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Exam 2 Review

Complex roots in characteristic eq.

\( y'' + 4y' + 5y = 0 \) \( \quad \) \( y(0) = 1, \) \( y'(0) = 0 \)

\( r^2 + 4r + 5 = 0 \)

\( r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i \)

\( e^{(-2 \pm i)t} = e^{-2t} e^{it} \)
\( e^{-2t} e^{-it} \)

\( y = c_1 e^{-2t} \cos(t) + c_2 e^{-2t} \sin(t) \)

\( y' = -c_1 e^{-2t} \sin(t) - 2c_1 e^{-2t} \cos(t) + c_2 e^{-2t} \cos(t) - 2c_2 e^{-2t} \sin(t) \)

\( y(0) = 1 \rightarrow 1 = c_1 \)

\( y'(0) = 0 \rightarrow 0 = -2c_1 + c_2 \rightarrow c_2 = 2c_1 = 2 \)

\( y = e^{-2t} \cos(t) + 2e^{-2t} \sin(t) \)

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repeated roots

\( 100y'' - 20y' + y = 0 \)

\( 100r^2 - 20r + 1 = 0 \) \( \quad \) \( (10r - 1)(10r - 1) = 0 \)

\( r = +\frac{1}{10}, +\frac{1}{10} \)

\( y = c_1 e^{\frac{1}{10}t} + c_2 t e^{\frac{1}{10}t} \)

Same idea for higher order — each repeat is one additional factor of \( t \)

this came out of reduction of order — if one solution is known \( (y_1) \) then the second solution is \( y_2 = v(t)y_1 \)

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Solving Second-Order Differential Equations

\[ t^2 y'' - t y' + y = 0, \quad t > 0 \]

Given that \( y_1 = t \) is a solution, find the second solution \( y_2 \).

Assume \( y_2 = v(t) y_1 = vt \). We find \( v \) by substituting \( y_2 \) into the equation.

\[ \begin{aligned} y_2' &= v + v't \\ y_2'' &= v' + v' + v''t = 2v' + v''t \end{aligned} \]

Substituting these into the original differential equation:

\[ t^2(2v' + v''t) - t(v + v't) + vt = 0 \]

Expanding the terms:

\[ 2t^2v' + v''t^3 - vt - v't^2 + vt = 0 \]

Simplifying (the \( vt \) terms cancel out):

\[ \begin{aligned} t^3 v'' + t^2 v' &= 0 \\ t v'' + v' &= 0 \\ v'' &= -\frac{v'}{t} \end{aligned} \]

Now, solve for \( v' \):

\[ \frac{d(v')}{dt} = -\frac{v'}{t} \]

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Separating variables to solve for \( v' \):

\[ \frac{1}{v'} d(v') = -\frac{1}{t} dt \]

Integrating both sides:

\[ \begin{aligned} \ln |v'| &= -\ln |t| = \ln |t^{-1}| \\ v' &= \frac{1}{t} \\ v &= \ln |t| = \ln t \end{aligned} \]

So, the second solution is:

\[ y_2 = v y_1 = t \ln t \]

Wronskian

The Wronskian of \( y_1, y_2 \) is defined as:

\[ W[y_1, y_2] = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \]

For three functions, the Wronskian is:

\[ W[y_1, y_2, y_3] = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{vmatrix} \]

If \( W \neq 0 \) for at least one \( t \) on some interval of \( t \), then we can always find \( c_1, c_2, \dots \) for any given initial condition.

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Nonhomogeneous Equations: Undetermined Coefficients

Variation of Parameters

\[ y'' - 4y' + 3y = 2t + e^t \]

Complementary: solve \( y'' - 4y' + 3y = 0 \)

\[ r^2 - 4r + 3 = 0 \quad \Rightarrow \quad (r - 3)(r - 1) = 0 \] \[ r = 1, \quad r = 3 \]

\[ y = c_1 e^t + c_2 e^{3t} + Y \]

Note: \( Y \) resembles the right side of the original equation.

\[ 2t + e^t \]

Where \( 2t \) is linear and \( e^t \) is exponential.

\[ Y = At + B + Ce^t \]

Check for duplication: \( Ce^t \) is copying \( e^t \) in the complementary solution.

Fix:

\[ Y = At + B + Cte^t \]

Substitute into the equation to find \( A, B, C \).

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If \( \cos(at) \) or \( \sin(at) \) appear on the right, always include BOTH in \( Y \).

\[ y'' - 4y' + 3y = 2t + e^t + t^2 \cos(3t) \]

Where \( t^2 \) is 2nd order and \( \cos(3t) \) is cosine or sine.

\[ Y = At + B + Cte^t + (Dt^2 + Et + F) \cos(3t) + (Gt^2 + Ht + I) \sin(3t) \]

Variation of Parameters

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

Complementary:

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

\[ y_c = c_1 e^t + c_2 t e^t \]

General:

\[ y = u_1 y_1 + u_2 y_2 \]

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Variation of Parameters: Solving for Coefficients

Find \( u_1, u_2 \) by solving the following system of equations:

\[ \begin{aligned} u_1' y_1 + u_2' y_2 &= 0 \\ u_1' y_1' + u_2' y_2' &= g(t) \end{aligned} \]

Where \( g(t) \) is the right side of the differential equation.

In matrix form, this is expressed as:

\[ \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ g(t) \end{bmatrix} \]

Example Calculation

Here, substituting the specific functions:

\[ \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \begin{bmatrix} e^t & t e^t \\ e^t & t e^t + e^t \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ \frac{e^t}{t} \end{bmatrix} \]

Using the formula for the inverse of a \( 2 \times 2 \) matrix:

\[ = \frac{1}{\begin{vmatrix} e^t & t e^t \\ e^t & t e^t + e^t \end{vmatrix}} \begin{bmatrix} t e^t + e^t & -t e^t \\ -e^t & e^t \end{bmatrix} \begin{bmatrix} 0 \\ \frac{e^t}{t} \end{bmatrix} \]

Integrate to find \( u_1, u_2 \).

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Mass-Spring-Damper Systems

Governing Equation

\[ m u'' + \gamma u' + k u = f(t) \]

\( m \): mass (NOT weight)
\( \gamma \): damping constant
\( k \): spring constant (Hooke's Law)
\( f(t) \): external force

Example Problem

If \( m = 2, k = 10, u(0) = 1, u'(0) = 0.15 \), and assuming no damping (\( \gamma = 0 \)) and no external force (\( f(t) = 0 \)):

\[ \begin{aligned} 2u'' + 10u &= 0 \\ u'' + 5u &= 0 \end{aligned} \]

The characteristic equation is:

\[ r^2 + 5 = 0 \implies r = \pm i \sqrt{5} \]

The general solution is:

\[ u = C_1 \cos(\sqrt{5} t) + C_2 \sin(\sqrt{5} t) \]

The derivative is:

\[ u' = -\sqrt{5} C_1 \sin(\sqrt{5} t) + \sqrt{5} C_2 \cos(\sqrt{5} t) \]

Applying Initial Conditions

\( u(0) = 1 \rightarrow 1 = C_1 \)

\( u'(0) = 0.15 \rightarrow 0.15 = \sqrt{5} C_2 \implies C_2 = \frac{0.15}{\sqrt{5}} \)

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\[ u = \cos(\sqrt{5}t) + \frac{0.15}{\sqrt{5}} \sin(\sqrt{5}t) \]

frequency: \( \sqrt{5} = \sqrt{\frac{k}{m}} \)

period: \( \frac{2\pi}{\text{freq}} \)

\[ u = R \cos(\sqrt{5}t - \delta) \]

\( R \): amplitude
\( \delta \): phase shift

\[ R^2 = \sqrt{c_1^2 + c_2^2} \]\[ \delta = \tan^{-1}\left(\frac{c_2}{c_1}\right) \]

Damping Cases

\[ mu'' + \gamma u' + ku = f(t) \]

\( \gamma^2 < 4km \) — underdamped

only case w/ cosines and sines

\( \gamma^2 = 4km \) — critically damped

\( \gamma^2 > 4km \) — overdamped

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Resonance

\[ mu'' + ku = f(t) \]

if \( f(t) \) is periodic w/ the same frequency as \( \sqrt{\frac{k}{m}} \rightarrow \) resonance

\[ u = c_1 \cos\left(\sqrt{\frac{k}{m}}t\right) + c_2 \sin\left(\sqrt{\frac{k}{m}}t\right) + At \cos\left(\sqrt{\frac{k}{m}}t\right) + Bt \sin\left(\sqrt{\frac{k}{m}}t\right) \]

t's make

\( |u| \rightarrow \infty \) as \( t \rightarrow \infty \)