First-Order Initial Value Problems: ( \displaystyle \frac{dy}{dt} = f(t,y), \,\, y(t{0}) = y{0} \in \mathbb{R} )

First-Order Initial Value Problems: $\displaystyle \frac{dy}{dt} = f(t,y), \,\, y(t_{0}) = y_{0} \in \mathbb{R}$

First-Order Initial Value Problems: $ \displaystyle \frac{dy}{dt} = f(t,y), \,\, y(t_{0}) = y_{0} \in \mathbb{R} $

Theory and Definitions

Quantitative Solution Techniques

Linear Problems: $ \displaystyle \frac{dy}{dt} + q(t) y = f(t) $

Integrating Factor Method
Formulae for solution procedure:

$ \displaystyle \mu(t) = e^{\int q(t)dt} $
$ \frac{d}{dt} \left[ \mu y \right] = \mu(t) f(t) $

Pro: General statement for all linear equations

Con: Requires integration

Derivation of these formulae

Examples

Example 2
$ y' - \frac{2}{t} y = 2t^{2}, \, \, y(-2) = 4 $

The Method of Undetermined Coefficients
$ q(t) = q \in \mathbb{R}$
and inhomogeneity that is the sum or product of sine, cosine, exponential and/or power functions.

Pro: Easy to use and only uses differentiation and algebra
Con: Not general. Cannot be used for all linear problems

Key Statements

We find from the integrating factor method that the general solution to any linear problem must always be the solution to the corresponding homogeneous problem added to a single particular solution to the inhomogeneous problem
$ y(t)= y_{h}(t) + y_{p}(t)$
Solving the corresponding homogeneous problem is straight-forward because of the constant coefficient
Since the inhomogeneous term is built from functions that return upon differentiation, the form of the particular solutions can be guessed up to some unknown constants. (See: Particular Forms)

Playlist of Examples

$ y' = 6e^{2x} -y $
$ y' = -3y +e^{2t}$ and $ y' = 4y -3 \cos(5t)$
$ y' = y + e^{-2t}$
$ y' = y +\cos(2t)$
$ y' = 4y -5 e^{4t}$ (example of $ f(t) \propto y_{h}(t) $
$ y' = -2y + e^{t/3}$
$ y' = -y + \cos(2t)$

Note: Disregard notice of office hours. Those are way old and he is not your TA.

Local Existence and Uniquness

Hypothesis verification and implications

To apply the existence and uniqueness theorem, you must first verify the hypotheses. Specifically, we must first make sure that $ f(t,y)$ and $ f_{y}(t,y)$ are continuous functions about $ (t_{0},y_{0}) $. Notes:

If the functions become infinite or undefined at the point, then they certainly cannot be continuous.
If either of the two functions were defined but experienced a jump, then they cannot be continuous.

Note: The theorem only states when the existence of a unique solution can be guaranteed. If the hypotheses are not satisfied, then two solutions may cross at $ (t_{0},y_{0}) $ or they might not.

Qualitative Geometric Techniques

Example 1
$ y' = 2- \frac{y}{1+t}, \, \, y(0)=3 $

Theoretical Results

Theorem

For the first-order initial value problem
$ \displaystyle \frac{dy}{dt} = f(t,y)\, \, y(t_{0}) = y_{0} \in \mathbb{R} $,
if $ f(t,y) $ and $ f_{y}(t,y) $ are continous functions about the point $ (t_{0}, y_{0}$, then there exists a nontrivial interval for which exactly one solution to the initial value problem exists.

Taxonomy

Nonlinear Problems

Slope fields

Pure time equation

Autonomous Problems

Separable Equations
$ \displaystyle \frac{dy}{dt} = f(t,y) = g(t) h(y) $

Relevant Equations

Definition of differential
$ \displaystyle dy = \frac{dy}{dt} dt $
Separation step omitting $ y $ for which $ h(y)$ vanishes
$ \displaystyle \frac{1}{h(y)}\frac{dy}{dt}=g(t)$
Integrating both sides with respect to $ t $
$ \displaystyle \int \frac{1}{h(y)} dy = \int g(t)dt$

Youtube playlist

$ \displaystyle \frac{dy}{dx} = \frac{7x}{y} $
$ (x^{2} +1)y' = xy $
$ \displaystyle \frac{dy}{dt} = \frac{t- e^{-t}}{y+e^{y}}$
$ y' = (1-2t)y^{2}, \, \, y(0)=-1/6 $

Bernoulli equations

Autonomy: In mathematics, an autonomous system or autonomous differential equation is a system of ordinary differential equations which does not explicitly depend on the independent variable.

$ f(t,y) = f(y) $
$ \displaystyle \frac{dy}{dt} = f(y), \, \, y(t_{0}) = y_{0} \in \mathbb{R} $

Notes:

The slope field for an autonomous problem is invariant with repsect to horizontal translations
Differential equation may posses constant/equilibrium solutions $ y(t)=y^{*} \in \mathbb{R}$ such that $ f(y^{*})=0$

Pure Time: A pure-time differential equation is a differential equation where the derivative of a function is given as an explicit function of the independent variable.

$ f(t,y) = f(t)$
$ \displaystyle \frac{dy}{dt} = f(t), \, \, y(t_{0}) = y_{0} \in \mathbb{R} $

Notes:

Pure-time problems are essentially statements of the fundamental theorem of calculus
If $ y' = f(t) $ then $ y(t) = \int f(t) dt$

Linear equations: A linear ODE is a linear with respect to the dependent variable and its derivatives.

The way the independent variable enters into the the equation is not relevant to the classification of linearity.
$ f(t,y) = m(t) y + b(t) $
Standard form: $ \displaystyle \frac{dy}{dt} + q(t) y = f(t), \, \, y(t_{0}) = y_{0} \in \mathbb{R} $

Homogeneous: A linear equation is said to be homogeneous if does not contain terms devoid of the dependent variable or its derivatives.

Standard form: $ \displaystyle \frac{dy}{dt} + q(t) y = 0, \, \, y(t_{0}) = y_{0} \in \mathbb{R} $

Notes:

The solution to any linear problem is the sum of the general homogeneous solution with some particular solution to the inhomogeneous problem.
First-order linear problems have general solutions expressible through integrating factors.
If $ q(t) = q \in \mathbb{R} $ and $ f(t) $ can be written as the sum and/or product of sine, cosine, exponential and power functions then the method of undetermined coefficients permits solutions to be found without the use of integration.

Relevant Equations

Bernoulli equation (standard form)

$ \displaystyle \frac{dy}{dt} + q(t) y = f(t)y^{n} $

Substitutionvariable

$ u = y^{1-n} $

Relationship between the derivatives

$ \displaystyle \frac{du}{dt} = (1-n) y^{-n} \frac{dy}{dt} $

Corresponding linear equation

$ \displaystyle \frac{du}{dt} + (1-n) q(t) u = (1-n)f(t) $

Youtube Playlist

Derivation of the method and $ x y' + 6y= 3xy^{4/3}$
$ y' + xy = xy^2 $
$ x y' +y = 1/y^2 $
$ y' + 2y/x=-x^9 y^{5}, \, \, y(1) = 1$

Autonomous equation

A mixed equation

Equilibrium solutions

Stable sinks

Unstable sources

Semi-stable nodes

Phase line analysis

Notes

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Meaning

When existence and uniqueness is guaranteed on a region of the $ (t,y)-$plane, we can conclude that solutions cannot intersect. This is important if we think about $ y=y(t)$ as a dynamic variable. If two trajectories crossed, then if we go back in time to the point where they intersect, then there is no way to decide which path to take post-intersection.

Generalizations to any order

It can be shown that an n-th order ODE can be written as a system of first-order ODE. In this case the IVP looks like
$ \displaystyle \frac{d\textbf{Y}}{dt} = \textbf{F}(t,\textbf{Y}), \, \, \textbf{Y}(t_{0}) = \textbf{Y}_{0} \in \mathbb{R}^{n} $
The hypotheses are now on the continuity of $ \textbf{F}$ about $ t_{0},\textbf{Y}_{0}) \in \mathbb{R}^{1+n} $ and the continuity of all the partitial derivatives of $ \textbf{F}$ with respect to $ \textbf{Y} = (y_1, y_2, y_3, \dots, y_n) $.