Choose a method from fingerprints

Chapter 3 - Recognition before execution

Solving is not only performing a method. It is deciding which method fits. Blocked homework hides that decision because every exercise under "Separable Equations" is separable. Real problems do not carry those labels.

This chapter builds a classifier you will retrieve and refine throughout the book.

Normalize, then inspect

Before naming a method:

identify the unknown function and independent variable;
identify the highest derivative and therefore the order;
move or divide terms to expose a familiar form; and
check whether a simpler qualitative answer should come first.

The same equation can wear a disguise. For example,

$(1 + t) y^{'} = 2 y$

looks different from (y'=a(t)b(y)) until division gives

$y^{'} = \frac{2}{1 + t} y .$

Now its separable fingerprint is visible. It also happens to be linear; either method works.

The first-course method map

Autonomousy' = f(y)

First draw the phase line. Equilibria and stability may answer the important question without a formula.

Separabley' = a(t)b(y)

Move all state factors beside $dy$ and all time factors beside $dt$.

First-order lineary' + p(t)y = q(t)

Use an integrating factor to create one product derivative.

Second-order constant coefficientay'' + by' + cy = g(t)

Use characteristic roots for natural motion; add a response shaped like the input.

Linear systemx' = Ax

Read eigen-directions and eigenvalues, then connect them to a phase portrait.

No elementary fingerprinty' = f(t,y)

Use a field, numerical approximation, bounds, or a qualitative argument.

These categories overlap. (y'=ty) is autonomous? No, because the rate contains (t). It is both separable and linear. Choose the shorter route and retain the other as a check.

Lab: classify before solving

Method trainerEight mixed equations; no chapter labels.

Contrast cases: train the boundary

Deep recognition grows when similar-looking examples require different moves.

Separable

$y'=t(1+y)$

$\dfrac{dy}{1+y}=t\,dt$

Why

one factor uses only $t$, the other only $y$

Linear, not separable

$y'=t+y$

$y'-y=t$

Why

the sum $t+y$ cannot split into $a(t)b(y)$

Neither elementary form

$y'=t+y^2$

use a field or numerical method

Why

nonlinear in $y$, and no product separates

Almost separable. Which equation fails the separable test?

Worked example: name two routes

Classify $ty'+2y=t^3$, $t>0$method selection

Normalize. Divide by $t$: $y'+(2/t)y=t^2$.
Test separability. Rewriting gives $y'=t^2-(2/t)y$, a sum whose terms mix $t$ and $y$. It does not split into $a(t)b(y)$.
Test linearity. It matches $y'+p(t)y=q(t)$ with $p(t)=2/t$ and $q(t)=t^2$.
State the next move. Use the integrating factor $\mu=e^{\int 2/t\,dt}=t^2$ on $t>0$. Chapter 5 develops the method.

Fade the classification

Classify each equation and state only the first move, not the full solution.

(y'=(t^2+1)e^{-y})
(y'-\frac{1}{t}y=t)
(y'=y^2-4)
(y''-5y'+6y=0)

Check the fingerprints

Separable; multiply by $e^y$ and pair $e^y dy$ with $(t^2+1)dt$.
First-order linear; compute an integrating factor on an interval excluding zero.
Autonomous and separable; first find equilibria $y=\pm2$ and draw a phase line.
Homogeneous second-order constant coefficient; form $r^2-5r+6=0$.

Method choice is a claim you can test

Before executing, point to the exact fingerprint. "It looks linear" is too vague. Say "after dividing by (t), the equation has (y'+p(t)y=q(t)), and (y) appears only to the first power."

Error clinic: distributing $dy$ as if it were an algebra tile

Separation is justified through differential or integral notation, but only after the rate factors correctly. From $y'=t+y$, writing $dy/y=t\,dt+dt$ does not separate anything: the right side still came from an invalid division of a sum.

Practice

For every problem, write a one-line fingerprint justification before any algebra.

Practice - classification
3.1 Classify: (a) $y'=t^2/y$, (b) $y'+e^t y=1$, (c) $y'=y\sin y$, (d) $y''+9y=0$.
Detailed solution
Practice - contrast
3.2 Explain why $y'=t y+t$ is separable, while $y'=t y+1$ is generally not. Identify a method that works for both.
Detailed solution
Integrated - multiple routes
3.3 Find every applicable early-course method for $y'=y/t$, $t>0$. Which route uses less machinery?
Detailed solution
Challenge - strategy boundary
3.4 For $y'=y^2+t$, explain precisely why the separable and first-order linear fingerprints fail. List three useful things you can still do without an elementary formula.
Detailed solution

Retrieve the map

On blank paper, recreate the six-box method map from memory. For each box, write a representative equation that did not appear on this page. Then compare.

I rebuilt the method map from memory and classified the practice before solving anything.

Next: Turn the separable fingerprint into a reliable sequence of moves.

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Differential Equations as a Second Language