_0, y_0)h \\ &\vdots \\ &y_n = y_{n-1} + f(t_{n-1}, y_{n-1})h \end{aligned} \]
PAGE 10

2nd-order homogeneous constant-coefficient

\[ ay'' + by' + cy = 0 \]

characteristic eq. \( ar^2 + br + c = 0 \)

on exam: only distinct real roots

(no complex, no repeated)

solutions:

\[ \begin{aligned} y_1 &= e^{r_1 t} \\ y_2 &= e^{r_2 t} \end{aligned} \]

general solution: \( y = c_1 y_1 + c_2 y_2 \)

no Wronskian on the exam