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PAGE 1

Section 1.3 #4

\[ \frac{dy}{dt} + ty^2 = 0 \]

First order

Note: The exponent on \( y \) is 2 (not 1), making this equation nonlinear. (\( y^2 = y \cdot y \))

Section 1.2 #8

\[ \frac{dP}{dt} = rP \]\[ \frac{1}{P} \frac{dP}{dt} = r \]\[ \frac{d}{dt} \ln |P| = r \]\[ \ln P = rt + A \]\[ P(t) = e^{rt+A} = e^{rt} e^A = Ce^{rt} \]

Note: \( e^A \) is replaced by the constant \( C \).

\[ P(0) = P_0 \]\[ P_0 = C \]\[ P(t) = P_0 e^{rt} \]\[ 2P_0 = P_0 e^{30r} \]\[ \ln 2 = 30r \]

\[ r = \frac{\ln 2}{30} \]

\[ \approx 0.0231 \]\[ (2.31\%) \]

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Supplementary Problem A

\[ t^2 y'' - 4ty' + 4y = 0 \]

Check if \( y = Be^{-2t} \) is a solution for the differential equation \( 2y' + 4y = 3e^{-2t} \):

\[ y = Be^{-2t} \]

\[ y' = -2Be^{-2t} \]

Substituting into the equation:

\[ 2(-2Be^{-2t}) + 4(Be^{-2t}) = 3e^{-2t} \]\[ -4Be^{-2t} + 4Be^{-2t} = 3e^{-2t} \]\[ 0 = 3e^{-2t} \]

Not true

No choice of \( B \) can make \( y = Be^{-2t} \) satisfy the DE \( \rightarrow \) so \( y = Be^{-2t} \) cannot be a solution of the DE.

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2.1 Linear equations, Integrating Factors

solve eqs of the form

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

first order linear DE

example:

\[ y' + \frac{2}{t}y = \frac{4}{t} \]

notice if we multiply both sides by \( t^2 \)

product rule
\[ t^2 y' + 2ty = 4t \] \[ \frac{d}{dt} (t^2 y) = 4t \]

integrate:

\[ t^2 y = 2t^2 + C \]
so
\[ y = 2 + Ct^{-2} \]
\( t \neq 0 \)

\( t^2 \) is called the integrating factor : different for every DE

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so what is the integrating factor for arbitrary

\[ y' + p(t)y = g(t) \, ? \]
\( \rightarrow \mu(t) \) (lowercase Greek "mu")

multiply by \( \mu(t) \)

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

want left side to be

\[ \frac{d}{dt} [\mu(t)y] = \mu(t)y' + \mu'(t)y \]
\[ \mu(t)y' + \mu(t)p(t)y = \mu(t)y' + \mu'(t)y \]
\[ \mu(t)p(t)y = \mu'(t)y \]
\( \hookrightarrow \)
\[ \frac{d\mu(t)}{dt} = \mu(t)p(t) \]
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Derivation of the Integrating Factor

Starting from the differential equation for the integrating factor \(\mu(t)\):

\[\frac{1}{\mu(t)} \frac{d\mu(t)}{dt} = p(t)\]

Using the chain rule for logarithms, we can rewrite the left side:

\[\frac{d}{dt} \ln |\mu(t)| = p(t)\]

Integrating both sides with respect to \(t\):

\[\ln |\mu(t)| = \int p(t) \, dt\]
\[\mu(t) = e^{\int p(t) \, dt}\]
Integrating Factor
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Example

\[y' - 2y = t^2 e^{2t}\]
\[y' + p(t)y = g(t)\]
"standard form"

Here, \(p(t) = -2\).

So, the integrating factor is:

\[\mu(t) = e^{\int -2 \, dt} = e^{-2t + C} = e^{-2t}\]
Note: \(C=0\) in the integrating factor.

Multiply DE by \(\mu(t)\)

\[e^{-2t} y' - 2 e^{-2t} y = e^{-2t} t^2 e^{2t}\]

The right side simplifies because \(e^{-2t} e^{2t} = 1\).

Left side is always \(\frac{d}{dt} [\mu(t)y]\)

\[\frac{d}{dt} (\underbrace{e^{-2t}}_{\mu(t)} y) = t^2\]

Integrate

\[e^{-2t} y = \frac{t^3}{3} + C\]
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\[ y(t) = \frac{t^3}{3} e^{2t} + C e^{2t} \]

general solution \( (C = ?) \)

\( C \) depends on initial condition

example: \( y(0) = 1 \)

\[ 1 = 0 + C \quad \text{so} \quad C = 1 \]
\[ y(t) = \frac{t^3}{3} e^{2t} + e^{2t} \]

particular solution \( (C \text{ is known}) \)

what happens if \( t \to \infty \)?

\( y \to \infty \)

PAGE 8

example

\[ y' - \frac{1}{2} y = 2 \cos t \]

Note: In the equation above, \( -\frac{1}{2} \) is identified as \( p(t) \).

see slope field

depending on \( y(0) \),

  • \( y \to \infty \) as \( t \to \infty \) or
  • \( y \to -\infty \) as \( t \to \infty \) or
  • \( y \) is oscillatory as \( t \to \infty \)

looks like the critical \( y(0) \) is \( \approx -0.75 \)

let's find it.

\[ \mu(t) = e^{\int -\frac{1}{2} dt} = e^{-\frac{1}{2}t} \]
\[ e^{-\frac{1}{2}t} y' - \frac{1}{2} e^{-\frac{1}{2}t} y = 2 e^{-\frac{1}{2}t} \cos t \]
\[ \frac{d}{dt} (e^{-\frac{1}{2}t} y) = 2 e^{-\frac{1}{2}t} \cos t \]
PAGE 9

Direction Field Analysis

Source: http://comp.uark.edu/~aeb019/dfield.html | Date: 6/14/2016

Differential Equation

\[ y' - \frac{1}{2}y = 2 \cos t \]

The following direction field plotter visualization shows the slope field and several solution curves for the given first-order linear differential equation.

Direction field for y' - 0.5y = 2 cos(t) showing a grid of slope arrows and several oscillating solution curves.
Figure 1: Direction field plotter output with solution trajectories.
PAGE 10

Solving by Integration by Parts

LIATE Rule Applied

\[ e^{-\frac{1}{2}t} y = \int 2e^{-\frac{1}{2}t} \cos t \, dt \]

Integration by Parts

\[ u = \cos t \] \[ dv = 2e^{-\frac{1}{2}t} \, dt \] \[ du = -\sin t \, dt \] \[ v = -4e^{-\frac{1}{2}t} \]

Using the formula: \( uv - \int v \, du \)

\[ e^{-\frac{1}{2}t} y = -4e^{-\frac{1}{2}t} \cos t + \int 4e^{-\frac{1}{2}t} \sin t \, dt \]

By parts again...

⋮
\[ = -\frac{4}{5} e^{-\frac{1}{2}t} (\cos t - 2 \sin t) + C \]
PAGE 11

General Solution and Initial Value Analysis

\[ y = -\frac{4}{5} (\cos t - 2 \sin t) + C e^{\frac{1}{2} t} \]

Assume the initial condition:

\[ y(0) = a \]

Substituting the initial condition into the general solution:

\[ a = -\frac{4}{5} + C \quad \text{so} \quad C = a + \frac{4}{5} \]

The specific solution for the initial value is:

\[ y = -\frac{4}{5} (\cos t - 2 \sin t) + \left( a + \frac{4}{5} \right) e^{\frac{1}{2} t} \]

Oscillation Component

\[ -\frac{4}{5} (\cos t - 2 \sin t) \]

This part of the solution represents the oscillation.

Exponential Behavior

\[ \left( a + \frac{4}{5} \right) e^{\frac{1}{2} t} \]

  • Goes to \( \infty \) if \( a > -\frac{4}{5} \)
  • Goes to \( -\infty \) if \( a < -\frac{4}{5} \)
  • \( a = -\frac{4}{5} \) to have just oscillations