ODE Cheat Sheet
First Order Equations
Separable
0
y
= f (x)g(y)
R (x)
R
dy
= f (x) dx + C
g(y)
Linear First Order
y 0 (x) + p(x)y(x)
R x = f (x)
µ(x) = exp
p(ξ) dξ Integrating factor.
(µy)0 = f µ
Exact Derivative.
R
1
Solution: y(x) = µ(x)
f (ξ)µ(ξ) dξ + C
Exact
0 = M (x, y) dx + N (x, y) dy
Solution: u(x, y) = const where
du = ∂u
dx + ∂u
dy
∂x
∂y
∂u
∂u
=
M
(x,
y),
= N (x, y)
∂x
∂y
Condition: My = Nx
Method of Undetermined Coefficients
Taylor Method
P∞
n
f (x)
yp (x)
an xn + · · · + a1 x + a0
A n xn + · · · + A 1 x + A 0
bx
ae
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) ∼
Special cases
R
M −N
If yM x = h(y), then µ(y) = exp h(y) dy
= −h(x), then µ(y) = exp
R
h(x) dx
Second Order Equations
Linear
a(x)y 00 (x) + b(x)y 0 (x) + c(x)y(x) = f (x)
y(x) = yh (x) + yp (x)
yh (x) = c1 y1 (x) + c2 y2 (x)
Constant Coefficients
ay 00 (x) + by 0 (x) + cy(x) = f (x)
y(x) = erx ⇒ ar2 + br + c = 0
Cases
Distinct, real roots: r = r1,2 , yh (x) = c1 er1 x + c2 er2 x
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 00 (x) + bxy 0 (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
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
, cn =
f (n) (0)
n!
1. Differentiate DE repeatedly.
2. Apply initial conditions.
3. Find Taylor coefficients.
4. Insert coefficients into series form for y(x).
Power Series Solution
P∞
c (x
n=0 n
− a)n .
2. Find y 0 (x), y 00 (x).
Homogeneous Case
3. Insert expansions in DE.
4. Collect like terms using reindexing.
5. Find recurrence relation.
Nonhomogeneous Case
Method of Variation of Parameters
yp (x) = c1 (x)y1 (x) + c2 (x)y2 (x)
c01 (x)y1 (x) + c02 (x)y2 (x) = 0
f (x)
c01 (x)y10 (x) + c02 (x)y20 (x) = a(x)
Applications
6. Solve for coefficients and insert in y(x) series.
Ordinary and Singular Points
y 00 + a(x)y 0 + 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.
Frobenius Method
P∞
Free Fall
x00 (t)
1. Let y(x) =
= −g
v 0 (t) = −g + f (v)
c (x
n=0 n
− x0 )n+r .
2. Obtain indicial equation r(r − 1) + a0 r + b0 .
Population Dynamics
3. Find recurrence relation based on types of roots of
indicial equation.
P 0 (t) = kP (t)
P 0 (t) = kP (t) − bP 2 (t)
4. Solve for coefficients and insert in y(x) series.
Newton’s Law of Cooling
Laplace Transforms
T 0 (t) = −k(T (t) − Ta )
Transform Pairs
Oscillations
c
mx00 (t)
+ kx(t) = 0
mx00 (t) + bx0 (t) + kx(t) = 0
mx00 (t) + bx0 (t) + kx(t) = F (t)
Types of Damped Oscillation
Overdamped, b2 > 4mk
Critically Damped, b2 = 4mk
Underdamped, b2 < 4mk
eat
tn
sin ωt
cos ωt
sinh at
Numerical Methods
cosh at
Euler’s Method
y0 = y(x0 ),
yn = yn−1 + ∆xf (xn−1 , yn−1 ),
c x
n=0 n
1. Let y(x) =
Reduction of Order
Given y1 (x) satisfies L[y] = 0, find solution of L[y] = f as
v(x) = v(x)y1 (x). z = v 0 satisfies a first order linear ODE.
µ(x, y) (M (x, y) dx + N (x, y) dy) = du(x, y)
My = Nx
N ∂µ
− M ∂µ
= µ ∂M
− ∂N
.
∂x
∂y
∂y
∂x
My −Nx
N
Series Solutions
Given y1 (x) satisfies L[y] = 0, find second linearly independent
solution as v(x) = v(x)y1 (x). z = v 0 satisfies a separable ODE.
Non-Exact Form
If
Nonhomogeneous Problems
H(t − a)
n = 1, . . . , N.
δ(t − a)
c
s
1
,
s−a
n!
,
sn+1
ω
s>a
s>0
s2 + ω 2
s
s2 + ω 2
a
s2 − a2
s
s2 − a2
e−as
, 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
L
= sF (s) − f (0)
dt 2
d f
L
= s2 F (s) − sf (0) − f 0 (0)
dt2
L[eat f (t)] = F (s − a)
L[H(t − a)f (t − a)] = e−as F (s)
Z
t
L[(f ∗ g)(t)] = L[
f (t − u)g(u) du] = F (s)G(s)
0
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
Γ(x) =
t
e
0
Γ(x + 1) = xΓ(x).
dt,
x > 0.
Systems of Differential Equations
Planar Systems
x0 = ax + by
y 0 = cx + dy.
x00 − (a + d)x0 + (ad − bc)x = 0.
Matrix Form
0 x
x
a b
=
y
c d
y0
Guess x = veλt ⇒ Av = λv.
x0 =
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 00 + xy 0 + (x2 − p2 )y = 0.
Matrix Solutions
Let x0 = Ax.
Find eigenvalues λi
≡ Ax.
Eigenvalue Problem
Av = λv.
Find Eigenvalues: det(A − λI) = 0
Special Functions
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
2 − 1)n
(x
Pn (x) = 2n1n! dx
n
Repeated Eigenvalue: x(t) = c1 eλt v1 + c2 eλt (v2 + tv1 ), where
(1 − x2 )y 00 − 2xy 0 + 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
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: Ψ0 = AΨ, Ψ(t0 ) = I.
Find eigenvectors vi =
Nonhomogeneous Matrix Solutions
R t −1
x(t) = Φ(t)C + Φ(t)
Φ (s)f (s) ds
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