Differential Equations Flashcards
direction fields/isoclines
set slope equal to a constant and then graph it
autonomous
not dependent on t (eg: y’ = y(2 - y))
equilibrium solution
where the slope = 0, attracts or repels solutions
first order/second order
highest order of slope is y’ or y’’
linear diff eq
highest power of y in the diff eq is 1 (just y)
ordinary diff eq
only 1 variable (typically t)
existence and uniqueness theorem for LINEAR diff eqs (first or second order)
y’ + p(t)y = g(t)
y(t0) = y
if p(t) & g(t) are continuous on (a,b) and if t0 is in (a,b), then the initial value problem has a uinque solution on (a,b)
homogeneous diff eq
diff eq that equals 0 (g(t) = 0)
integrating factor
u(t) = e^(∫p(t)dt)
1) multiply both sides of the equation by this factor
2) the left side can use reverse power rule with derivatives
3) take integral of this statement
4) solve for y and constants of integration
mixture problems
Q(t) = amount of salt [lb]
V(t) = amount of liquid [gal]
ri(t) = rate in [gal/min]
ro(t) = rate out [gal/min]
ci(t) = concentration in [lb/gal]
co(t) = concentration out [lb/gal]
co = Q(t)/V(t)
dQ/dt = rici - roco = rici - ro[Q(t)/V(t)]
+ integrate using integrating factor u(t)
Newton’s Law of Cooling
θ(t) = temp of object
s(t) = temp of surroundings
θ’(t) = k[s(t) - θ(t)]
+ integrate as normal, possibly with integrating factor or separation of variables, use initial conditions to solve for k and constants of integration
Malthusian Model
dP/dt = kP
P = P0e^(kt)
+ k is the growth rate
OR
dP/dt = kP - R
+ R is the decrease rate
Radioactive Decay
dQ/dt = -kQ
Q = Ce^(-kt) = Q0e^(-kt)
+ h = half-life, when Q0/2 = Q
+ h = ln(2)/k
Existence and Uniqueness Theorem for Non-Linear ODE’s
y’ = f(t,y)
y(t0) = y0
a ≤ t ≤ b
if f(t,y) and ∂f/∂y are continuous on R, then there is an open interval (c,d) contained in (a,b) in which the initial value problem has a unique solution
(not necessarily all of (a,b) contains a unique solution)
Separable First Order Diff Eq’s
1) Separate y’s from t’s
2) Integrate
3) Plug & Chug to find constant(s) of integration
Logistic Model
dP/dt = r(1 - P(t)/Pe)*P(t)
+ Pe = equlibrium population, where dP/dt = 0
+ r ≠ 0 since it’s a constant
+ lim p(t) as t -> ∞ = Pe
P(t) = P0Pe/[P0 - (P0 - Pe)e^(-rt)]
Falling Object w/Air Resistance
m*dv/dt = -mg - kv
0 = -g - kv/m
v = -gm/k = termminal velocity, top velocity
dv/dt + kv/m = -g
v’ + kv/m = -g
…
v = -gm/k + [v0 + gm/k]*e^(-kt/m)
Euler’s Method (approximation method)
y(k+1) = y(k) + f(t(k),y(k))*h
+ h = step size
+ y’ = f(t,y)
∫dx/x =
ln|x| + C
∫a^(x)dx =
e^(x)/ln(a) + C
∫ln(x)dx =
x(ln(x)) - x + C
∫sin(x)dx =
-cos(x) + C
∫cos(x)dx =
sin(x) + C
∫tan(x)dx =
ln|sec(x)| + C
or
-ln|cos(x)| + C
∫cot(x)dx =
ln|sin(x)| + C
∫sec(x)dx =
ln|sec(x) + tan(x)| + C
∫csc(x)dx =
ln|csc(x) - cot(x)| + C
∫sec^2(x)dx =
tan(x) + C
∫sec(x)tan(x)dx =
sec(x) + C
∫csc^2(x)dx =
-cot(x) + C
∫tan^2(x)dx =
tan(x) - x + C
∫dx/√(a^2 - x^2) =
sin^-1(x/a) + C
∫dx/(a^2 + x^2) =
(1/a) tan^-1(x/a) + C
∫dx/[|x|√(x^2 - 1)] =
sec^-1(x) + C
integration by parts formula
∫udv = uv - ∫vdu
partial fractions
given:
[x^2 + 3]/[x^2 - 3x - 4] =
A/[x - 4] + B/[x + 1]
Set A(x + 1) + B(x - 4) = x^2 + 3
Principle of Superposition
If y1 & y2 are both soutions to y” + p(t)y’ + q(t)y = 0, then c1y1 + c2y2 is also a solution
Fundamental Set of Solutions
If any solution can be written as a linear combo of y1 & y2, then for any solution:
y(t) = c1y1(t) + c2y2(t)
to prove something is a _________ of a diff eq, just plug it into the equation and simplify
solution
Wronskian
W(y1,y2) = det(matrix formed by y1, y2, and their derivatives)
if y1 & y2 form a fundamental set of solutions, W(y1,y2) ≠ 0
Characteristic Polynomial (and general solution)
ay” + by’ + cy = 0
aλ^2 + bλ + c = 0
y = C1e^(λ1t) + C2e^(λ2t)
Repeated roots (w/characteristic polynomial)
just add a t
y = C1e^(λt) + C1te^(λt)
reduction of order
(given one solution)
y2(t) = u*y1
y2(t)’ = …
y2(t)” = …
plug above values into the diff eq
simplify until you get one u” and u’
set u’ = v
integrate to find v
integrate v to find u
the constant of integration & fractions out front don’t matter because it’s for a fundamental set of solutions
imaginary roots (for characteristic polynomial)
λ = α ± βi
y = e^(αt)[Acos(βt) + Bsin(βt)]
OR
y = Re^(αt)cos(βt - δ)
R = √(A^2 + B^2)
δ = tan^-1(B/A)
General Solution of a Linear, Non-Homogeneous Equation
y = yc + yp
+ to find yc, set diff eq = 0, find characteristic polynomial and lambda
+ to find yp (particular solution) follow method of undetermined coefficients
Method of Undetermined Coefficients
+ way to find yp given non-homogeneous constant coefficient linear diff eq
accepted forms:
1) = 7e^(2t) –> yp = Ae^(2t)
2) = 5t^2 –> yp = At^2 + Bt + C
3) = sin(2t) –> asin(2t) + bcos(2t)
+ if a lambda root matches particular solution, must multiply yp part corresponding to that by t
+ to find the vars (a,b,etc), plug yp inot the diff eq and solve
ex 1: = t^2e^(3t) –> [A2t^2 + A1t + A0]e^(3t)
ex 2: = te^(2t)cos(t) –> yp = [A1t + A0]e^(2t)sin(t) + [B1t + B0]e^(2t)*cos(t)
Method of Variation of Parameters
what if the diff eq isn’t a constant coefficient equation (generally just means = g(t) isn’t a constant)
yp = -y1∫[y2g(t)/W]dt + y2∫[y1g(t)/W]dt
Abel’s Theorem (for second order diff eq)
W(t) = W0*e^[-∫p(t)dt]
(integral from t0 –> t)
Unforced Mechanical Systems (no damping)
F = -kx
y = mg/k
ω^2 = k/m
y” + (ω^2)y = 0
y = Acos(ωt) + Bsin(ωt)
Unforced Mechanical Systems (dampening force added)
General Equation
Fd = -γy’
F(t) = my” + γy’ + ky
Unforced Mechanical Systems (dampening force added)
Case: damping is strong
γ^2 > 4mk –> two real roots
y = c1e^(λ1t) + c2e^(λ2t)
λ1 & λ2 < 0
+ no oscillation
Unforced Mechanical Systems (dampening force added)
Case: critically damped
γ^2 = 4mk –> λ = -γ/[2m]
y = c1e^(-γt/[2m]) + c2te^(-γt/[2m])
+ no oscillation
+ approaches 0 fastest
Unforced Mechanical Systems (dampening force added)
Case: underdamped
γ^2 < 4mk –> imaginary roots
y(t) = e^(αt)[c1cos(βt) + c2sin(βt)]
α = -γ/[2m]
Mechanical Systems + External Force
my” + γy’ + ky = Fα(t)
+ assume γ = 0 (no damping)
y” + ω0^2*y = fα(t) = Fcos(ω1t)
Mechanical Systems + External Force
Case 1: ω0 ≠ ω1
y = (F/[(ω1)^2 - (ω0)^2])cos(ω0t) + F/[(ω0)^2 - (ω1)^2]cos(ω1t)
Mechanical Systems + External Force
Case 2: ω0 = ω1
y = (Ft/[2ω0])*sin(ω0t)
Higher Order Homogeneous Constant Coefficient Diff Eq
+ use characteristic polynomial, long division if necessary
+ algebraic multiplicity comes into play here too (if going in the reverse direction)
Ask him about the process for imaginary roots with roots greater than 2 or 3
A^-1 =
(if n = 2)
1/[ad-bc] [d -b, -c a]
[A(t) + B(t)]’ =
A’(t) + B’(t)
[f(t)A(t)]’ =
+ f(t) = scalar function
+ A(t) = matrix function
f’(t)A(t) + f(t)A’(t)
[A(t)B(t)]’ =
A’(t)B(t) + A(t)B’(t) ≠ [B(t)A(t)]’
Existence and Uniqueness Theorem for Matrices
Each element of the matrix must be continuous on (a,b) for a unique solution to exist on (a,b)
Rewriting an nth order scalar linear diff eq as a matrix
y1 = y
y2 = y’
y3 = y”
y1’ = y2
y2’ = y3
y3 = y’’’
Place in format of y’ = Ay + g(t)
Homogenous Lineaer Systems
y1,…,yn is a fundamental set of solutions if every solution of y = Ay’ can be written as a linear combo of y1(t),…,yn(t)
Ψ =
solution matrix of y’ = Ay
= [y1, … , yn]
det Ψ =
W (wronskian)
if detΨ ≠ 0, we can find c1, c2, … , cn
Abel’s Theorem (for matrices)
Let y1, y2,…,yn be a set of solutions of y’ = P(t)y, a < t < b, then:
W(t) = W0e^(∫[tr(P(s))]ds)
integral from t0 –> t
trace of a matrix
sum of the diagonal elements
eigenvectors and eigenvalues (general process)
1) det(A - λI) to find λ
2) plug λ into [A - λI]x = 0
3) find x to get the eigenvector
y = c1e^(λ1t)x1 + c2e^(λ2t)x2
How to determine if we have a fundamental set of solutions (for a matrix system)
Use W(t0) to determine if we have a fundamental set of solutions (W(t0) ≠ 0)
In these two cases, A has a full set of eigenvectors:
1) if A has n distinct eigenvalues, then A has a full set of eigenvectors
2) if A is a symmetric, real matrix, then A has a full set of eigenvectors
Note: A^T = A –> symmetric (change cols to rows, rows to cols)
complex eigenvalues
(how to find the corresponding eigenvectors)
λ = α ± βi
1) plug into [A - λI]x = 0 to find x
2) y = e^[(α + βi)t] = e^(αt)e^(βit)x = e^(αt)[cos(βt) + isin(βt)]x
3) multiply into x, separate into 2 vectors (1 real, one imaginary, then pull out i)
4) y1 + iy2
only one eigenvalue is necessary to find x
repeated eigenvalues (only for n = 2)
general process
|A - λI| = |(α - λ) β, 0 (α - λ)| = (α - λ)^2 –> λ = α
case 1:
β = 0
+ x1 can be anything
+ x2 can be anything
y = c1e^(αt)[1, 0] + c2e^(αt)[0, 1]
case 2:
β ≠ 0
+ x2 = 0
+ x1 can be anything
y1 = e^(λt)v1
y2 = te^(λt)v1 + e^(λt)v2
how to find the generalized eigenvector (v2) with repeated roots
[A - λI]v2 = v1
Non-homogeneous linear systems
y(t) = Ψ(t)Ψ^-1(t0)y0 + Ψ(t)∫Ψ^-1(s)g(s)ds
for y’ = P(t)y + g(t)
integral from t0 –> t
equilibrium solutions & direction fields (phase plane)
ye is the equilibrium solution that makes all y1’,y2’,…,yn’ = 0
if n = 2:
x’ = f(x,y)
y’ = g(x,y)
m = g(x,y)/f(x,y)
+ if g(x,y) = 0 –> horizontal tangent line
+ if f(x,y) = 0 –> vertical tangent line
+ if g(x,y) & f(x,y) = 0 –> equilibrium point (exist where the nullclines cross)
stability test
the autonomous linear system y’ = Ay (A is an invertible, 2x2 matrix) has a unique equilibrium point which is ye = 0
the equilibrium point is:
a) asymptotically stable if the eigenvalues of A have negative, real parts
b) stable but not asymptotically stable if the eigenvalues of A are purely imaginary
c) unstable if at least one eigenvalue has a positive, real part
Jacobian =
given:
x’ = f(x,y)
y’ = g(x,y)
ye = (xe, ye)
J = [∂f(xe,ye)/∂x ∂f(xe,ye)/∂y, ∂g(xe,ye)/∂x ∂g(xe,ye)/∂y]
linearization and equilibrium point classification
let y’ = f(y) be a 2-dimensional, autonomous system that is almost linear at an equilibrium point y = ye
let z’ = Jz be the corresponding lienarized system
a) if z = 0 is an asymptotically stable equilibrium point of z’ = Az, then ye = y is an asymptotically stable equilibrium point of y’ = f(y)
b) if z = 0 is an unstable equilibrium point of z’ = Az, then y = ye is an unstable equilibrium point of y’ = f(y)
c) if z = 0 is a stable equilibrium point of z’ = Az, then no conclusion on stability of y = ye
graphing cases
purely imaginary eigenvalues
center, circle
graphing cases
both eigenvalues are complex w/negative, real part
spiral point (spirals into the point)
graphing cases
both eigenvalues are copmlex with positive, real parts
spiral point (spirals away from the point)
graphing cases
both eigenvalues are real & negative
node (lines pointing toward it)
graphing cases
both eigenvalues are real & positive
node (lines pointing away from it)
grahping cases
λ1 > 0
λ2 < 0
(or vice versa)
saddle point
