Ordinary Differential Equations Cheat sheet PDF

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Formula sheet for ordinary differential equations Math 140...

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ODE Cheat Sheet First Order Equations Separable ′

=

f (x) dx + C

Linear First Order y ′ (x) + p(x)y(x) R x = f (x) µ(x) = exp p(ξ) dξ Integrating factor. ′ (µy) = f µ Exact Derivative. R  1 Solution: y(x) = µ(x) f (ξ )µ(ξ) dξ + C

Condition: My = Nx

= −h(x), then µ(y) = exp

R

Applications

Linear

Ordinary and Singular Points y ′′ + a(x)y ′ + b(x)y = 0. x0 is a Ordinary point: a(x), b(x) real analytic in |x − x0 | < R Regular singular point: (x − x0 )a(x), (x − x0 )2 b(x) have convergent Taylor series about x = x0 . Irregular singular point: Not ordinary or regular singular point.

1. Let y(x) =

Distinct, real roots: r = r1,2 , yh (x) = c1 + c2 One real root: yh (x) = (c1 + c2 x)erx Complex roots: r = α ± iβ, yh (x) = (c1 cos βx + c2 sin βx)eαx

Cauchy-Euler Equations ax2 y ′′ (x) + bxy ′ (x) + cy (x) = f (x) y(x) = xr ⇒ ar(r − 1) + br + c = 0 Cases Distinct, real roots: r = r1,2 , yh (x) = c1 xr1 + c2 xr2

Laplace Transforms Transform Pairs c

mx′′ (t)

+ kx(t) = 0 mx′′ (t) + bx′ (t) + kx(t) = 0 mx′′ (t) + bx′ (t) + kx(t) = F (t) Types of Damped Oscillation Overdamped, b2 > 4mk Critically Damped, b2 = 4mk Underdamped, b2 < 4mk

eat n

t

sin ωt cos ωt sinh at

Numerical Methods

cosh at

Euler’s Method y0 = y(x0 ), yn = yn−1 + ∆xf(xn−1 , yn−1 ),

− x0 )n+r .

4. Solve for coefficients and insert in y(x) series.

Oscillations e r2 x

c (x n=0 n

3. Find recurrence relation based on types of roots of indicial equation.

T ′ (t) = −k(T (t) − Ta )

One real root: yh (x) = (c1 + c2 ln |x|)xr Complex roots: r = α ± iβ, yh (x) = (c1 cos(β ln |x|) + c2 sin(β ln |x|))xα

6. Solve for coefficients and insert in y(x) series.

2. Obtain indicial equation r(r − 1) + a0 r + b0 .

Newton’s Law of Cooling

ay ′′ (x) + by ′ (x) + cy (x) = f (x) y(x) = erx ⇒ ar2 + br + c = 0 Cases

− a)n .

4. Collect like terms using reindexing.

P ′ (t) = kP (t) P ′ (t) = kP (t) − bP 2 (t)

Constant Coefficients

c (x n=0 n

y ′′ (x).

3. Insert expansions in DE.

Population Dynamics

a(x)y ′′ (x) + b(x)y ′ (x) + c(x)y (x) = f (x) y(x) = yh (x) + yp (x) yh (x) = c1 y1 (x) + c2 y2 (x)

y ′ (x),

Frobenius Method P∞

Free Fall x′′ (t) = −g v ′ (t) = −g + f (v)

e r1 x

4. Insert coefficients into series form for y(x).

Power Series Solution P∞

5. Find recurrence relation.

yp (x) = c1 (x)y1 (x) + c2 (x)y2 (x) c′1 (x)y1 (x) + c2′ (x)y2 (x) = 0 f (x) c′1 (x)y 1′ (x) + c2′ (x)y 2′ (x) = a(x)

Second Order Equations

f (n) (0) n!

3. Find Taylor coefficients.

Nonhomogeneous Case

Method of Variation of Parameters

h(x) dx

, cn =

2. Apply initial conditions.

2. Find

Given y1 (x) satisfies L[y] = 0, find solution of L[y ] = f as v (x) = v (x)y1 (x). z = v ′ satisfies a first order linear ODE.

Special cases R M −N If yM x = h(y), then µ(y) = exp h(y) dy

c x n=0 n

1. Differentiate DE repeatedly.

Given y1 (x) satisfies L[y] = 0, find second linearly independent solution as v(x) = v(x)y1 (x). z = v ′ satisfies a separable ODE.

µ(x, y ) (M (x, y ) dx + N (x, y ) dy ) = du(x, y ) M y = Nx   − M ∂µ − ∂N . = µ ∂M N ∂µ ∂x ∂y ∂y ∂x My −Nx N

f (x) ∼

1. Let y(x) =

Homogeneous Case

Non-Exact Form

If

Taylor Method P∞ n

Reduction of Order

Exact

0 = M (x, y) dx + N (x, y) dy Solution: u(x, y) = const where dx + ∂u dy du = ∂u ∂x ∂y ∂u = N (x, y) = M (x, y), ∂u ∂y ∂x

Series Solutions

Method of Undetermined Coefficients f (x) yp (x) an xn + · · · + a1 x + a0 An xn + · · · + A1 x + A0 aebx Aebx a cos ωx + b sin ωx A cos ωx + B sin ωx Modified Method of Undetermined Coefficients: if any term in the guess yp (x) is a solution of the homogeneous equation, then multiply the guess by xk , where k is the smallest positive integer such that no term in xk yp (x) is a solution of the homogeneous problem.

= f (x)g( R y) Ry (x) dy g(y)

Nonhomogeneous Problems

n = 1, . . . , N.

H(t − a)

δ(t − a)

c s

1 , s−a n! , sn+1 ω s2

ω2

s>a s>0

+ s s2 + ω 2 a s2 − a2 s s2 − a2 −as e , s>0 s e−as , a ≥ 0, s > 0

Laplace Transform Properties L[af(t) + bg(t)] = aF (s) + bG(s) d L[tf(t)] = − F (s) ds h i df = sF (s) − f (0) L  dt  d2 f = s2 F (s) − sf(0) − f ′ (0) L dt2 at L[e f (t)] = F (s − a) L[H(t − a)f (t − a)] = e−as F (s)

Z

L[(f ∗ g)(t)] = L[

t

0

f (t − u)g(u) du] = F (s)G(s)

Solve Initial Value Problem 1. Transform DE using initial conditions. 2. Solve for Y (s). 3. Use transform pairs, partial fraction decomposition, to obtain y(t).

Bessel Functions, Jp (x), Np (x) Gamma R ∞Functions x−1 −t e Γ(x) = 0 t Γ(x + 1) = xΓ(x).

dt,

x > 0.

Systems of Differential Equations x′

= ax + by y ′ = cx + dy. x′′ − (a + d)x′ + (ad − bc)x = 0.



x a b x = y′ c d y λt Guess x = ve ⇒ Av = λv. x′ =



≡ Ax.

Eigenvalue Problem





vi1 vi2 Form the Fundamental Matrix Solution:   v11 eλ1 t v21 eλ2 t Φ= v12 eλ1 t v22 eλ2 t General Solution: x(t) = Φ(t)C for C Find C: x0 = Φ(t0 )C ⇒ C = Φ−1 (t0 )x0 Particular Solution: x(t) = Φ(t)Φ−1 (t0 )x0 . Principal Matrix solution: Ψ(t) = Φ(t)Φ−1 (t0 ). Particular Solution: x(t) = Ψ(t)x0 . Note: Ψ′ = AΨ, Ψ(t0 ) = I. Find eigenvectors vi =

Av = λv. Find Eigenvalues: det(A − λI) = 0 Find Eigenvectors (A − λI)v = 0 for each λ. Cases Legendre Polynomials Real, Distinct Eigenvalues: x(t) = c1 eλ1 t v1 + c2 eλ2 t v2 dn (x2 − 1)n Pn (x) = 2n1n! dx n Repeated Eigenvalue: x(t) = c1 eλt v1 + c2 eλt (v2 + tv1 ), where 2 ′′ ′ (1 − x )y − 2xy + n(n + 1)y = 0. Av2 − λv2 = v1 for v2 . (n + 1)Pn+1 (x) = (2n +P 1)xPn (x) − nPn−1 (x), n = 1, 2, . . . . Complex Conjugate Eigenvalues: x(t) = ∞ g(x, t) = √ 1 P (x)tn , |x| ≤ 1, |t| < 1. = c1 Re(eαt (cos βt + i sin βt)v) + c2 Im(eαt (cos βt + i sin βt)v). n=0 n 1−2xt+t2

Special Functions

Matrix Solutions Let x′ = Ax. Find eigenvalues λi

Planar Systems

Matrix    Form ′

Solution Behavior Stable Node: λ1 , λ2 < 0. Unstable Node: λ1 , λ2 > 0. Saddle: λ1 λ2 < 0. Center: λ = iβ. Stable Focus: λ = α + iβ, α < 0. Unstable Focus: λ = α + iβ, α > 0.

x2 y ′′ + xy ′ + (x2 − p2 )y = 0.

Nonhomogeneous Matrix Solutions R t −1 Φ (s)f (s) ds x(t) = Φ(t)C + Φ(t) Rt0t −1 x(t) = Ψ(t)x0 + Ψ(t)

t0

Ψ

2 × 2 Matrix Inverse   −1 a c

b d

=

1 det A

d −c

(s)f (s) ds

−b a

...