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Problem I

Part a)

Show that \( y(t) = \int_{0}^{t} e^{-u^2} du \) is a solution to:

\[ \frac{dy}{dt} = e^{-t^2}, \quad y(0) = 0 \]

Verification of the derivative:

\[ y'(t) = \frac{d}{dt} \int_{0}^{t} e^{-u^2} du = e^{-t^2} \]

Verification of the Initial Condition (IC):

\[ y(0) = 0 \implies y(0) = \int_{0}^{0} e^{-u^2} du = 0 \]

Part b)

Evaluate \( y(2) = \int_{0}^{2} e^{-u^2} du \) by solving \( y' = e^{-t^2} \) using Euler's method with step size \( h = 1/2 \).

Start at \( t = 0 \), then calculate for \( t = 1/2, t = 1, t = 3/2, t = 2 \).

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Problem 20

Given the differential equation and initial condition:

\[ y' = 1 - t + y, \quad y(t_0) = y_0 \]

Part a)

Show that the exact solution is:

\[ y = (y_0 - t_0)e^{t-t_0} + t \]

Part b)

Show that the Euler update formula results in:

\[ y_k = (1+h)y_{k-1} + h - ht_{k-1} \]

Derivation

Euler update formula:

\[ y_{n+1} = y_n + f(t_n, y_n)h \]

Let \( n+1 = k \), then:

\[ y_k = y_{k-1} + f(t_{k-1}, y_{k-1})h \]

Given \( f(t, y) = 1 - t + y \), we have:

\[ f(t_{k-1}, y_{k-1}) = 1 - t_{k-1} + y_{k-1} \]

Substituting back into the update formula:

\[ y_k = y_{k-1} + (1 - t_{k-1} + y_{k-1})h \]\[ y_k = y_{k-1} + h - ht_{k-1} + hy_{k-1} \]\[ y_k = (1+h)y_{k-1} + h - ht_{k-1} \]
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d).

\[ y_n = (1+h)^n (y_0 - t_0) + t_n \]

let \( h = \frac{t - t_0}{n} \)

\[ y_n = \left( 1 + \frac{t - t_0}{n} \right)^n (y_0 - t_0) + t_n \]
\[ \lim_{n \to \infty} y_n = \left[ \lim_{n \to \infty} \left( 1 + \frac{t - t_0}{n} \right)^n \right] (y_0 - t_0) + t_n \]
\[ y(t) = e^{t - t_0} (y_0 - t_0) + t \]
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3.1 2nd Order Homogeneous Eqs w/ Constant Coefficients

last lesson on exam 1.

2nd order linear: \( y'' + p(t)y' + q(t)y = g(t) \)

if \( g(t) = 0 \) for all \( t \), then equation is homogeneous

does not mean function of \( \frac{y}{x} \) as in first order

today, focus on:

\[ \left. \begin{aligned} p(t) &= a \\ q(t) &= b \end{aligned} \right\} \text{ constants.} \]
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What kind of solutions?

Simple one:

\[ y'' - y = 0 \rightarrow y'' = y \]

function equal to its own second derivative

\[ y = e^t, \quad y = e^{-t} \]

or any constant-multiple of these

Another one:

\[ y'' - y' = 0 \rightarrow y'' = y' \]
\[ y = e^t, \quad y = 1 = e^{0t} \]

note: solutions are in the form of \( y = e^{rt} \)

\( r = \text{constant} \)

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Example

\[ y'' + y' - 2y = 0 \]

assume solution \( y = e^{rt} \), \( y' = re^{rt} \), \( y'' = r^2e^{rt} \)

plug into DE

\[ r^2e^{rt} + re^{rt} - 2e^{rt} = 0 \]\[ e^{rt}(r^2 + r - 2) = 0 \]

since \( e^{rt} \neq 0 \),

\[ r^2 + r - 2 = 0 \]

characteristic polynomial

note similarity to DE.

\[ (r + 2)(r - 1) = 0 \]

so \( r = -2, \quad r = 1 \)

two solutions are: \( y_1 = e^{-2t}, \quad y_2 = e^t \)

general solution:

\[ y(t) = c_1y_1 + c_2y_2 \]\[ = c_1e^{-2t} + c_2e^t \]

\( c_1, c_2 \) constants

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Initial Conditions for Second-Order Linear ODEs

\( c_1, c_2 \) depend on two initial conditions, typically \( y(t_0) = y_0 \) and \( y'(t_0) = y'_0 \).

Example

\[ 6y'' - 5y' + y = 0, \quad y(0) = 4, \quad y'(0) = 0 \]

Characteristic Equation

\[ 6r^2 - 5r + 1 = 0 \]

\[ (2r - 1)(3r - 1) = 0 \]

\[ r = 1/2, \quad r = 1/3 \]

General Solution

\[ y = c_1 e^{1/2 t} + c_2 e^{1/3 t} \]

\[ y' = \frac{1}{2} c_1 e^{1/2 t} + \frac{1}{3} c_2 e^{1/3 t} \]

Applying Initial Conditions (IC)

IC: \( y(0) = 4 \)

\[ 4 = c_1 + c_2 \quad \text{①} \]

IC: \( y'(0) = 0 \)

\[ 0 = \frac{1}{2} c_1 + \frac{1}{3} c_2 \quad \text{②} \]

\[ 0 = c_1 + \frac{2}{3} c_2 \quad \text{③} \]

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Solving for Constants

① - ③: \[ 4 = \frac{1}{3} c_2 \implies c_2 = 12 \]

From ①: \[ c_1 = -8 \]

Particular Solution

\[ y = -8 e^{1/2 t} + 12 e^{1/3 t} \]

\[ \lim_{t \to \infty} y = \text{unbound (to } \pm \infty) \]

Asymptotic Behavior of Solutions

  • If both roots of characteristic equations are positive, then all non-zero solutions become unbounded as \( t \to \infty \).

  • If both negative, all non-zero solutions go to 0 as \( t \to \infty \).

  • If one positive, one negative, then the solution with the positive exponent becomes unbounded, the other goes to 0. The general solution is unbounded, too.

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Example: Second-Order Linear ODE with Initial Conditions

Consider the differential equation:

\[ y'' - y' - 2y = 0 \]

With initial conditions:

\[ y(0) = \alpha, \quad y'(0) = 2 \]

Find \( \alpha \) so that:

\[ \lim_{t \to \infty} y = 0 \]

Note: \( y \) refers to the general solution.

Characteristic Equation

\[ r^2 - r - 2 = 0 \]\[ (r - 2)(r + 1) = 0 \]

The roots are:

\[ r = -1, \quad r = 2 \]

The root \( r = 2 \) leads to an unbounded solution.

General Solution

\[ y = C_1 e^{-t} + C_2 e^{2t} \]\[ y' = -C_1 e^{-t} + 2C_2 e^{2t} \]

Applying Initial Conditions (IC)

IC: \( y(0) = \alpha \)

\[ \alpha = C_1 + C_2 \quad \text{--- (1)} \]

IC: \( y'(0) = 2 \)

\[ 2 = -C_1 + 2C_2 \quad \text{--- (2)} \]
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Solving the System of Equations

Adding equations (1) and (2):

\[ \alpha + 2 = 3C_2 \]\[ C_2 = \frac{1}{3}\alpha + \frac{2}{3} \]

From equation (1):

\[ C_1 = \alpha - C_2 = \frac{2}{3}\alpha - \frac{2}{3} \]

Particular Solution

\[ y = \left( \frac{2}{3}\alpha - \frac{2}{3} \right)e^{-t} + \left( \frac{1}{3}\alpha + \frac{2}{3} \right)e^{2t} \]

As \( t \to \infty \):

  • The term \( e^{-t} \to 0 \).
  • The term \( e^{2t} \) is unbounded.

Therefore, setting \( \alpha = -2 \) will ensure that the coefficient of the unbounded term is zero, resulting in:

\[ \lim_{t \to \infty} y = 0 \]

Exam 1 Information

  • 11 questions total
  • 8 multiple-choice questions
  • Location: BRNG 2280
  • Time: During class time on Tuesday, 6/28