Exam 1 Review (10/07/26)

Exam 1 Review (10/07/26)

Tomorrow, Thursday 10/8/26 8pm–9pm
in
WTHR 200 Balcony

10 questions: 7 Multiple choice, 3 free response

Topics: Lesson 1–16 (upto 3.3 section)

Office Hours: Today

• 1:30–2:30   \(\Big\}\) Math 842
• 4:30–5:30   /

Methods:

1. \( y' = f(x) \)

\( y' = f(x) \rightsquigarrow \) integrate both sides

2. Seperable:

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

integrate both sides

3. Homogeneous:

\[ y' = \phi\left(\frac{y}{x}\right) \rightsquigarrow v = y/x \quad\text{OR}\quad y = xv \]

\[ y' = v + xv' \]

get seperable eq.

4. Linear:

\[ y' + p(x)y = q(x) \]

Int. factor: \( \mu = e^{\int p(x)\,dx} \rightsquigarrow \mu' = p(x)\mu(x) \)

\[ \mu y' + \underbrace{\mu p(x)}_{\mu'} y = \mu(x)q(x) \rightsquigarrow (\mu(x)y(x))' = \mu(x)q(x) \]

integrate both sides

5. Bernouli:

\[ y' + p(x)y = q(x)y^n, \quad n \neq 0, 1 \]

\[ v = y^{1-n}, \quad v' = (1-n)y^{-n}y' \]

\[ \frac{1}{1-n}v' + p(x)v = q(x) \rightsquigarrow \text{Linear} \]

6. Exact equations:

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

or

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

is exact if \( M_y = N_x \)

Look for \( \phi(x,y) \) such that \( \phi_x = M \), \( \phi_y = N \)

then \( \phi(x,y) = c \) is general sol.

7. Numerical: Euler, Improved Euler.

Applications

1. Velocity-acceleration Models

\[ v = s'(t), \quad a = v'(t) = s''(t) \]

2. Mixing Problems: Mixing "Salt Water" in Water

Calculate Volume, Amount, Concentration

\[ \text{Concentration} = \frac{\text{Amount}}{\text{Volume}} \]

Adding at a rate \( r_1 \), & concentration \( c_1 \)

Leaking at rate \( r_2 \), \( c_2 = \left.\frac{A(t)}{V(t)}\right\} \text{Not given.} \)

\[ \frac{dV}{dt} = r_1 - r_2 \implies V(t) = (r_1 - r_2)t + V_0 \]

\[ \frac{dA}{dt} = c_1 r_1 - c_2 r_2 \implies A' = c_1 r_1 - \frac{A(t)}{V(t)} r_2 \]

3. Population models, especially logistic model

\[ P' = P(M - P) \]

Other topics:

1. Slope fields, solution curves
2. Equilibrium solutions
3. Critical points, stability
4. Existence + Uniqueness

2^nd Order Linear Homogeneous equations

1. Ch. equation w/ \( y = e^{rx} \) solution
2. Reduction of order w/ if \( y_1 \) is solution, Look for \( y_2 = v(x) y_1 \)

eg ①! Solve \( y' - y = 4y^3, \quad y > 0 \)

\[ y^{-3}y' - y^{-2} = 4 \]

Let:

\[ \begin{aligned} v &= y^{-2} \\ v' &= -2y^{-3}y' \end{aligned} \]

Substituting into the equation:

\[ \frac{v'}{-2} - v = 4 \] \[ v' + 2v = -8 \]

Linear: \( p(x) = 2 \implies \mu(x) = e^{\int 2\,dx} = e^{2x} \)

\[ e^{2x}v' + 2e^{2x}v = -8e^{2x} \] \[ (e^{2x}v)' = -8e^{2x} \] \[ e^{2x}v = -4e^{2x} + K \implies v = -4 + Ke^{-2x} \] \[ y = \frac{1}{\sqrt{-4 + Ke^{-2x}}} \]

Example

\( y' = \sqrt{x-y}, \quad y(2) = y_0 \)

\( y' = f(x, y) \), want \( \left. \begin{array}{l} f(x, y) \\ f_y(x, y) \end{array} \right\} \text{Continuous} \)

\[ \left. \begin{aligned} f(x, y) &= \sqrt{x-y} \quad \text{is } x \ge y \\ f_y &= \frac{1}{2\sqrt{x-y}} \cdot (-1) \quad \text{is } x > y \end{aligned} \right\} \quad \text{if } x > y \]

Want: \( x_0 > y_0 \quad \text{is } 2 > y_0 \implies \underline{y_0 < 2} \quad \text{CAN Apply} \)

\( y_0 \ge 2 \quad \underline{\text{CANNOT Apply}} \)

\[ y' = f(x, y) \]

\[ y_1 = y_0 + f(x_0, y_0) \cdot h \]

Repeat.

\[ y' = y + t^2, \quad h = 1/2 \]

\[ f(t, y) = y + t^2, \quad (t_0, y_0) = (0, 1) \]

\[ f(t_0, y_0) = 1 \]

\[ y_1 = 1 + 1 \cdot 1/2 \]

\[ y_1 = 3/2, \quad t_1 = 1/2 \]

\[ y_2 = y_1 + f(t_1, y_1)h \]

\[ f(t_1, y_1) = \frac{3}{2} + \frac{1}{4} = 7/4 \]

\[ y_2 = \frac{3}{2} + \frac{7}{4} \cdot \frac{1}{2} = 19/8 \]

\[ y_1^* = y_0 + \overbrace{f(t_0, y_0)}^{k_1} h \]

\[ = 3/2 \]

\[ \left. \begin{aligned} k_1 &= f(t_0, y_0) = 1/2 \\ k_2 &= f(t_1, y_1^*) = 7/4 \end{aligned} \right\} k = \frac{1}{2}\left(\frac{1}{2} + \frac{7}{4}\right) = 9/8 \]

\[ y_1 = y_0 + \frac{1}{2}(k_1 + k_2)h = 1 + 9/8(1/2) = 1 + 9/16 \]

\(\text{initial volume} = 100\text{ gal},\)
\(\text{initial amount} = 50\text{ lb}\)

Adding: \(\text{rate} = r_{\text{in}} = 5\text{ gal/min}\)
\(\text{concentration} = C_{\text{in}} = 1\text{ lb/gal.}\)

Removing/Leaking: \(\text{rate} = r_{\text{out}} = 5\text{ gal/min}\)
\[ C_{\text{out}} = ? = \frac{A(t)}{V(t)} \]

\[ \frac{dV}{dt} = r_{\text{in}} - r_{\text{out}} = 0 \implies V(t) = 100\text{ gal}. \]

\[ \frac{dA}{dt} = C_{\text{in}}\,r_{\text{in}} - C_{\text{out}}\,r_{\text{out}} = 5 - \frac{A(t)}{V(t)}\,5 = 5 - \frac{A(t)}{20} \]

\[ A'(t) = 5 - \frac{1}{20}A(t) \implies A'(t) + \frac{1}{20}A(t) = 5 \]

\[ A' + \frac{1}{20} A = 5, \quad A(0) = 50 \]

Find \( A(20) \).

\[ \mu(t) = e^{\int \frac{1}{20} dt} = e^{\frac{1}{20} t} \]

\[ \left(e^{\frac{1}{20} t} A\right)' = 5 e^{\frac{1}{20} t} \]

\[ e^{\frac{1}{20} t} A = 100 e^{\frac{1}{20} t} + k \]

\[ A(t) = 100 + k e^{-\frac{1}{20} t} \]

\[ 50 = 100 + k \implies k = -50. \]

\[ A(t) = 100 - 50 e^{-\frac{1}{20} t} \]

\[ A(20) = 100 - 50 e^{-1} \]