Study Guide - Pre-MFE Differential Equations (Spring 2026)


1) Course Scope and Exam Target #

This study guide follows the seminar syllabus and is designed to make you exam-ready for:

  • first-order ODE techniques
  • higher-order linear ODEs
  • PDE classification and heat equation methods
  • Fourier transform methods
  • core financial engineering PDE applications (especially Black-Scholes and perpetual American options)

Target outcome: you can classify problems quickly, choose the right method, solve with clean steps, and explain financial meaning.

On the web. Prose math is rendered with KaTeX as you scroll each h2 section. This page stays text-focused for fast review. lecture-notes.md (same folder) has the full interactive chart set—chart headings and legends use \(...\) in plot JSON (KaTeX); Plotly axis labels use $...$ (MathJax). See recap_guide.md §3.5 for escaping rules.


2) Method-Selection Map (Use First on Any Problem) #

ODE triage #

  1. First-order?
    • check separable
    • then linear
    • then exact (or integrating factor to make exact)
    • then Bernoulli / homogeneous substitution
  2. Higher-order linear with constant coefficients?
    • solve homogeneous via characteristic polynomial
    • add a particular solution for non-homogeneous part
  3. Variable coefficients but homogeneous form?
    • test transformation to constant coefficients (commonly Cauchy-Euler type)

PDE triage #

  1. Classify second-order PDE: $ Au_{xx}+Bu_{xt}+Cu_{tt}+\cdots=0 $ using $ \Delta = B^2-4AC $
  • Δ < 0: elliptic
  • Δ = 0: parabolic
  • Δ > 0: hyperbolic
  1. If heat-type parabolic PDE, choose:
    • fundamental solution + convolution on full line
    • Fourier transform method
    • separation/eigenfunction expansions with boundary conditions

3) First-Order ODEs #

3.1 Separable ODEs #

Idea

Separate all $y$ terms on one side and all $x$ terms on the other, then integrate.

Definition / Rule $ \frac{dy}{dx}=f(x)g(y) \quad\Rightarrow\quad \frac{1}{g(y)}\,dy=f(x)\,dx $

When to use

  • right side factors into a function of $x$ times a function of $y$

When not to use

  • mixed terms cannot be algebraically separated

Typical mistake

  • forgetting absolute values after integrating logarithms

Example:

Solve $\dfrac{dy}{dx}=xy$ with $y(0)=2$.

$$ \frac{1}{y}\,dy = x\,dx $$
$$ \ln|y| = \frac{x^2}{2} + C $$
$$ y = C e^{x^2/2}, \quad 2 = C \;\Rightarrow\; y = 2 e^{x^2/2} $$

Check: derivative gives $xy$ exactly.

3.2 Linear First-Order ODEs #

Idea

We are not guessing $y$ directly. We multiply the equation by a smart factor so the left side becomes one exact derivative, then integrate once.

In other words, we force: $ y'+p(x)y \quad\longrightarrow\quad \frac{d}{dx}\left[\mu(x)y(x)\right]. $ Once this happens, the problem is straightforward integration.

Definition / Rule $ y'+p(x)y=q(x),\qquad \mu(x)=e^{\int p(x)\,dx} $ $ \frac{d}{dx}\left[\mu(x)y(x)\right]=\mu(x)q(x) $

What we are exactly doing (algorithm)

  1. Identify $p(x)$ and $q(x)$ in $y' + p(x)y = q(x)$.
  2. Compute integrating factor $\mu(x)=e^{\int p(x)\,dx}$.
  3. Multiply the whole ODE by $\mu(x)$.
  4. Rewrite left side as $\frac{d}{dx}(\mu y)$.
  5. Integrate both sides and solve for $y$.

How the answer splits (homogeneous vs particular)
After step 5 you always have $\mu(x)\,y=\int \mu(x)\,q(x)\,dx + C$, hence $ y(x)=\underbrace{\frac{1}{\mu(x)}\int \mu(x)\,q(x)\,dx}_{y_p}+\underbrace{\frac{C}{\mu(x)}}_{y_h}. $ The piece $C/\mu$ is the general homogeneous solution: it solves $y'+p(x)y=0$ (equivalently $y=C e^{-\int p\,dx}=C/\mu$). In practice it often looks like $C$ times a single exponential (or one power of $x$ if you have already normalized differently). The integral term is a particular solution $y_p$. This is the same $y=y_h+y_p$ picture as in §4.2, already visible in first order.

Typical mistake

  • computing integrating factor with wrong sign

Example:

Solve $y' + 2y = e^{-x}$.

$$ \mu=e^{\int 2\,dx}=e^{2x} $$
$$ e^{2x}y'+2e^{2x}y=e^x \Rightarrow \frac{d}{dx}(e^{2x}y)=e^{x} $$
$$ e^{2x}y=e^x+C \Rightarrow y=e^{-x}+Ce^{-2x} $$

Here $y_h = C e^{-2x}$ (the $C$ term after dividing by $\mu$) and $y_p = e^{-x}$. Multiplying by $e^{2x}$ turns the left side into one product derivative, which is why the equation becomes easy to integrate.

Family plot ($y'+2xy=x^3$, $\mu=e^{x^2}$): each curve is a different homogeneous constant $C$; scroll to load the chart.

3.3 Exact ODEs #

Idea

An exact ODE means the left side is already a total differential of some potential function. So instead of solving for derivatives step-by-step, we recover that potential directly.

Definition / Rule $ M(x,y)\,dx+N(x,y)\,dy=0 $ is exact if $ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $ Then find potential $\Phi(x,y)$ with $ \Phi_x=M,\quad \Phi_y=N,\quad \Phi(x,y)=C $

When to use

  • the ODE is written naturally as $M(x,y)\,dx + N(x,y)\,dy = 0$
  • cross-partial test $M_y = N_x$ passes

When not to use

  • cross-partial test fails and no integrating factor is provided

Typical mistake

  • integrating $M$ and forgetting the "constant" can depend on $y$

Example:

Solve $\dfrac{dy}{dx}=-\dfrac{2xy+3}{x^2+4y}$ by converting to exact form.

$$ \frac{dy}{dx}=-\frac{2xy+3}{x^2+4y} \Rightarrow (2xy+3)\,dx + (x^2+4y)\,dy = 0 $$
$$ M(x,y)=2xy+3,\quad N(x,y)=x^2+4y $$
$$ M_y=2x,\quad N_x=2x\Rightarrow \text{ exact} $$
$$ \Phi_x=2xy+3\Rightarrow \Phi=x^2y+3x+h(y) $$
$$ \Phi_y=x^2+h'(y)=x^2+4y\Rightarrow h'(y)=4y\Rightarrow h(y)=2y^2 $$
$$ \Phi(x,y)=x^2y+3x+2y^2=C $$

Interpretation: once exactness is verified, the problem is “find the potential” rather than “solve for $y'$ directly.”

3.4 Bernoulli ODE #

Idea

Bernoulli is “almost linear”: the nonlinear term is a power of $y$. A substitution turns it into a linear first-order ODE.

Definition / Rule $ y'+p(x)y=q(x)y^n,\quad n\neq 0,1 $ Use substitution $ v=y^{1-n} $ to reduce it to a linear ODE in $v$.

When to use

  • use this when the equation matches $y' + p(x)y = q(x)y^n$ with $n \neq 0,1$.

When not to use

  • $n = 0$ or $n = 1$ (it is already linear / directly solvable)

Typical mistake

  • using the substitution but forgetting to transform $y'$ correctly

Example:

Solve $y'+y=xy^3$.

$$ n=3,\quad v=y^{1-n}=y^{-2} $$
$$ v'=-2y^{-3}y' \quad\Rightarrow\quad y'=-\frac{y^3}{2}\,v'. $$

Substitute into $y'+y=xy^3$:

$$ -\frac{y^3}{2}\,v'+y=xy^3. $$

Multiply by $-2y^{-3}$ (for $y\neq 0$):

$$ v'-2y^{-2}=-2x \quad\Rightarrow\quad v'-2v=-2x. $$
$$ \mu=e^{\int -2\,dx}=e^{-2x},\quad \frac{d}{dx}(e^{-2x}v)=-2xe^{-2x} $$
$$ v=x+\frac12+Ce^{2x} $$
$$ y=v^{-1/2}=\left(x+\frac12+Ce^{2x}\right)^{-1/2} $$

Interpretation: Bernoulli looks nonlinear, but after substitution it becomes a standard linear ODE in $v$.

Class example plot ($y'+2xy=x^3y^3$, same as lecture notes):

3.5 Homogeneous First-Order ODE #

Idea

Here “homogeneous” does not mean “zero right-hand side” (as in linear ODEs). It means: if you package the slope as a function of the single dimensionless ratio $u=y/x$, the equation closes. Writing $y=vx$ sets that ratio to $v$ and turns the ODE into one where only $v$ and $x$ appear in a separable way.

Terminology (avoid confusion)

  • Linear homogeneous: $y'+p(x)y=0$ (different topic).
  • Homogeneous in the $y/x$ sense (this section): $ \frac{dy}{dx}=F\!\left(\frac{y}{x}\right). $ The right-hand side depends on $x$ and $y$ only through the combination $y/x$.

Definition / Rule

If $ \frac{dy}{dx}=F\!\left(\frac{y}{x}\right), $ set $ y=vx,\qquad \frac{dy}{dx}=v+x\frac{dv}{dx}\quad(\text{product rule}). $ Then $y/x=v$, so the ODE becomes $ v+xv'=F(v) \quad\Rightarrow\quad xv'=F(v)-v. $ For values of $v$ where $F(v)\neq v$, separate variables: $ \frac{dv}{F(v)-v}=\frac{dx}{x}. $ Integrate, solve for $v(x)$, then back-substitute $y(x)=x\,v(x)$. (If $F(v)=v$ for some constant $v=v_0$, check whether $y=v_0 x$ is a singular solution.)

Why $y=vx$ is the right move

You are not “guessing” at random. The structure $\dfrac{dy}{dx}=F(y/x)$ is scale-invariant: if $(x,y)$ lies on a solution direction field, then $(kx,ky)$ sees the same slope, because $(ky)/(kx)=y/x$. The substitution $y=vx$ is the standard way to encode that symmetry: $v$ is exactly the quantity the slope law already depends on, and the extra $xv'$ term is the chain-rule price of letting $v$ vary with $x$.

How to recognize and simplify larger expressions

Many exam problems hide $F(y/x)$ inside a rational expression. A practical recipe:

  1. Write the equation as $\dfrac{dy}{dx}=\dfrac{N(x,y)}{D(x,y)}$ (or solve for $\dfrac{dy}{dx}$).
  2. Try to write $N$ and $D$ as sums of monomials of the same total degree in $(x,y)$. Example: $x^2+xy+y^2$ is degree $2$ in every term.
  3. Divide numerator and denominator by the same power of $x$ (usually $x^k$ where $k$ matches that degree), treating regions $x>0$ or $x<0$ separately if needed. After division, every surviving piece is a function of $y/x$ alone.
  4. If you can isolate $\dfrac{dy}{dx}$ as something that depends only on $y/x$, you are in this class.

Procedure (repeatable checklist)

  1. Confirm (possibly after algebra) that $\dfrac{dy}{dx}=F(y/x)$.
  2. Substitute $y=vx$ and $y'=v+xv'$.
  3. Replace $y/x$ by $v$ everywhere; simplify to $xv'=F(v)-v$.
  4. Separate: $\displaystyle \int \frac{dv}{F(v)-v}=\int \frac{dx}{x}$ (watch for $F(v)=v$).
  5. Solve for $v$, then $y=xv$. Apply initial conditions at the end if given.

When to use

  • $\dfrac{dy}{dx}$ can be reduced to a function of $y/x$ only (often after dividing a rational expression by $x^k$).

When not to use

  • The slope genuinely depends on $x$ and $y$ not only through $y/x$ (e.g. $\dfrac{dy}{dx}=x+y$, or $\dfrac{dy}{dx}=y/x+x$).
  • You need a different substitution (Bernoulli, exact, integrating factor on a linear equation, etc.).

Typical mistakes

  • Using $y'=v'$ instead of $y'=v+xv'$ (missing the product-rule term).
  • Dividing by $x$ without stating a domain (often $x>0$ vs $x<0$ changes $\ln|x|$ details).
  • Dividing by $F(v)-v$ and losing constant solutions where $F(v_0)=v_0$.

Example:

Solve $\dfrac{dy}{dx}=1+\dfrac{y}{x}$ for $x>0$.

Here $F(v)=1+v$ directly.

$$ y=vx,\quad y'=v+xv' $$
$$ v+xv'=1+v\Rightarrow xv'=1 $$
$$ v'=\frac1x\Rightarrow v=\ln x + C $$
$$ y=xv=x(\ln x + C) $$

Interpretation: after $y=vx$, the $v$ terms cancel on the right and you are left with a simple separable equation in $v$ and $x$.

Example:

Rational RHS. Solve $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{xy}$ for $x>0$, $y>0$.

Divide numerator and denominator by $x^2$:

$$ \frac{dy}{dx}=\frac{1+(y/x)^2}{y/x}. $$

Let $v=y/x$. Then $y=vx$, $y'=v+xv'$, and

$$ v+xv'=\frac{1+v^2}{v} \Rightarrow xv'=\frac{1+v^2}{v}-v=\frac{1}{v}. $$

Separate:

$$ v\,dv=\frac{dx}{x} \Rightarrow \frac{v^2}{2}=\ln x + C \Rightarrow v=\sqrt{2\ln x + C_1}\quad(C_1=2C). $$

Hence

$$ y=xv=x\sqrt{2\ln x + C_1}. $$

Interpretation: the “messy” rational form was only hiding a clean $F(v)$; the substitution is the same every time once $y/x$ is exposed.


4) Higher-Order Linear ODEs #

This chapter is about equations where unknown $y$ and its derivatives appear linearly (no $y^2$, $\sin(y)$, etc.). The order $n$ tells you how many independent constants appear in the general solution.

Big picture

  • Homogeneous ($g=0$): solutions form a vector space of dimension $n$; find $n$ independent basis solutions and take linear combinations.
  • Non-homogeneous ($g\neq 0$): superposition of forcing — if $L[y_1]=g_1$ and $L[y_2]=g_2$, then $L[c_1 y_1+c_2 y_2]=c_1 g_1+c_2 g_2$. So solve homogeneous once ($y_h$), then add one particular $y_p$ for each forcing piece (or one combined $y_p$ if you prefer).
  • Constant coefficients (§4.1–4.2): exponentials $e^{rx}$ are the natural “modes” because $\dfrac{d}{dx}$ acts multiplicatively on them.
  • Cauchy–Euler (§4.3): coefficients are powers of $x$ matched to derivative order; power laws $x^m$ (or $t=\ln x$) replace exponentials in $x$.

4.1 Homogeneous with Constant Coefficients #

Idea

Write $D=\dfrac{d}{dx}$. A constant-coefficient equation $a_n y^{(n)}+\cdots+a_0 y=0$ is $P(D)y=0$ for a polynomial $P$. Because $D$ commutes with itself, exponentials $e^{rx}$ are eigenfunctions: $D e^{rx}=r e^{rx}$, hence $P(D)e^{rx}=P(r)e^{rx}$. So $y=e^{rx}$ solves the ODE exactly when $P(r)=0$. That is why the whole method is: find roots of the characteristic polynomial, then build real-valued combinations when roots are complex, and multiply by $x^k$ when a root repeats — those extra powers keep solutions independent while still killing $P(D)$.

Why the ansatz $y=e^{rx}$

For $y''+a y'+b y=0$, plug $y=e^{rx}$:

$$ r^2 e^{rx}+a r e^{rx}+b e^{rx}=0 \Rightarrow r^2+a r+b=0. $$

Each root $r$ gives one solution $e^{rx}$. For order $n$, the same idea gives $a_n r^n+\cdots+a_0=0$.

Definition / Rule

$$ a_n y^{(n)}+\cdots+a_1 y'+a_0 y=0 $$

Characteristic equation:

$$ a_n r^n+\cdots+a_1r+a_0=0. $$

Build a real fundamental set from the roots:

  1. Distinct real $r_i$: use $e^{r_i x}$.
  2. Repeated real $r$ of multiplicity $m$: use $e^{rx},\,x e^{rx},\,\ldots,\,x^{m-1}e^{rx}$.
  3. Complex pair $\alpha\pm i\beta$: use $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ (Euler: $e^{(\alpha+i\beta)x}=e^{\alpha x}(\cos\beta x+i\sin\beta x)$).

How to use (checklist)

  1. Divide through so the leading coefficient is nonzero (often normalize to $1$).
  2. Write the characteristic polynomial carefully (watch signs: $y''+y=0\Rightarrow r^2+1=0$, not $r^2-1$).
  3. Factor or use the quadratic formula; classify every root (real distinct / repeated / complex).
  4. Write $y_h$ as a sum of the basis pieces with arbitrary constants.
  5. If an initial value problem is given, impose conditions on $y_h$ (and later on full $y=y_h+y_p$ for §4.2) to solve for constants.

When to use

  • Linear ODE, constant coefficients, zero right-hand side.

When not to use

  • Variable coefficients in the standard form (unless you transform first — §4.3).
  • Nonlinear ODE.

Typical mistakes

  • Wrong multiplicity: a repeated root needs all powers $1,x,\ldots,x^{m-1}$ times $e^{rx}$.
  • Forgetting damping $e^{\alpha x}$ when $\alpha\neq 0$ (complex roots are $\alpha\pm i\beta$, not only $\pm i\beta$).
  • Using $e^{i\beta x}$ in the final answer instead of real $\cos/\sin$ when the problem asks for real-valued $y$.

Example:

Distinct real roots + IVP. Solve $y'' - 3y' + 2y = 0$ with $y(0)=0$, $y'(0)=1$.

$$ r^2-3r+2=(r-1)(r-2)=0 \Rightarrow r=1,\;2,\qquad y_h=C_1 e^{x}+C_2 e^{2x}. $$
$$ y(0)=C_1+C_2=0,\quad y'(0)=C_1+2C_2=1 \Rightarrow C_1=-1,\; C_2=1. $$
$$ y(x)=-e^{x}+e^{2x}. $$

Check: $y(0)=0$, $y'(0)=-1+2=1$.

Example:

Pure oscillation (complex roots on imaginary axis). Solve $y'' + 4y = 0$.

$$ r^2+4=0 \Rightarrow r=\pm 2i \quad(\alpha=0,\;\beta=2). $$
$$ y(x)=C_1\cos(2x)+C_2\sin(2x). $$

Physical picture: undamped harmonic oscillator with angular frequency $2$.

Example:

Damped oscillator (complex roots with nonzero real part). Solve $y''+2y'+5y=0$.

$$ r^2+2r+5=0 \Rightarrow r=-1\pm 2i. $$
$$ y(x)=e^{-x}\big(C_1\cos(2x)+C_2\sin(2x)\big). $$

Interpretation: oscillation $\cos(2x),\sin(2x)$ multiplied by decay $e^{-x}$ — typical of damped springs or RLC circuits with underdamping.

Example:

Repeated real root (third order). Solve $y'''-3y''+3y'-y=0$.

$$ r^3-3r^2+3r-1=(r-1)^3=0 \Rightarrow r=1 \text{ multiplicity } 3. $$
$$ y(x)=(C_1+C_2 x+C_3 x^2)\,e^{x}. $$

Interpretation: one exponential mode, but three independent ways to “dress” it ($1,x,x^2$) before the next differentiation would collapse the span.

4.2 Non-Homogeneous with Constant Coefficients #

Idea

Write $L[y]=g(x)$ with the same linear operator $L$ as in §4.1. If $y_p$ is any one solution of $L[y_p]=g$, then for any homogeneous solution $y_h$ you have $L[y_h+y_p]=0+g=g$. So the general solution is $y=y_h+y_p$ — all freedom is still in $y_h$ (the $n$ constants). Think of $y_p$ as a particular response to the forcing, and $y_h$ as transient / homogeneous modes the initial data can excite.

Method options

  • Undetermined coefficients (UC): fast when $g(x)$ is built from polynomials, exponentials, and sines/cosines (finite-dimensional trial family). You guess a form with unknown coefficients, substitute into $L[y]=g$, and match like terms.
  • Variation of parameters (VoP): always works in principle on intervals where the homogeneous fundamental solutions have nonzero Wronskian; heavier algebra, but handles $g(x)=\sec x$, rational functions, etc.

Undetermined coefficients — practical rules

Forcing term in $g(x)$ Start trial $y_p$ (then fix overlaps)
polynomial of degree $d$ general polynomial of degree $d$
$e^{ax}$ $A e^{ax}$
$\cos(\omega x)$ or $\sin(\omega x)$ $A\cos(\omega x)+B\sin(\omega x)$
products products of the above trials

Resonance / overlap: if a trial term solves the homogeneous equation, multiply the whole trial by $x$; if it still overlaps, multiply by $x$ again, and so on, until no term lies in $\ker(L)$.

How to use

  1. Solve $L[y]=0$ completely → $y_h$.
  2. Inspect $g(x)$. If UC applies, pick trial $y_p$ from the table and apply resonance bumps.
  3. Substitute $y_p$ into $L[y]=g$; solve linear equations for the undetermined coefficients.
  4. If $g=g_1+\cdots+g_k$, either find one combined trial or solve for each piece and add (superposition).
  5. Add: $y=y_h+y_p$; use initial conditions on the sum.

When to use

  • Constant coefficients and $g$ suitable for UC, or any $g$ on an interval if you accept VoP.

When not to use

  • Variable coefficients (try §4.3, series, or other transforms).
  • Expecting UC to work on $g(x)=\ln x$ or $g(x)=1/x$ without changing method (use VoP or rewrite).

Typical mistakes

  • Resonance: trial $A e^{rx}$ when $e^{rx}$ is already in $y_h$ — must use $A x e^{rx}$, or $A x^k e^{rx}$ for higher multiplicity.
  • Dropping terms from the trial (e.g. using only $\cos$ when both $\cos$ and $\sin$ are needed).
  • Applying initial conditions to $y_h$ only instead of $y=y_h+y_p$.

Example:

Polynomial forcing (no exponential overlap). Solve $y''-3y'+2y=2x+1$.

Homogeneous: $y_h=C_1 e^{x}+C_2 e^{2x}$ as before.

Trial $y_p=ax+b$. Then $y_p'=a$, $y_p''=0$:

$$ -3a+2(ax+b)=2x+1 \Rightarrow 2a=2,\;-3a+2b=1 \Rightarrow a=1,\;b=2. $$
$$ y=C_1 e^{x}+C_2 e^{2x}+x+2. $$

Example:

Trig forcing without resonance. Solve $y''+4y=\cos x$.

$$ y_h=C_1\cos(2x)+C_2\sin(2x),\qquad \omega_g=1\neq 2. $$

Trial $y_p=A\cos x+B\sin x$. Substitution gives $-A+4A=\frac12$ on $\cos$ terms if we match $\cos x$ coefficient — systematically:

$$ y_p''+4y_p=(-A+4A)\cos x+(-B+4B)\sin x=3A\cos x+3B\sin x=\cos x, $$

so $A=\frac13$, $B=0$:

$$ y=C_1\cos(2x)+C_2\sin(2x)+\frac13\cos x. $$

Example:

Trig resonance. Solve $y''+4y=\sin(2x)$.

Here $y_h$ already contains $\sin(2x)$ and $\cos(2x)$, so bump:

$$ y_p=x\big(A\cos(2x)+B\sin(2x)\big). $$

One finds $A=-\frac14$, $B=0$ (compute $y_p''+4y_p$ and match $\sin(2x)$), hence

$$ y=C_1\cos(2x)+C_2\sin(2x)-\frac14 x\cos(2x). $$

Interpretation: driving at the natural frequency produces growth in amplitude like $x$ (beating / resonance in finite-dimensional analog).

Example:

Exponential resonance. Solve $y'' - 3y' + 2y = e^{x}$.

$$ y_h = C_1 e^{x} + C_2 e^{2x} $$
$$ \text{trial } y_p = A x e^{x}\quad(\text{not }Ae^{x}\text{, since }e^{x}\text{ is in }y_h) $$
$$ y_p' = A(x+1)e^{x},\quad y_p'' = A(x+2)e^{x} $$
$$ y_p''-3y_p'+2y_p = A\big((x+2)-3(x+1)+2x\big)e^{x} = -A e^{x} = e^{x} \Rightarrow A=-1 $$
$$ y = C_1 e^{x} + C_2 e^{2x} - x e^{x} $$

Interpretation: forcing matches a homogeneous mode, so the particular solution needs an extra factor $x$.

4.3 ODEs with Homogeneous Coefficients (Cauchy–Euler type) #

Idea

The unknown is still $y$ as a function of $x$, but the coefficients are powers of $x$ matched to the order of the derivative ($x^2$ with $y''$, $x$ with $y'$). That pattern is called equidimensional (or homogeneous in $x$): if you replace $x$ by $c x$, the equation keeps the same shape. Because of that scaling symmetry, power functions $y=x^m$ and, after a change of variable, exponentials in $t=\ln x$ play the same role that $e^{rx}$ plays for constant-coefficient ODEs in $x$.

Compared to §4.1: there you had translation symmetry in $x$ (constant coefficients → exponentials in $x$). Here you have scale symmetry in $x$ → power laws in $x$, or equivalently exponentials in $\ln x$.

Why $x=e^t$ (take $x>0$ so $t=\ln x$ is defined)
For constant coefficients you guess $y=e^{rx}$ because $\frac{d}{dx}$ multiplies by $r$ each time. Here $y''$ comes with $x^2$ and $y'$ with $x$, so the naive exponential in $x$ is not the right object. Instead set $ x=e^t,\qquad u(t):=y(x)=y(e^t). $ Then the chain rule packages $x$-derivatives into $t$-derivatives with constant coefficients: $ \frac{du}{dt}=\frac{dy}{dx}\frac{dx}{dt}=y'\,e^t = x\,y', $ $ \frac{d^2u}{dt^2}=\frac{d}{dt}(x y')=\frac{dx}{dt}y'+x\frac{d}{dt}(y')=e^t y'+e^t\frac{dy'}{dx}\frac{dx}{dt}=x y'+x^2 y''. $ Solve the last display for $x^2 y''$: $ x^2 y''=\frac{d^2u}{dt^2}-\frac{du}{dt},\qquad x y'=\frac{du}{dt}. $ Substitute into $x^2 y''+\alpha x y'+\beta y=0$ to get an ODE for $u(t)$ with constant coefficients: $ \left(\frac{d^2}{dt^2}-\frac{d}{dt}\right)u+\alpha\frac{du}{dt}+\beta u=0 \quad\Longleftrightarrow\quad u''+(\alpha-1)u'+\beta u=0. $ That is the whole point of $x=e^t$: it turns the equidimensional equation into the type you already solve with a characteristic polynomial in $\dfrac{d}{dt}$.

Ansatz $y=x^m$ (same algebra, faster on homework)
If $y=x^m$, then $y'=m x^{m-1}$ and $y''=m(m-1)x^{m-2}$. Every term in the homogeneous equation carries a factor $x^m$; dividing by $x^m$ (for $x\ne0$) leaves a polynomial in $m$, the indicial equation. Roots $m$ work like exponents $r$ in the $t$-picture because $x^m=e^{m t}$ when $x=e^t$.

Repeated / complex roots (what to write for $y$)
Real roots: $m_1\ne m_2$ give $x^{m_1}$ and $x^{m_2}$. Repeated real $m$ gives $x^{m}$ and $x^{m}\ln x$ (parallel to $e^{mt}$ and $t e^{mt}$ in $t$). Complex pair $m=\alpha\pm i\beta$ gives $x^{\alpha}\cos(\beta\ln x)$ and $x^{\alpha}\sin(\beta\ln x)$.

Definition / Rule $ x^2y''+\alpha xy'+\beta y = f(x) $ Use $x=e^t$ and $u(t)=y(e^t)$ with the formulas above, or use the indicial equation from $y=x^m$ on the homogeneous part. Then translate back to $x$ if you worked in $t$.

How to use (checklist)

  1. Confirm the standard Cauchy–Euler pattern: $x^2y''$, $xy'$, constant multiples, plus $y$ (possibly with forcing $f(x)$ on the right).
  2. Pick $x>0$ or $x<0$ when using $t=\ln x$ or $t=\ln|x|$; the indicial method with $y=x^m$ avoids logs until you interpret $\ln x$.
  3. Substitute $y=x^m$, divide by $x^m$, solve the indicial polynomial for $m$.
  4. Assemble real / repeated / complex formulas for $y_h$; if $f\neq 0$, find $y_p$ (UC/VoP in $t$ after $x=e^t$ is often simplest).

When to use

  • You see the pattern $x^2y''$, $xy'$, and $y$ with constant $\alpha,\beta$ (possibly with forcing $f(x)$).

When not to use

  • Coefficients are not matched to derivative order (e.g. $x y'' + y'$ without the $x^2$ structure in standard form — may need a different substitution or rewriting).
  • You try to plug $y=e^{rx}$ in the variable $x$ without passing to $t=\ln x$ first (usually the wrong ansatz for this symmetry).

Typical mistake

  • Applying the constant-coefficient characteristic method directly in $x$ on $x^2y''+\alpha xy'+\beta y=0$ without the $x=e^t$ map or the $x^m$ indicial equation.

Example:

Complex conjugate roots (oscillation in $\ln x$). Solve $x^2 y'' + x y' + y = 0$ for $x>0$.

Full method ($u$-substitution).

(1) Match $x^2y''+\alpha xy'+\beta y=0$: here $\alpha=1$, $\beta=1$.

(2) Set $x=e^t$ ($t=\ln x$) and $u(t):=y(e^t)$. Use (from §4.3):

$$ x y'=\frac{du}{dt},\qquad x^2 y''=\frac{d^2u}{dt^2}-\frac{du}{dt}. $$

(3) Substitute into the ODE ($y=u$ at $x=e^t$):

$$ \bigl(u''-u'\bigr)+u'+u=u''+u=0. $$

(4) Characteristic equation $r^2+1=0$, so $r=\pm i$. Real solutions in $t$:

$$ u(t)=C_1\cos t+C_2\sin t. $$

(5) Back-substitute $t=\ln x$:

$$ y(x)=C_1\cos(\ln x)+C_2\sin(\ln x). $$

Interpretation: the solution oscillates in log-time $t=\ln x$, not in $x$ itself.

Short version (indicial $y=x^m$). Try $y=x^m$ → $m(m-1)+m+1=m^2+1=0$ → $m=\pm i$ → same $y(x)=C_1\cos(\ln x)+C_2\sin(\ln x)$.

Example:

Double real root. Solve $x^2 y'' - 3x y' + 4y = 0$ for $x>0$.

Full method ($u$-substitution).

(1) Cauchy–Euler form: $x^2 y''$, $-3x y'$, $+4y$ → $\alpha=-3$, $\beta=4$.

(2) $x=e^t$, $u(t)=y(e^t)$, and

$$ x y'=\frac{du}{dt},\qquad x^2 y''=\frac{d^2u}{dt^2}-\frac{du}{dt}. $$

(3) Substitute ($y=u$):

$$ \bigl(u''-u'\bigr)-3u'+4u =u''-4u'+4u=0. $$

(4) Characteristic equation $r^2-4r+4=(r-2)^2=0$ → double root $r=2$. From §4.1 (repeated root):

$$ u(t)=(C_1+C_2 t)\,e^{2t}. $$

(5) Back-substitute $t=\ln x$, $e^t=x$:

$$ y(x)=(C_1+C_2\ln x)\,x^2. $$

Two independent solutions in $x$: $x^2$ and $x^2\ln x$.

Short version (indicial $y=x^m$). $y=x^m$ gives $m(m-1)-3m+4=m^2-4m+4=(m-2)^2=0$ → double root $m=2$ → $y(x)=x^2(C_1+C_2\ln x)$ (same answer).


5) PDE Foundations #

5.1 PDE Classification #

Idea

Classification predicts behavior and suitable boundary/initial data.

For second-order PDE in two variables: $ Au_{xx}+Bu_{xt}+Cu_{tt}+\cdots=0,\quad \Delta=B^2-4AC $

  • elliptic: smooth equilibrium-like behavior
  • parabolic: diffusion/smoothing over time
  • hyperbolic: wave-like propagation

5.2 Heat PDE (Core Parabolic Model) #

Definition / Rule $ u_t=\kappa u_{xx} $ for diffusion coefficient $\kappa > 0$.

Interpretation

  • temperature diffusion
  • in finance, transformed pricing PDEs often map to heat-equation structure

6) Solving the Heat PDE #

6.1 Fundamental Solution and Convolution #

For full-line problem with initial data $u(x,0)=\varphi(x)$: $ G(x,t)=\frac{1}{\sqrt{4\pi\kappa t}} \exp\!\left(-\frac{x^2}{4\kappa t}\right),\quad t>0 $ $ u(x,t)=(G(\cdot,t)*\varphi)(x) =\int_{-\infty}^{\infty}G(x-\xi,t)\varphi(\xi)\,d\xi $

Typical mistake

  • missing normalization factor in Gaussian kernel

6.2 Fourier Transform Method #

Take Fourier transform in space: $ \widehat{u}_t(\omega,t)=-\kappa\omega^2\widehat{u}(\omega,t) $ so $ \widehat{u}(\omega,t)=\widehat{\varphi}(\omega)e^{-\kappa\omega^2 t} $ then invert transform.

When useful

  • infinite domain
  • convolution-friendly or transform-friendly initial data

7) Financial Engineering PDE Core #

7.1 Black-Scholes PDE #

For derivative price $V(S,t)$ under standard assumptions: $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+rSV_S-rV=0 $ with terminal condition at maturity: $ V(S,T)=\text{payoff}(S) $

Boundary-condition intuition

  • call option: growth like $S$ for large $S$
  • put option: tends toward discounted strike as $S \to 0$ (European setting)

7.2 Black-Scholes Formula (European Call/Put) #

$$ C=S_0N(d_1)-Ke^{-rT}N(d_2) $$
$$ P=Ke^{-rT}N(-d_2)-S_0N(-d_1) $$
$$ d_1=\frac{\ln(S_0/K)+(r+\sigma^2/2)T}{\sigma\sqrt{T}}, \quad d_2=d_1-\sigma\sqrt{T} $$

Typical mistake

  • confusing current time-to-maturity $\tau = T - t$ with final maturity $T$ in formulas

7.3 Perpetual American Options #

Idea

No maturity means time-homogeneous ODE in continuation region.

Typical structure for value $V(S)$ in continuation region: $ \frac{1}{2}\sigma^2S^2V''+rSV'-rV=0 $ Use value matching + smooth pasting at optimal exercise boundary $S^*$.

Key conditions $ V(S^*)=\text{intrinsic}(S^*),\quad V'(S^*)=\text{intrinsic}'(S^*) $

7.4 Extensions of Black-Scholes PDE #

Know how PDE changes under:

  • continuous dividend yield $q$: $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+(r-q)SV_S-rV=0 $
  • modified dynamics / local volatility / other model assumptions

8) Error Map (High-Probability Exam Traps) #

  • Method mismatch: forcing linear ODE method on non-linear equation
  • Assumption omission: using Black-Scholes formulas without checking model assumptions
  • Boundary-condition errors: wrong condition at $S = 0$ or as $S \to \infty$
  • Transform mistakes: dropping coefficients during variable changes
  • Finance interpretation errors: mathematically valid expression but economically inconsistent sign/limit

20-second self-check:

  • Are conditions and domain stated?
  • Does equation type match method?
  • Are constants/signs correct?
  • Does limit behavior make sense?
  • If substituted back, does it satisfy ODE/PDE and conditions?

9) Practice Set #

A) Direct Application #

  1. Solve separable ODE: $ y'=(1+x^2)y,\quad y(0)=1 $
  2. Solve linear ODE: $ y'-3y=e^{2x} $
  3. Classify: $ u_{xx}+4u_{xt}+3u_{tt}=0 $
  4. Write Black-Scholes PDE with continuous dividend yield q.

B) Structured Multi-Step #

  1. Solve: $ y'+\frac{2}{x}y=x^3,\quad x>0 $
  2. Solve Cauchy-Euler: $ x^2y''-3xy'+4y=0 $
  3. For $ u_t=\kappa u_{xx},\quad u(x,0)=e^{-x^2} $ set up solution using convolution with heat kernel.

C) Mixed Exam-Style #

  1. Derive and solve the pricing ODE in continuation region for a perpetual American put, then apply value-matching and smooth-pasting conditions conceptually.
  2. Starting from Black-Scholes PDE, outline transformation steps to reduce to heat equation form.
  3. Given a candidate solution for option price, verify PDE and boundary behavior.

10) Answer Key (Final Results) #

A) Direct Application #

$$ y=e^{x+x^3/3} $$
$$ y=Ce^{3x}-e^{2x} $$
$$ \Delta=4^2-4(1)(3)=4>0 $$

Hyperbolic. 4. $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+(r-q)SV_S-rV=0 $

B) Structured Multi-Step #

$$ y=\frac{x^4}{6}+\frac{C}{x^2} $$
  1. Characteristic in transformed variable: $ m(m-1)-3m+4=0 \Rightarrow (m-2)^2=0 $ $ y=x^2(C_1+C_2\ln x) $
  2. Setup: $ u(x,t)=\int_{-\infty}^{\infty} \frac{1}{\sqrt{4\pi\kappa t}} \exp\!\left(-\frac{(x-\xi)^2}{4\kappa t}\right)e^{-\xi^2}\,d\xi $

11) Fast Review Plan (6 Sessions + Exam) #

  • Session 1-2 focus: first-order ODE mastery and rapid classification drills
  • Session 3 focus: higher-order linear ODEs + resonance traps
  • Session 4 focus: PDE classification + heat equation fundamentals
  • Session 5 focus: Fourier transform and heat PDE solving
  • Session 6 focus: Black-Scholes derivation/interpretation + perpetual American options
  • Final-exam prep: mixed set under time pressure, then full error-map review

Recommended timing split:

  • 40% mathematics technique drills
  • 40% financial PDE applications
  • 20% mixed synthesis and explanation practice

12) What to Memorize vs What to Derive #

Memorize:

  • canonical forms of first-order ODE classes
  • characteristic-equation solution templates
  • PDE classification discriminant
  • Black-Scholes PDE and call/put formulas
  • value matching and smooth pasting conditions

Derive confidently:

  • integrating factor workflow
  • convolution solution setup for heat equation
  • transformation path from pricing PDE to heat-equation structure
  • continuation-region ODE solving for perpetual American options

13) Deep Dive: Term Structure of Interest Rates #

13.1 Core Objects #

Idea

The term structure tells you how discounting depends on maturity.

Definition / Rule

  • Zero-coupon bond price at time $t$ maturing at $T$: $ P(t,T) $
  • Continuously compounded spot rate: $ R(t,T)=-\frac{1}{T-t}\ln P(t,T) $
  • Instantaneous forward rate: $ f(t,T)=-\frac{\partial}{\partial T}\ln P(t,T) $

Relation: $ P(t,T)=\exp\!\left(-\int_t^T f(t,u)\,du\right) $

If short rate is deterministic r(t), risk-neutral discounting gives: $ P(t,T)=\exp\!\left(-\int_t^T r(u)\,du\right) $

If r_t is stochastic Markov process, bond value P = P(t,r) typically solves: $ P_t+\mu_r(t,r)P_r+\frac{1}{2}\sigma_r^2(t,r)P_{rr}-rP=0, \quad P(T,r)=1 $

This is a parabolic pricing PDE, structurally similar to Black-Scholes.

13.3 Affine Intuition (Exam Concept) #

In many models (Vasicek/CIR class), bond prices are exponential-affine: $ P(t,T)=\exp\!\left(A(t,T)-B(t,T)r_t\right) $ and A, B satisfy ODE systems from PDE substitution.

Typical mistake

  • mixing physical drift and risk-neutral drift in pricing PDE.

14) Deep Dive: Bond Default Probability Models #

14.1 Reduced-Form (Intensity/Hazard) Model #

Idea

Default time $\tau$ is random with intensity $\lambda_t$; survival drives credit bond pricing.

If intensity is deterministic: $ \mathbb{Q}(\tau>T\mid \mathcal{F}_t) =\exp\!\left(-\int_t^T \lambda(u)\,du\right) $

With short rate r(u) and zero recovery, defaultable zero-coupon bond: $ P^{\text{def}}(t,T) =\exp\!\left(-\int_t^T (r(u)+\lambda(u))\,du\right) $

With fractional recovery of treasury delta in [0,1] (simplified deterministic setup): $ P^{\text{def}}(t,T)\approx \delta P(t,T)+(1-\delta)\exp\!\left(-\int_t^T (r(u)+\lambda(u))\,du\right) $

14.2 Structural vs Reduced-Form #

  • Structural model: default triggered when firm value crosses barrier.
  • Reduced-form model: default modeled directly by hazard rate.

When to use

  • reduced-form for tractable calibration to credit spreads
  • structural for balance-sheet/economic mechanism intuition

Typical mistake

  • interpreting hazard rate as physical default frequency without measure/context.

15) Deep Dive: Black-Scholes PDE Derivation and Interpretation #

15.1 Derivation Skeleton (Must Know Cold) #

Under risk-neutral dynamics: $ dS_t=rS_t\,dt+\sigma S_t\,dW_t $ For V(S,t), Ito gives: $ dV=\left(V_t+rSV_S+\frac{1}{2}\sigma^2S^2V_{SS}\right)dt+\sigma SV_S\,dW_t $ Self-financing replication with stock and cash removes dW_t risk. No-arbitrage implies portfolio earns r, giving: $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+rSV_S-rV=0 $

15.2 Boundary/Terminal Data #

  • Terminal condition: $ V(S,T)=\Phi(S) $
  • Call asymptotics: $ V(0,t)=0,\qquad V(S,t)\sim S-Ke^{-r(T-t)}\ \text{as}\ S\to\infty $
  • Put asymptotics: $ V(S,t)\sim Ke^{-r(T-t)}\ \text{as}\ S\to 0,\qquad V(S,t)\to 0\ \text{as}\ S\to\infty $

15.3 Financial Interpretations of PDE Terms #

In $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+rSV_S-rV=0 $

  • Time decay effect (theta-type contribution) comes from $V_t$.
  • Convexity/gamma effect under diffusion comes from $\frac{1}{2}\sigma^2 S^2 V_{SS}$.
  • Financing drift from holding delta in stock comes from $r S V_S$.
  • Discounting at the risk-free rate comes from $-rV$.

16) Deep Dive: Solving Black-Scholes PDE and Extensions #

16.1 PDE-to-Heat Transformation (High-Level) #

Set time-to-maturity: $ \tau=T-t $ log-price variable: $ x=\ln(S/K) $ and exponential rescaling: $ V(S,t)=K e^{\alpha x+\beta \tau}u(x,\tau) $ Choose $\alpha$, $\beta$ so first-derivative and zeroth-order terms simplify, reducing to: $ u_\tau = \frac{1}{2}\sigma^2 u_{xx} $ Then solve via heat kernel/Fourier methods and map back, yielding Black-Scholes formulas.

16.2 Extension Patterns You Should Recognize #

  1. Dividend yield $q$: $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+(r-q)SV_S-rV=0 $
  2. Time-dependent coefficients: $ V_t+\frac{1}{2}\sigma(t)^2S^2V_{SS}+r(t)SV_S-r(t)V=0 $
  3. Local volatility: $ V_t+\frac{1}{2}\sigma(S,t)^2S^2V_{SS}+rSV_S-rV=0 $
  4. American exercise feature:
  • free-boundary problem with early exercise boundary $S^*(t)$
  • continuation region satisfies PDE, stopping region satisfies payoff condition
  • smooth pasting at boundary

16.3 Perpetual American Put (Canonical Result Form) #

For perpetual case (no $t$), continuation ODE gives power-law solution: $ V(S)=A S^{m_1}+B S^{m_2} $ where $m_1$, $m_2$ solve characteristic quadratic from: $ \frac{1}{2}\sigma^2S^2V''+rSV'-rV=0 $ Impose boundedness/economic conditions plus: $ V(S^*)=K-S^*,\qquad V'(S^*)=-1 $ to identify boundary $S^*$ and constant(s).

Typical mistake

  • applying European terminal-condition logic to perpetual/American setting.

17) Master Detailed Guide (Clean Notation, Expert Level) #

This section is the canonical, high-detail version with cleaner reading flow:

  • short symbols stay inline (for example $P(t,T)$, $R(t,T)$, $f(t,T)$, $\tau$)
  • display math is used only for derivations and key formulas
  • every topic includes intuition, assumptions, method logic, and exam traps

17.1 Differential Equations and the Term Structure of Interest Rates #

Core objects

  • Zero-coupon bond price: $P(t,T)$ (value at time $t$ of receiving $1$ at time $T$).
  • Spot (zero) rate: $R(t,T) = -\frac{1}{T-t}\ln P(t,T)$.
  • Instantaneous forward rate: $f(t,T) = -\partial_T \ln P(t,T)$.

Key identity $ P(t,T)=\exp\!\left(-\int_t^T f(t,u)\,du\right). $

Deterministic short-rate benchmark
If short rate is deterministic $r(u)$, then $P(t,T) = \exp\left(-\int_t^T r(u)\,du\right)$.

Stochastic short-rate pricing PDE (template)
For Markov short rate $r_t$ under risk-neutral measure $\mathbb{Q}$: $ P_t+\mu_r^{\mathbb{Q}}(t,r)P_r+\frac{1}{2}\sigma_r^2(t,r)P_{rr}-rP=0,\qquad P(T,r)=1. $

Interpretation
This is structurally parallel to Black-Scholes: diffusion term, drift term, and discount term $-rP$.

Exam traps

  • Mixing real-world drift with risk-neutral drift.
  • Treating $f(t,T)$ as a literal forecast without model assumptions.

17.2 Bond Default Probability Model #

Reduced-form (intensity) setup

  • Default time: $\tau$.
  • Hazard/intensity process: $\lambda_t$.
  • Survival probability (deterministic $\lambda$ case): $ \mathbb{Q}(\tau>T\mid\mathcal{F}_t)=\exp\!\left(-\int_t^T \lambda(u)\,du\right). $

Simplified risky zero-coupon pricing (zero recovery heuristic) $ P^{\mathrm{def}}(t,T)\approx \mathbb{E}^{\mathbb{Q}}\!\left[e^{-\int_t^T r(u)\,du}\mathbf{1}_{\tau>T}\right] \approx \exp\!\left(-\int_t^T (r(u)+\lambda(u))\,du\right). $

Structural vs reduced-form

  • Structural: default from firm-value barrier crossing.
  • Reduced-form: default arrives as a jump time with intensity.

Exam traps

  • Interpreting $\lambda$ as physical default frequency without measure clarification.
  • Ignoring recovery assumptions when applying formulas.

17.3 Perpetual American Options #

Why this is special
No maturity removes explicit time-dependence, so continuation value is a function of $S$ only.

Continuation ODE (Black-Scholes assumptions) $ \frac{1}{2}\sigma^2S^2V''+rSV'-rV=0. $

Boundary logic

  • Value matching at free boundary $S^*$.
  • Smooth pasting at $S^*$: $ V(S^*)=\text{intrinsic}(S^*),\qquad V'(S^*)=\text{intrinsic}'(S^*). $

Method

  1. Solve homogeneous power-law form in continuation region.
  2. Apply boundedness/economic asymptotics.
  3. Use value matching + smooth pasting to solve constants and $S^*$.

Exam trap
Using finite-maturity terminal-condition logic on perpetual problems.

17.4 Derivation of the Black-Scholes PDE + Boundary Conditions + Financial Meaning #

Model under risk-neutral measure $\mathbb{Q}$ $ dS_t=rS_t\,dt+\sigma S_t\,dW_t. $

Itô expansion for $V(S,t)$ $ dV=\left(V_t+rSV_S+\frac{1}{2}\sigma^2S^2V_{SS}\right)dt+\sigma SV_S\,dW_t. $

Choose hedged portfolio with $\Delta = V_S$ to cancel diffusion risk, impose risk-free growth on hedge, and obtain: $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+rSV_S-rV=0. $

Data required for a well-posed pricing problem

  • Terminal condition: $V(S,T)=\Phi(S)$.
  • Boundary/asymptotic behavior (depends on payoff type).

Financial interpretation of each term

  • Time decay contribution comes from $V_t$.
  • Convexity/gamma contribution comes from $\frac{1}{2}\sigma^2 S^2 V_{SS}$.
  • Financing drift of a delta-held stock position comes from $r S V_S$.
  • Discounting of option value comes from $-rV$.

17.5 Solving Black-Scholes PDE and Deriving Black-Scholes Formulas #

Variable choices

  • Time-to-maturity: $\tau = T - t$.
  • Log-moneyness-like variable: $x = \ln(S/K)$.

Transformation idea Use rescaling $V = K e^{\alpha x+\beta\tau}u(x,\tau)$ to reduce BS to heat equation: $ u_\tau=\frac{1}{2}\sigma^2u_{xx}. $

Solve heat equation (kernel/Fourier), map back, and get: $ C=S_0N(d_1)-Ke^{-r\tau}N(d_2),\qquad P=Ke^{-r\tau}N(-d_2)-S_0N(-d_1), $ with $ d_1=\frac{\ln(S_0/K)+(r+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2=d_1-\sigma\sqrt{\tau}. $

Quick checks

  • Put-call parity (no dividends): $C - P = S_0 - Ke^{-r\tau}$.
  • Limit $\tau \to 0$ recovers payoff shape.

17.6 Extensions of Black-Scholes PDE #

Continuous dividend yield $q$ $ V_t+\frac{1}{2}\sigma^2S^2V_{SS}+(r-q)SV_S-rV=0. $

Time-dependent coefficients $ V_t+\frac{1}{2}\sigma(t)^2S^2V_{SS}+r(t)SV_S-r(t)V=0. $

Local volatility $ V_t+\frac{1}{2}\sigma(S,t)^2S^2V_{SS}+rSV_S-rV=0. $

American exercise extension

  • Free-boundary formulation.
  • Continuation region solves PDE.
  • Stopping region equals intrinsic value.
  • Smooth pasting at boundary.

17.7 First-Order ODEs (Separable, Linear, Homogeneous, Exact, Bernoulli) #

Classification-first workflow

  1. Test separable form $y' = f(x)g(y)$.
  2. Test linear form $y' + p(x)y = q(x)$.
  3. Test homogeneous-in-ratio form $y' = F(y/x)$ with substitution $y = vx$.
  4. Test exact form $M\,dx + N\,dy = 0$ with condition $M_y = N_x$.
  5. Test Bernoulli $y' + p(x)y = q(x)y^n$ with substitution $v = y^{1-n}$.

Linear integrating factor $ \mu(x)=e^{\int p(x)\,dx},\qquad \frac{d}{dx}(\mu y)=\mu q. $

Main trap
Choosing method by habit instead of equation structure.

17.8 Homogeneous and Non-homogeneous ODEs with Constant Coefficients #

For $a_n y^{(n)} + \cdots + a_0 y = g(x)$:

  • Solve homogeneous part using characteristic polynomial.
  • Build particular solution and combine $y = y_h + y_p$.

Root dictionary

  • Distinct real roots: exponential modes.
  • Repeated roots: multiply by powers of $x$.
  • Complex roots: oscillatory sine/cosine envelope.

Resonance rule
If the trial particular solution overlaps with $y_h$, multiply the trial by a sufficient power of $x$.

17.9 ODEs with Homogeneous Coefficients (Cauchy-Euler) #

Prototype: $x^2 y'' + \alpha x y' + \beta y = f(x)$.

Homogeneous part
Try $y = x^m$, giving indicial equation: $ m(m-1)+\alpha m+\beta=0. $

Repeated root gives $x^m \ln x$ term.

Alternative method
Set $x = e^t$ to convert into constant-coefficient equation in $t$.

17.10 PDE Classification: Parabolic, Elliptic, Hyperbolic #

For second-order form $A u_{xx} + B u_{xt} + C u_{tt} + \cdots = 0$ use: $ \Delta=B^2-4AC. $

  • Δ < 0: elliptic
  • Δ = 0: parabolic
  • Δ > 0: hyperbolic

Finance diffusion pricing PDEs (including BS-type) are parabolic evolution equations.

17.11 Heat PDE: Fundamental Solution, Convolution, Boundary Conditions #

Equation on the full line $ u_t=\kappa u_{xx}. $

Kernel $ G(x,t)=\frac{1}{\sqrt{4\pi\kappa t}} \exp\!\left(-\frac{x^2}{4\kappa t}\right). $

Full-line IVP solution $ u(x,t)=\int_{-\infty}^{\infty}G(x-\xi,t)\varphi(\xi)\,d\xi=(G*\varphi)(x). $

Boundary-condition note

  • Dirichlet: boundary value prescribed.
  • Neumann: boundary flux prescribed.
  • Robin: mixed linear boundary relation.

17.12 Fourier Transform and Heat Equation Solution #

With Fourier transform in space: $ \widehat{u}_t(\omega,t)=-\kappa\omega^2\widehat{u}(\omega,t), \qquad \widehat{u}(\omega,t)=\widehat{\varphi}(\omega)e^{-\kappa\omega^2 t}. $

Invert to recover $u(x,t)$. This is exactly equivalent to kernel convolution; it is the same mathematics in frequency coordinates.

17.13 Final Integration Logic (How to Study This Like an Expert) #

For any new exercise, run this exact sequence:

  1. Classify object (ODE/PDE/finance pricing/credit).
  2. Write assumptions explicitly (measure, coefficients, boundary/terminal data).
  3. Choose method by structure, not memory.
  4. Solve with one primary method.
  5. Run at least one independent check:
    • substitute back into equation,
    • verify boundary/terminal conditions,
    • verify asymptotics/limits,
    • verify sign/economic intuition.

If you want, I can now do one more pass on this same file to add:

  • a complete theorem/proof-sketch appendix,
  • a 40-problem graded set (easy/medium/hard),
  • and fully worked solutions for each topic.