Higher-Order & Series Methods Cheat Sheet
QUICK DECISION TREE — Which method?
Look at your ODE and ask:
- Variable is \(t\) and form is \(mu'' + \gamma u' + ku = F(t)\)?
→ Spring/mass. Solve char eq for \(u_h\), then method of undetermined coefficients for \(u_p\). (Sec 3.7–3.8)
- Variable is \(x\), form is \(x^2 y'' + \alpha x y' + \beta y = 0\)?
→ Euler equation. Substitute \(y = x^r\), get indicial eq, split by \(\Delta\). (Sec 5.4)
- Variable is \(x\), form is \(P(x)y'' + Q(x)y' + R(x)y = 0\) with polynomial \(P, Q, R\)?
→ Check \(P(x_0)\): if \(\neq 0\), Power series method at ordinary point \(x_0\) (Sec 5.3, 6-Step Method). If \(= 0\), classify singular point as regular or irregular (Sec 5.4 limits test).
- Problem asks "find radius of convergence of \(\sum a_n x^n\)"?
→ Ratio Test on the series itself. (Sec 5.1–5.2)
ESSENTIAL TOOLS & FORMULAS
Quadratic Formula
For \(ax^2 + bx + c = 0\): \(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\). Discriminant: \(\Delta = b^2 - 4ac\). If \(\Delta < 0\), roots are complex: \(x = \dfrac{-b \pm i\sqrt{|\Delta|}}{2a}\)
Extracting \(\lambda\) and \(\mu\) from Complex Roots
If \(r = \alpha + \beta i\) (or roots \(\alpha \pm \beta i\)): \(\lambda = \alpha\) (real part), \(\mu = |\beta|\) (magnitude of imaginary part)
Method of Undetermined Coefficients (Particular Solutions)
For: \(ay'' + by' + cy = f(t)\).
Guess form for \(y_p\) based on \(f(t)\):
| \(f(t)\) looks like | Try \(y_p =\) |
|---|
| \(P_n(t)\) — poly of degree \(n\) | \(A_n t^n + \ldots + A_1 t + A_0\) |
| \(P_n(t) e^{rt}\) | (poly degree \(n\))\(\cdot e^{rt}\) |
| \(A\cos(\omega t)\) or \(B\sin(\omega t)\) | \(C\cos(\omega t) + D\sin(\omega t)\) (BOTH!) |
| \(e^{rt}[P\cos\omega t + Q\sin\omega t]\) | \(e^{rt}[C\cos\omega t + D\sin\omega t]\) (same form) |
⚠ Duplication rule (critical!): If your guessed \(y_p\) duplicates part of the homogeneous solution \(y_h\), multiply the guess by \(t\) (or \(t^2\) for double duplication).
Examples:
- \(y_h = c_1 e^{-t}\) and \(f(t) = e^{-t}\) → try \(y_p = At\,e^{-t}\) (not \(Ae^{-t}\))
- \(y_h = (c_1 + c_2 t)e^{-t}\) and \(f(t) = e^{-t}\) → try \(y_p = At^2\,e^{-t}\)
- \(y_h = c_1\cos(2t) + c_2\sin(2t)\) and \(f(t) = 3\cos(2t)\) → try \(y_p = t[A\cos 2t + B\sin 2t]\) (resonance!)
Procedure: (1) Find \(y_h\) first (you need it to check for duplication). (2) Guess \(y_p\) from the table, multiply by \(t\) or \(t^2\) if it duplicates \(y_h\). (3) Compute \(y_p', y_p''\). (4) Substitute into the ODE. (5) Match coefficients of each function type (\(\cos, \sin, e^{rt}, t^k\)). (6) Solve the system for the unknown constants.
Note: Index shifting for series is in the 5.1–5.2 section below — see the Pre-Computed Power Series Table. Do not re-derive; just pick off the row you need.
3.7–3.8 — Damped & Forced Oscillations
Spring Equation
\(mu'' + \gamma u' + ku = F(t)\), where \(m\) = mass, \(\gamma\) = damping, \(k\) = spring const, \(F(t)\) = external force.
Homogeneous: \(mu'' + \gamma u' + ku = 0\)
Characteristic equation: \(mr^2 + \gamma r + k = 0\). Solve with quadratic formula: \(r = \dfrac{-\gamma \pm \sqrt{\gamma^2-4mk}}{2m}\). The sign of the discriminant \(\gamma^2 - 4mk\) decides the case.
Step-by-step:
- Read \(m, \gamma, k\) off the ODE.
- Compute discriminant \(\Delta = \gamma^2 - 4mk\) to pick the case (table below).
- Solve the quadratic for the roots \(r\).
- Plug into the matching solution template.
- If initial conditions given, substitute \(t=0\) into \(u(t)\) and \(u'(t)\) to solve for \(c_1, c_2\).
| \(\Delta = \gamma^2 - 4mk\) | Case | Roots | Solution template |
|---|
| \(> 0\) | Overdamped | Two distinct real \(r_1, r_2\) (both negative) | \(u = c_1 e^{r_1 t} + c_2 e^{r_2 t}\) |
| \(= 0\) | Critically damped | Repeated real \(r = -\gamma/(2m)\) | \(u = (c_1 + c_2 t)\,e^{rt}\) |
| \(< 0\) | Underdamped | Complex \(r = \lambda \pm \mu i\), with \(\lambda = -\gamma/(2m)\), \(\mu = \sqrt{4mk-\gamma^2}/(2m)\) | \(u = e^{\lambda t}\,(c_1 \cos\mu t + c_2 \sin\mu t)\) |
Ex (Underdamped): \(u'' + 0.5\,u' + 5u = 0\)
1. Read: \(m=1, \gamma=0.5, k=5\).
2. Discriminant: \(0.25 - 20 = -19.75 < 0\) ⇒ underdamped.
3. Roots: \(r = \dfrac{-0.5 \pm \sqrt{-19.75}}{2} = -0.25 \pm 2.236\,i\). So \(\lambda = -0.25\), \(\mu = 2.236\).
4. Plug template: \(u = e^{-0.25 t}(c_1 \cos 2.236\,t + c_2 \sin 2.236\,t)\).
Ex (Critically damped): \(u'' + 4u' + 4u = 0\)
1. Read: \(m=1, \gamma=4, k=4\).
2. Discriminant: \(16 - 16 = 0\) ⇒ critically damped.
3. Root: \(r = -\gamma/(2m) = -2\) (repeated).
4. Plug template: \(u = (c_1 + c_2\,t)\,e^{-2t}\).
Ex (Overdamped): \(u'' + 5u' + 6u = 0\)
1. Read: \(m=1, \gamma=5, k=6\).
2. Discriminant: \(25 - 24 = 1 > 0\) ⇒ overdamped.
3. Roots: \(r = \dfrac{-5 \pm 1}{2} = -2\) or \(-3\).
4. Plug template: \(u = c_1 e^{-2t} + c_2 e^{-3t}\).
Applying Initial Conditions
Given: \(u(0) = u_0\), \(u'(0) = u_1\). Procedure: (1) Substitute \(t=0\) into \(u(t)\) to find \(c_1\), (2) Differentiate \(u(t)\), substitute \(t=0\) into \(u'(t)\) to find \(c_2\).
Ex with IC: \(u'' + 2u' + u = 0\), \(u(0)=1\), \(u'(0)=0\)
Char: \(r^2+2r+1=(r+1)^2=0 \Rightarrow r=-1\) (repeated)
Gen sol: \(u=(c_1+c_2 t)e^{-t}\)
IC1: \(u(0)=c_1=1\)
IC2: \(u'(t)=c_2 e^{-t}-(c_1+c_2 t)e^{-t}\), so \(u'(0)=c_2-c_1=0 \Rightarrow c_2=1\)
Ans: \(u=(1+t)e^{-t}\)
Forced: \(mu'' + \gamma u' + ku = F_0\cos(\omega t)\)
General solution: \(u = u_h + u_p\). Terminology: \(u_h\) = “transient solution” (dies out); \(u_p\) = “steady-state solution” (persists).
Steady-state amplitude: \(A = \dfrac{F_0}{\sqrt{(k-m\omega^{2})^{2} + (\gamma\omega)^{2}}}\); \(\tan\phi = \dfrac{\gamma\omega}{k-m\omega^{2}}\)
Resonance: When \(\omega = \sqrt{k/m}\) (natural frequency) with \(\gamma \approx 0\), amplitude \(\to \infty\).
Ex (forced, full walkthrough): \(m=1, k=6, \gamma=2, F(t)=2\cos t\)
Step 1 — ODE: \(mu'' + \gamma u' + ku = F(t)\) ⇒ \(u'' + 2u' + 6u = 2\cos t\)
Step 2 — Homogeneous char eq: \(r^2 + 2r + 6 = 0\)
Step 3 — Quadratic formula: \(r = \dfrac{-2\pm\sqrt{4-24}}{2} = \dfrac{-2\pm\sqrt{-20}}{2} = \dfrac{-2\pm 2\sqrt{5}\,i}{2} = -1\pm\sqrt{5}\,i\). So \(\lambda=-1\), \(\mu=\sqrt 5\).
Step 4 — Transient (homogeneous): \(u_h = e^{\lambda t}(c_1\cos(\mu t) + c_2\sin(\mu t)) = e^{-t}(c_1\cos(\sqrt 5\,t) + c_2\sin(\sqrt 5\,t))\)
Step 5 — Steady-state ansatz (RHS is \(2\cos t\) with \(\omega = 1\), which is NOT a root of char eq, so no duplication): \(U = A\cos t + B\sin t\).
Step 6 — Derivatives of ansatz: \(U' = -A\sin t + B\cos t\); \(U'' = -A\cos t - B\sin t\)
Step 7 — Build \(u''+2u'+6u\) piece by piece:
\(U'' = -A\cos t - B\sin t\)
\(2U' = -2A\sin t + 2B\cos t\)
\(6U = 6A\cos t + 6B\sin t\)
Sum: \((-A + 2B + 6A)\cos t + (-B - 2A + 6B)\sin t = (5A+2B)\cos t + (-2A+5B)\sin t\)
Step 8 — Set equal to \(2\cos t + 0\sin t\) and match:
\(\cos t\) equation: \(5A + 2B = 2\)
\(\sin t\) equation: \(-2A + 5B = 0\) ⇒ \(B = \tfrac{2A}{5}\)
Substitute: \(5A + \tfrac{4A}{5} = 2\) ⇒ \(\tfrac{29A}{5} = 2\) ⇒ \(A = \tfrac{10}{29}\), \(B = \tfrac{4}{29}\)
Step 9 — Steady-state: \(U = \tfrac{10}{29}\cos t + \tfrac{4}{29}\sin t\)
Step 10 — Full solution: \(u(t) = e^{-t}(c_1\cos(\sqrt 5\,t) + c_2\sin(\sqrt 5\,t)) + \tfrac{10}{29}\cos t + \tfrac{4}{29}\sin t\)
Ex: Forced with UC — \(u'' + 2u' + u = 5\cos(t)\)
Homogeneous: \((r+1)^2=0 \Rightarrow r=-1\) (repeated) \(\Rightarrow u_h=(c_1+c_2 t)e^{-t}\)
Particular: Try \(u_p = A\cos t + B\sin t\). Then \(u_p'=-A\sin t+B\cos t\), \(u_p''=-A\cos t-B\sin t\)
Substitute: \((-A\cos t - B\sin t) + 2(-A\sin t + B\cos t) + (A\cos t + B\sin t) = 5\cos t\)
Collect: \(\cos t\) terms: \(-A + 2B + A = 2B = 5\). \(\sin t\) terms: \(-B - 2A + B = -2A = 0\)
Solution: \(A=0, B=\frac{5}{2} \Rightarrow u_p = \frac{5}{2}\sin t\)
General: \(u = (c_1+c_2 t)e^{-t} + \frac{5}{2}\sin t\)
5.1–5.2 — Power Series & Convergence
⭐ Anatomy of a Power Series (read once, then you know forever)
Standard form: \(\displaystyle y = \sum_{n=0}^{\infty} a_n\,(x - x_0)^n\)
| Piece | What it is |
|---|
| \(n\) | counter (dummy variable). Runs \(0, 1, 2, 3, \ldots\) |
| starts at \(n=0\) | always, unless the problem says otherwise |
| \(a_n\) | the coefficient of the \(n\)-th term. Subscript matches \(n\). |
| \(x - x_0\) | the “center” is \(x_0\). For \(x_0 = 0\), this is just \(x\). |
| \((x - x_0)^n\) | the \(n\)-th power of the shift. Exponent matches \(n\). |
Unfolded: \(y = a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + a_3(x-x_0)^3 + \cdots\)
Key facts: \(y(x_0) = a_0\) and \(y'(x_0) = a_1\). So \(a_0, a_1\) ARE the initial values \(y(x_0), y'(x_0)\).
Why subscript and exponent match: By convention \(a_n\) is defined as “the coefficient attached to \((x-x_0)^n\).” After you do derivatives or multiply by \(x\), this matching breaks — you will see \(a_{n+1} x^n\) or \(a_n x^{n+2}\). That’s NORMAL; the subscript is just an address.
Convergence via Ratio Test
\(\sum a_n(x-x_0)^n\) converges for \(|x-x_0| < R\) where \(R = \dfrac{1}{\lim|a_{n+1}/a_n|}\)
Differentiation: \(\dfrac{d}{dx}\sum a_n x^n = \sum na_n x^{n-1} = \sum (n+1)a_{n+1}x^n\)
Ex: Radius of convergence — \(\sum \dfrac{n!}{2^{n}}x^{n}\)
\(\dfrac{a_{n+1}}{a_{n}} = \dfrac{(n+1)!}{2^{n+1}} \cdot \dfrac{2^{n}}{n!} = \dfrac{n+1}{2}\)
\(\lim_{n\to\infty}\dfrac{n+1}{2} = \infty \Rightarrow R = 0\)
Ex 2: Ratio test — \(\sum \dfrac{x^n}{n!}\)
\(\dfrac{a_{n+1}}{a_n} = \dfrac{1/(n+1)!}{1/n!} = \dfrac{1}{n+1} \to 0\), so \(R = \infty\) (converges everywhere)
⭐⭐ PRE-COMPUTED POWER SERIES TABLE — use directly, do not re-derive
Start: ansatz \(y = \sum_{n=0}^{\infty} a_n (x-x_0)^n\). For \(x_0 = 0\) this is \(\sum_{n=0} a_n x^n\). All entries below are for \(x_0 = 0\); shift \(x \to (x - x_0)\) if your \(x_0 \neq 0\).
| Expression | Raw form (from diff/mult rules) | Relabeled to \(x^n\) — USE THIS |
|---|
| \(y\) | \(\sum_{n=0}^{\infty} a_n\,x^n\) | \(\sum_{n=0}^{\infty} a_n\,x^n\) |
| \(y'\) | \(\sum_{n=1}^{\infty} n\,a_n\,x^{n-1}\) | \(\sum_{n=0}^{\infty} (n+1)\,a_{n+1}\,x^n\) |
| \(y''\) | \(\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}\) | \(\sum_{n=0}^{\infty} (n+2)(n+1)\,a_{n+2}\,x^n\) |
| \(xy\) | \(\sum_{n=0}^{\infty} a_n\,x^{n+1}\) | \(\sum_{n=1}^{\infty} a_{n-1}\,x^n\) |
| \(x^2 y\) | \(\sum_{n=0}^{\infty} a_n\,x^{n+2}\) | \(\sum_{n=2}^{\infty} a_{n-2}\,x^n\) |
| \(xy'\) | \(\sum_{n=1}^{\infty} n\,a_n\,x^n\) | \(\sum_{n=1}^{\infty} n\,a_n\,x^n\) |
| \(x^2 y'\) | \(\sum_{n=0}^{\infty}(n+1)\,a_{n+1}\,x^{n+2}\) | \(\sum_{n=2}^{\infty} (n-1)\,a_{n-1}\,x^n\) |
| \(xy''\) | \(\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-1}\) | \(\sum_{n=1}^{\infty} n(n+1)\,a_{n+1}\,x^n\) |
| \(x^2 y''\) | \(\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^n\) | \(\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^n\) |
How to use: pick off whichever row matches each term of the ODE, write the relabeled-form sum, then combine. No substitution needed.
Shortcut rules (if you forget the table):
- Each derivative bumps subscript of \(a\) UP by 1, coefficient grows by a factor of the index; start stays at \(n=0\).
- Each multiplication by \(x\) bumps subscript of \(a\) DOWN by 1 AND bumps starting index UP by 1.
- Apply derivatives first, then multiplications. (e.g. \(xy''\) = "do \(y''\) then multiply by \(x\)".)
Why: derivative drops exponent by 1, multiplying by \(x\) raises it by 1; subscript moves opposite direction to keep the actual series unchanged — only its address changes.
⭐ Pulling Out Low-\(n\) Terms (to combine mixed-start sums)
Problem: you have sums starting at different indices (e.g. one at \(n=0\), one at \(n=1\)). Cannot combine yet.
Fix (3 steps):
- Find the HIGHEST starting index among all your sums (call it \(N\)).
- From every sum that starts below \(N\), PEEL OFF the low terms (plug in \(n = 0, 1, \ldots, N-1\)) and write them as standalone numbers out front.
- Now every sum starts at \(N\), so they combine: \(\bigl(\text{peeled terms}\bigr) + \sum_{n=N}^{\infty}[\cdots]x^n = 0\).
After combining: each power of \(x\) (including the peeled standalone pieces) must have zero coefficient. Standalone pieces give special cases; the combined sum gives the recurrence.
Even/Odd Decoupling
If the recurrence jumps by 2 (i.e., links \(a_n\) to \(a_{n+2}\)): even-indexed \(a\)’s depend ONLY on \(a_0\); odd-indexed ONLY on \(a_1\). So \(a_0\) and \(a_1\) are two independent “seeds” — pick \((1,0)\) to get \(y_1\), pick \((0,1)\) to get \(y_2\).
⭐ Spotting the Closed-Form Pattern for \(a_k\)
The method (do all 4 steps, in order):
- Compute 5+ coefficients, NOT 3. Three is rarely enough to see the pattern. Go to \(a_8\) or \(a_{10}\).
- Identify the SIGN pattern: all positive, alternating \((-1)^k\), all negative, ...
- Identify the NUMERATOR pattern: usually just \(1\), \(a_0\), or \(a_1\). Sometimes powers like \(2^k\), \(3^k\).
- Identify the DENOMINATOR pattern (the hard one — use the table below).
Then VERIFY: plug \(k = 0, 1, 2\) into your formula and check it reproduces the coefficients you computed.
Common denominator patterns to memorize:
| What you see | Closed form | Why |
|---|
| \(1, 2, 6, 24, 120, 720\) | \(k!\) | factorial |
| \(2, 24, 720, 40320\) | \((2k)!\) | double-step factorial |
| \(2, 4, 8, 16, 32\) | \(2^k\) | geometric, base 2 |
| \(3, 9, 27, 81\) | \(3^k\) | geometric, base 3 |
| \(2 \cdot 4 \cdot 6 \cdots (2k)\) | \(2^k\,k!\) | factor one 2 from each → \(k\) twos, leaves \(1\cdot 2\cdots k\) |
| \(1 \cdot 3 \cdot 5 \cdots (2k-1)\) | \(\dfrac{(2k)!}{2^k\,k!}\) | odd double factorial |
| \(3 \cdot 5 \cdot 7 \cdots (2k+1)\) | \(\dfrac{(2k+1)!}{2^k\,k!}\) | same, shifted |
Worked pattern example (from Problem 5.3 #1, Step 3c):
Computed: \(a_0=1,\; a_2=-\tfrac{1}{2},\; a_4=\tfrac{1}{8},\; a_6=-\tfrac{1}{48},\; a_8=\tfrac{1}{384},\; a_{10}=-\tfrac{1}{3840}\)
1. Sign: \(+, -, +, -, +, -\) → alternating → \((-1)^k\)
2. Numerator: always \(1\)
3. Denominator: \(1, 2, 8, 48, 384, 3840\). Factor: \(1, 2, 2\cdot 4, 2\cdot 4\cdot 6, 2\cdot 4\cdot 6\cdot 8, 2\cdot 4\cdot 6\cdot 8\cdot 10\) → matches \(2 \cdot 4 \cdots (2k)\) → \(2^k\,k!\)
4. Formula: \(a_{2k} = \dfrac{(-1)^k}{2^k\,k!}\)
Verify: \(k=0\): \(1/(1\cdot 1) = 1\) ✓. \(k=1\): \(-1/(2\cdot 1)= -\tfrac{1}{2}\) ✓. \(k=2\): \(1/(4\cdot 2) = \tfrac{1}{8}\) ✓.
Worked pattern example 2 (from Problem 5.3 #1, Step 4):
Computed: \(a_1=1,\; a_3=-\tfrac{1}{3},\; a_5=\tfrac{1}{15},\; a_7=-\tfrac{1}{105},\; a_9=\tfrac{1}{945}\)
1. Sign: \(+, -, +, -, +\) → \((-1)^k\) (with \(k=0\) corresponding to \(a_1\))
2. Numerator: always \(1\)
3. Denominator: \(1, 3, 15, 105, 945\). Factor: \(1, 3, 3\cdot 5, 3\cdot 5\cdot 7, 3\cdot 5\cdot 7\cdot 9\) → odd number product
4. Formula: \(a_{2k+1} = \dfrac{(-1)^k}{3 \cdot 5 \cdots (2k+1)}\) (empty product = 1 for \(k=0\))
Pro tips for factoring tricky denominators:
- If every factor has a 2 in it (\(2, 4, 6, 8, \ldots\)), pull one 2 from each factor. Count how many 2's you pulled (that is \(k\)) → \(2^k \cdot (\text{smaller product})\).
- If denominators grow faster than powers (like \(6, 24, 120, \ldots\)), suspect factorial \(k!\) or \((2k)!\).
- If the original denominator contains products of consecutive integers, think factorial. Divide \((2k)!\) by the missing pieces: \(1\cdot 3\cdots(2k-1) = (2k)! / [2\cdot 4\cdots (2k)] = (2k)!/(2^k k!)\).
- Check for \(x^2/2\) appearing naturally: many ODEs produce series where \(e^{x^2/2}\) or \(e^{-x^2/2}\) is the closed form. If you see \(2^k k!\) in the denominator with \(x^{2k}\) in the numerator, you are looking at \(\sum (x^2/2)^k / k!\) which IS the exponential series.
5.3 — Series at Ordinary Points
Ordinary Point Test
For \(P(x)y'' + Q(x)y' + R(x)y = 0\): point \(x_0\) is ordinary if \(P(x_0) \neq 0\)
Solution form: \(y = \sum a_n(x-x_0)^n\) where \(a_0, a_1\) free, higher via recurrence.
When \(x_0 = 0\): \((x-x_0)^n = x^n\), so this reduces to \(y = \sum a_n x^n\).
⭐ 6-Step Series Method
The algorithm:
- Substitute & re-index. Start with \(y = \sum a_n(x-x_0)^n\) (if \(x_0 = 0\), this is \(\sum a_n x^n\)). Write \(y', y''\) as series in the same power. Reindex \(xy, xy',\) etc. so every sum has the same power.
- Get the recurrence. Combine sums, pull out lowest-\(n\) terms that do not align, set each coefficient to zero.
- Build \(y_1\): plug in \(a_0 = 1, a_1 = 0\).
- Build \(y_2\): plug in \(a_0 = 0, a_1 = 1\).
- Radius \(R\): apply Ratio Test to \(y_1\) (or \(y_2\)).
- Wronskian \(W[y_1, y_2](0)\): by construction, \(=\det\begin{bmatrix}1&0\\0&1\end{bmatrix}=1\neq 0\), so \(\{y_1, y_2\}\) fundamental.
First-order exception: If the ODE is first order (only \(a_0\) free), skip Steps 3, 4, 6. Do Steps 1, 2, 5 only.
⭐ Full worked example: \(y'' - xy' - y = 0\), \(x_0=0\)
Step 1: \(y = \sum a_n(x-x_0)^n = \sum a_n x^n\) (since \(x_0=0\)); \(y'' = \sum(n+2)(n+1)a_{n+2}x^n\); \(xy' = \sum na_n x^n\) (\(n\geq 1\)).
Step 2: Combine → \((2a_2 - a_0) + \sum_{n\geq 1}[(n+2)(n+1)a_{n+2} - na_n - a_n]x^n = 0\). \(\;a_2 = \dfrac{a_0}{2};\;\) for \(n\geq 1:\; a_{n+2} = \dfrac{a_n}{n+2}\).
Step 3 (\(a_0=1, a_1=0\)): \(y_1 = 1 + \dfrac{x^2}{2} + \dfrac{x^4}{2\cdot 4} + \dfrac{x^6}{2\cdot 4\cdot 6} + \cdots = \sum\dfrac{x^{2k}}{2^k k!} = e^{x^2/2}\)
Step 4 (\(a_0=0, a_1=1\)): \(y_2 = x + \dfrac{x^3}{3} + \dfrac{x^5}{3\cdot 5} + \dfrac{x^7}{3\cdot 5\cdot 7} + \cdots\)
Step 5: For \(y_1\), \(\lim\left|\dfrac{x^2}{2(k+1)}\right| = 0 < 1\) ⇒ \(R=\infty\).
Step 6: \(W[y_1, y_2](0) = \det\begin{bmatrix}1&0\\0&1\end{bmatrix} = 1 \neq 0\) ⇒ fundamental set.
Ex (sign-flip variant): \(y'' + xy' + y = 0\) at \(x_0=0\)
Same as above but with sign: recurrence \(a_{n+2} = -\dfrac{a_n}{n+2}\). \(y_1 = e^{-x^2/2}\); \(y_2 = x - \dfrac{x^3}{3} + \dfrac{x^5}{3\cdot 5} - \cdots\). \(R = \infty\), \(W(0) = 1\).
Ex (first order, one constant): \(y' - xy = 0\)
\(xy = \sum a_{n-1}x^n\) (\(n\geq 1\)). Recurrence: \(a_1 = 0,\; a_{n+1} = \dfrac{a_{n-1}}{n+1}\).
Odd vanish; even: \(a_{2k} = \dfrac{a_0}{2^k k!}\). Ans: \(y = a_0 e^{x^2/2}\).
Solving IVP with Series Method
Given: \(y(0)=y_0\), \(y'(0)=y_1\). Then: \(a_0 = y_0\), \(a_1 = y_1\) directly. Use recurrence for the rest.
Wronskian Shortcut (Step 6)
By construction: \(y_1\) has \(y_1(0)=1, y_1'(0)=0\); \(y_2\) has \(y_2(0)=0, y_2'(0)=1\). So \(W[y_1,y_2](0) = \det\begin{bmatrix}1&0\\0&1\end{bmatrix} = 1\) automatically ⇒ fundamental set.
5.4 — Singular Points & Euler Equations
Singular Point Classification
Singular point: \(P(x_0)=0\) and at least one of \(Q(x_0), R(x_0) \neq 0\)
Regular singular: Both \(\lim_{x\to x_0}(x-x_0)\dfrac{Q}{P}\) and \(\lim(x-x_0)^2\dfrac{R}{P}\) are finite.
⭐ Ex: Classify \(x^2 y'' + xy' + (x^2-4)y = 0\)
\(P = x^2\), \(Q = x\), \(R = x^2-4\). Singular at \(x=0\).
Normalize: \(y'' + \dfrac{x}{x^2}y' + \dfrac{x^2-4}{x^2}y = 0\). Test limits:
\(\lim_{x\to 0}x\cdot\dfrac{x}{x^2} = \lim\dfrac{x^2}{x^2} = 1\) ✓
\(\lim_{x\to 0}x^2\cdot\dfrac{x^2-4}{x^2} = \lim(x^2-4) = -4\) ✓
Both finite ⇒ \(x=0\) is regular singular.
Ex (Problem 5.4 #14): \(x^2(1-x)y'' + (x-2)y' - 3xy = 0\)
Singular at \(x=0, 1\). At \(x=0\): \(x\cdot p = \dfrac{x-2}{x(1-x)} \to \pm\infty\) ⇒ irregular.
At \(x=1\): \((x-1)p = \dfrac{2-x}{x^2} \to 1\); \((x-1)^2 q = \dfrac{3(x-1)}{x} \to 0\). Both finite ⇒ regular.
Euler Equations: \(x^2y'' + \alpha xy' + \beta y = 0\)
Indicial equation: \(r^2 + (\alpha-1)r + \beta = 0\), \(\Delta = (\alpha-1)^2 - 4\beta\)
Case I (\(\Delta>0\)): \(y = c_1 x^{r_1} + c_2 x^{r_2}\) (distinct real \(r_1, r_2\))
Case II (\(\Delta=0\)): \(y = c_1 x^r + c_2 x^r\ln x\) (repeated root \(r\))
Case III (\(\Delta<0\)): \(y = c_1 x^\lambda\cos(\mu\ln x) + c_2 x^\lambda\sin(\mu\ln x)\) (\(r=\lambda \pm \mu i\))
Ex (Case I): \(x^2y'' - xy' - 3y = 0\)
\(\alpha=-1, \beta=-3 \Rightarrow r^2 - 2r - 3 = (r-3)(r+1) = 0 \Rightarrow r=3, -1\)
Ans: \(y = c_1 x^3 + c_2 x^{-1}\) or \(c_1 x^3 + \dfrac{c_2}{x}\)
Ex (Case II): \(x^2y'' - 5xy' + 9y = 0\)
\(\alpha=-5, \beta=9 \Rightarrow r^2 - 6r + 9 = (r-3)^2 = 0 \Rightarrow r=3\) (repeated)
Ans: \(y = (c_1 + c_2\ln x)x^3\)
Ex (Case III, full steps): \(x^2y'' + 3xy' + 2y = 0\)
Step 1: \(\alpha = 3, \beta = 2\)
Step 2: Indicial: \(r^2 + (\alpha-1)r + \beta = r^2 + 2r + 2 = 0\)
Step 3: \(\Delta = (\alpha-1)^2 - 4\beta = 4 - 8 = -4 < 0\) ⇒ Case III
Step 4: \(r = \dfrac{-2 \pm \sqrt{-4}}{2} = \dfrac{-2 \pm 2i}{2} = -1 \pm i\). So \(\lambda=-1, \mu=1\).
Step 5 (plug into Case III form): \(y = x^{\lambda}[c_1\cos(\mu\ln x) + c_2\sin(\mu\ln x)]\)
Ans: \(y = x^{-1}[c_1\cos(\ln x) + c_2\sin(\ln x)] = \dfrac{c_1\cos(\ln x) + c_2\sin(\ln x)}{x}\), \(x > 0\)
⚠ Case III extraction pitfalls:
- \(\lambda\) = REAL part of the root. Including the sign. (If \(r = -1 \pm 2i\), then \(\lambda = -1\).)
- \(\mu\) = MAGNITUDE of imaginary part (always positive). (If \(r = 3 \pm 4i\), then \(\mu = 4\), NOT \(-4\).)
- The argument of cos/sin is \(\mu\ln x\), NOT \(\mu x\). Euler equations produce \(\ln x\) inside trig; characteristic-eq equations produce \(t\) inside trig.
- Solution valid for \(x > 0\). If asked about \(x < 0\), replace \(x\) with \(|x|\) and \(\ln x\) with \(\ln|x|\).
Domain reminder: All Euler solutions are valid for \(x > 0\) (standard). Singular point at \(x = 0\). For \(x < 0\), replace \(x\) with \(|x|\) in the solution.
Euler Cases — Quick Decision Table
| \(\Delta\) | Roots | Case | Solution Form |
|---|
| \(>0\) | Real distinct \(r_1, r_2\) | I | \(c_1 x^{r_1} + c_2 x^{r_2}\) |
| \(=0\) | Repeated \(r\) | II | \((c_1+c_2\ln x)x^r\) |
| \(<0\) | Complex \(\lambda \pm \mu i\) | III | \(x^\lambda(c_1\cos(\mu\ln x) + c_2\sin(\mu\ln x))\) |
COMMON CONFUSIONS
#1. The \((n+1), (n+2)\) in series derivatives
These come from relabeling, NOT from the derivative itself.
Raw derivative: \(y' = \sum_{n=1} n\,a_n\,x^{n-1}\) (exponent drops — classic calc).
After relabeling (\(m=n-1\), rename back to \(n\)): \(y' = \sum_{n=0} (n+1)\,a_{n+1}\,x^n\).
Same series, different addressing. The \((n+1)\) is what \(m+1\) became after renaming.
#2. When sums start at different \(n\), align by peeling
If one sum starts at \(n=1\) and another at \(n=0\), evaluate the \(n=0\) summand of the \(n=0\)-starting sum and place it out front as a standalone. Then both sums start at \(n=1\) and can merge. Result: \((n=0 \text{ term}) + \sum_{n\geq 1}[\cdots]x^n = 0\).
#3. Recurrence that jumps by 2 → even/odd decouple
If your recurrence says \(a_{n+2} = f(n)\,a_n\), the even indices depend ONLY on \(a_0\) and odd indices ONLY on \(a_1\). That is why Steps 3 & 4 (plug in \((a_0,a_1)=(1,0)\) and \((0,1)\)) cleanly isolate \(y_1\) and \(y_2\).
#4. Why Wronskian \(=1\) automatically at 0
Because \(y(0)=a_0\) and \(y'(0)=a_1\) directly (plug \(x=0\) into the series). So choosing \((a_0,a_1)=(1,0)\) for \(y_1\) FORCES \(y_1(0)=1, y_1'(0)=0\), and \((0,1)\) for \(y_2\) FORCES \(y_2(0)=0, y_2'(0)=1\). The Wronskian is then \(\det\begin{bmatrix}1&0\\0&1\end{bmatrix}=1\) by construction.
#5. Ratio Test — what to divide by what
For a power series \(\sum c_k x^{2k}\) (or \(x^k\)), include the \(x\) in \(c_k\). Compute \(L(x) = \lim |c_{k+1}/c_k|\) as a function of \(x\). The radius \(R\) is the critical \(|x|\) at which \(L = 1\). If \(L = 0\) for all \(x\), then \(R = \infty\). If \(L = \infty\) for all \(x \neq 0\), then \(R = 0\).
Rule of thumb: factorial downstairs ⇒ \(R = \infty\); factorial upstairs ⇒ \(R = 0\).
#6. Spring vs. Euler — don't mix \(t\) and \(\ln x\)
Spring/mass problems: variable is \(t\), solutions involve \(\cos(\omega t), \sin(\omega t), e^{rt}\).
Euler equations: variable is \(x\), solutions involve \(x^r\), \(x^r\ln x\), \(x^{\lambda}\cos(\mu\ln x)\).
If you see \(\ln x\) inside a cosine, you are in Euler-Case-III territory; if you see \(t\), you are in char-eq territory.
#7. When to check for duplication in UC
After guessing \(y_p\), compare each term with \(y_h\). If ANY part of \(y_p\) appears in \(y_h\), multiply \(y_p\) by \(t\) (or \(t^2\) if the char eq had that root twice). Forgetting this gives contradictions when matching coefficients (e.g. \(0 = 5\)).
#8. Singular point: regular vs. irregular — how to think fast
Multiply \(p(x) = Q/P\) by \((x-x_0)\) and \(q(x) = R/P\) by \((x-x_0)^2\). If BOTH products have a finite limit at \(x_0\), regular. If either blows up, irregular. In practice: count how hard \(P\) vanishes at \(x_0\) vs. how hard \(Q, R\) vanish. If the \((x-x_0)\) and \((x-x_0)^2\) factors are enough to absorb \(P\)'s vanishing, regular.
#9. When a problem says "Verify \(a_0\) is arbitrary"
This is just asking you to confirm that nothing in the substitution/reindexing forces \(a_0\) to a specific value. In first-order problems, \(a_0\) is ALWAYS the free constant (because first-order ODEs have one arbitrary constant). In second-order problems both \(a_0\) and \(a_1\) are free.
QUICK CHECKS — Before You Finish
Red Flags:
- Series: Did you use \(\dfrac{}{}\) not \(\frac{}{}\)? Braced exponents like \(\omega^{2}\) not \(\omega^2\)?
- Euler: Extracted \(\lambda\) and \(\mu\) correctly from complex roots? Check: \(\lambda =\) real part, \(\mu = |imag|\)
- IC application: \(a_0 = y(0)\), \(a_1 = y'(0)\)? Then find higher coefficients from recurrence?
- Particular solution: Did you plug \(y_p\) back into the ODE and match all coefficients?
- Oscillations: Domain: Euler only for \(x>0\) OR \(x<0\), not both. Spring: check discriminant for damping case.
- Convergence: \(R = \dfrac{1}{\text{limit}}\), not just the limit. Check: \(|x-x_0| < R\)