$$ \newcommand{\RR}{\mathbb{R}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\FF}{\mathbb{F}} \renewcommand{\epsilon}{\varepsilon} % ALTERNATE VERSIONS % \newcommand{\uppersum}[1]{{\textstyle\sum^+_{#1}}} % \newcommand{\lowersum}[1]{{\textstyle\sum^-_{#1}}} % \newcommand{\upperint}[1]{{\textstyle\smallint^+_{#1}}} % \newcommand{\lowerint}[1]{{\textstyle\smallint^-_{#1}}} % \newcommand{\rsum}[1]{{\textstyle\sum_{#1}}} \newcommand{\uppersum}[1]{U_{#1}} \newcommand{\lowersum}[1]{L_{#1}} \newcommand{\upperint}[1]{U_{#1}} \newcommand{\lowerint}[1]{L_{#1}} \newcommand{\rsum}[1]{{\textstyle\sum_{#1}}} % extra auxiliary and additional topic/proof \newcommand{\extopic}{\bigstar} \newcommand{\auxtopic}{\blacklozenge} \newcommand{\additional}{\oplus} \newcommand{\partitions}[1]{\mathcal{P}_{#1}} \newcommand{\sampleset}[1]{\mathcal{S}_{#1}} \newcommand{\erf}{\operatorname{erf}} $$

36  Differential Equations

What are differential equations etc.

36.1 Quadrature

Solving equations \(y^\prime = f\) with the fundamental theorem of calculus. Existence of darboux integrals proves the existence of solutions. Uniqueness given by corollary to the MVT about antidifferentiation.

Theorem 36.1 (Solving \(y^\prime =f\))  

Can compute solution via antidifferentiation but this won’t always work (in terms of known functions). Otherwise can actually use the definition of integration to get the function: hard to work with - but not so bad with a computer!

36.2 First Order Linear

Homogeneous + Non-Homogeneous

Procedure to ‘guess’ solution to first order homogeneous:

  • Differentiating results in multiplication by \(P\): a chain rule?
  • Getting a ‘log’ to show up, and reducing to quadratrue.

Theorem 36.2 (First Order Linear Homogeneous)  

Procedure to guess the solution here: getting product rule to show up and reducing to quadrature. Doing the example of \(y^\prime+xy=Q\) first, where its already a product rule.

Theorem 36.3 (First Order Linear)  

Proof. Existence is a calculation. For uniqueness, assume two solutions and consider their difference. This solves homogeneous equation and has initial condition \(y(a)=0\). But we know existence and uniqueness here implies the only such solution is the zero function. Thus their difference is zero.

Exercise 36.1 Prove the existence portion of this theorem via a direct calculation.

36.3 Beyond

Beyond this its in general very difficult to solve differential equations exactly: techniques come down to using algebra + calculus to put into this form. Maybe do a reduction of order thing for 2nd order given one solution?

36.4 Applications

Find some cool applications to calculus!