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PAGE 1

ODE Refresher

1st-order linear

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

solved by using an integrating factor

\[\mu(t) = e^{\int p(t) dt}\]
leading coefficient is 1

multiply both sides by that

left side is the derivative of \(\mu(t)\) and \(y(t)\)

\[\mu y' + \mu p y = \mu q\]
\[\frac{d}{dt}(\mu y) = \mu q\]

integrate and solve for \(y\)

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example

\[y' + \frac{2}{t}y = \frac{\cos(t)}{t^2} \quad (t > 0)\]
\[\mu = e^{\int \frac{2}{t} dt} = e^{2 \ln t} = e^{\ln t^2} = t^2\]

multiply both sides by that

\[t^2 y' + 2t y = \cos(t)\]
\[\frac{d}{dt}(t^2 y) = \cos(t)\]

integrate

\[t^2 y = \sin(t) + C\]
\[y = \frac{\sin(t)}{t^2} + \frac{C}{t^2}\]

1st-order linear will make an appearance later when we solve the heat equation:

\[\frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2} \quad u(x,t)\]
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First-Order Differential Equations in the Heat Equation

1st-order shows up in the context of the heat eq. as

\[ T' + \alpha^2 \lambda T = 0 \quad \quad T = T(t) \quad \quad \alpha, \lambda \text{ constants, positive} \]

this can be solved by using an integrating factor or as a separable eq.

\[ \frac{dT}{dt} + \alpha^2 \lambda T = 0 \]\[ \frac{dT}{dt} = -\alpha^2 \lambda T \]\[ \frac{dT}{T} = -\alpha^2 \lambda dt \quad \text{integrate both sides} \]\[ \ln |T| = -\alpha^2 \lambda t + C \]\[ T = e^{-\alpha^2 \lambda t + C} = e^{-\alpha^2 \lambda t} \cdot e^C \]

Note: \( e^C \) is a constant.

\[ T(t) = C e^{-\alpha^2 \lambda t} \]
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this governs the time part of the solution of heat eq.

for a given \( \lambda \), the larger \( \alpha^2 \) is, the faster \( T \) decays

Graph of exponential decay curves for T(t) showing faster decay for larger alpha squared.

2nd-order homogeneous

\[ y'' + ky = 0 \quad \quad k : \text{constant} \]

Solutions are of the form \( e^{rt} \)

where \( r \) is the solution to the characteristic eq.

\[ r^2 + k = 0 \]

Solution depends on what \( k \) is

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Case Analysis for Differential Equation Solutions

Case 1: If \( K < 0 \)

The characteristic roots are given by:

\[ r = \pm \sqrt{K} \]

The individual solutions are \( y = e^{\pm \sqrt{K}t} \). Note that this solution grows w/o bound.

The general solution is:

\[ y = c_1 e^{\sqrt{K}t} + c_2 e^{-\sqrt{K}t} \]

Alternatively, the general solution can be expressed using hyperbolic functions:

\[ y = d_1 \cosh(\sqrt{K}t) + d_2 \sinh(\sqrt{K}t) \]

\( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

\( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

Case 2: If \( K = 0 \)

The equation becomes:

\[ y'' = 0 \]

So, the general solution is (just a line):

\[ y = c_1 + c_2 t \]

Case 3: If \( K > 0 \)

The characteristic roots are imaginary:

\[ r = \pm i \sqrt{K} \]

The individual solutions are \( y = \cos(\sqrt{K}t) \) and \( y = \sin(\sqrt{K}t) \).

The general solution is:

\[ y = c_1 \cos(\sqrt{K}t) + c_2 \sin(\sqrt{K}t) \]

bounded solution

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We will see that when solving the heat, wave, and Laplace's eqs. it shows up as:

\[ X''(x) + \lambda X(x) = 0 \quad \text{or} \quad Y''(y) + \lambda Y(y) = 0 \]

Note: uppercase \( X \)

Solutions:

  • \( X(x) = c_1 \cos(\sqrt{\lambda}x) + c_2 \sin(\sqrt{\lambda}x) \)if \( \lambda > 0 \)
  • \( X(x) = c_1 \cosh(\sqrt{\lambda}x) + c_2 \sinh(\sqrt{\lambda}x) \)if \( \lambda < 0 \)
  • \( X(x) = c_1 + c_2 x \)if \( \lambda = 0 \)

In the context of the heat eq., this is the space part of the solution (it tells us how temperature varies w/ position with time fixed).

Next, let's review nonhomogeneous 2nd-order eqs.

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Mass-Spring Systems with Sinusoidal Input

In the context of a mass-spring system:

\[ mx'' + kx = f(t) \]

We will focus on sinusoidal \( f(t) \):

\[ f(t) = F_0 \sin(\omega t) \]

Where \( F_0 \) is a constant and \( \omega \) is the frequency of input.

General Solution

\[ x(t) = x_c + x_p \]
  • \( x_c \) (complementary): The solution when the right side is 0.
  • \( x_p \) (particular): The solution due to the right side.

For \( mx'' + kx = 0 \):

\[ x_c = C_1 \cos\left(\sqrt{\frac{k}{m}}t\right) + C_2 \sin\left(\sqrt{\frac{k}{m}}t\right) \]
\[ \sqrt{\frac{k}{m}} = \omega_0 \quad \text{(natural frequency)} \]

Method of Undetermined Coefficients

If \( f(t) = F_0 \sin(\omega t) \) and \( \omega \neq \omega_0 \):

Assume:

\[ x_p = A \cos(\omega t) + B \sin(\omega t) \]

Note: Always include both cosine and sine terms.

PAGE 8

Substitute into \( mx'' + kx = F_0 \sin(\omega t) \):

\[ A = 0, \quad B = \frac{F_0}{k - m\omega^2} \]

Trouble occurs if \( k = m\omega^2 \) or \( \omega = √(k/m) \).

The full solution is:

\[ x(t) = C_1 \cos\left(\sqrt{\frac{k}{m}}t\right) + C_2 \sin\left(\sqrt{\frac{k}{m}}t\right) + \frac{F_0}{k - m\omega^2} \sin(\omega t) \]

Resonance

If \( \sqrt{\frac{k}{m}} = \omega = \omega_0 \) (natural frequency = input frequency), we have resonance.

Adjustment to the particular solution:

\[ x_p = A t \cos(\omega t) + B t \sin(\omega t) \]

Substitute into \( mx'' + kx = F_0 \sin(\omega t) \):

\[ x_p = -\frac{F_0}{2m\omega} t \cos(\omega t) \]

This solution blows up as \( t \to \infty \).