Download Study Sheet: Solving & Analyzing First Order Equations & Systems and more Lecture notes Mathematics in PDF only on Docsity!
Differential Equations Study Sheet
Matthew Chesnes
It’s all about the Mathematics!
Kenyon College
Exam date: May 11, 2000
6:30 P.M.
1 First Order Differential Equations
- Differential equations can be used to explain and predict new facts for about everything that changes continuously.
d^2 x dt^2
dx dt
- t is the independent variable, x is the dependent variable, a and k are parameters.
- The order of a differential equation is the highest deriviative in the equation.
- A differential equation is linear if it is linear in parameters such that the coefficients on each deriviative of y term is a function of the independent variable (t).
- Solutions: Explicit → Written as a function of the independent variable. Implicit → Written as a function of both y and t. (defines one or more explicit solutions.
1.1 Population Model
- Model: dP dt = kP.
- Equilibrium solution occurs when dP dt
- Solution: P (t) = Ae(kt).
- If k > 0, then limt→∞ P (t) = ∞. If k < 0, then limt→∞ P (t) = 0.
- Redefine model so it doesn’t blow up to infinity.
- dP dt = kP (1 −
P
N
- N is the carrying capacity of the population.
1.2 Seperation of Variables Technique
dy dt = g(t)h(y).
h(y) dy = g(t)dt.
- Integrate both sides and solve for y.
- You might lose the solution h(y) = 0.
1.7 Bifurcations
- Bifurcations occur at parameters where the equilibrium profile changes.
- Draw phase lines (y) for several values of a.
1.8 Linear Differential Equations and Integrating Factors
- Properties of Linear DE: If yp and yh are both solutions to a differential equation, (particular and homogeneous), then yp + yh is also a solution.
- Using the integrating factor to solve linear differential equations such that dy dt
+P (t)y = f (t).
- The integrating factor is therefore e(
∫ (^) P (t)dt) .
- Multiply both sides by the integrating factor.
- e(
∫ (^) P (t)dt) dy dt
∫ (^) P (t)dt) P (t)y = e(
∫ (^) P (t)dt) f (t).
- then via chain rule ...
- d{e(
∫ (^) P (t)dt) y} dt
((Integrating factor * y))= e(
∫ (^) P (t)dt) f (t).
- Then integrate to find solution.
1.9 Integration by Parts
udv = uv −
vdu.
2 Systems
dx dt = ax − bxy,
dy dt = −cy + dxy.
- Equilibrium occurs when both differential equations are equal to zero.
- a and c are growth effects and b and d are interaction effects.
- To verify that x(t), y(t) is a solution to a system, take the deriviative of each and compare them to the originial differerential equations with x and y plugged in.
- Converting a second order differential equation, d^2 y dt^2 = y. Let v = dy dt . Thus dv = d^2 y dt
2.1 Vector Notation
- A system of the form dx dt = ax + bxy and dy dt = cy + exy can be written in vector notation.
- d dt P(t) =
dx dt dy dt
[
ax + bxy cy + exy
]
2.2 Decoupled System
- Completely decoupled: dx dt
= f (x), dy dt
= g(y).
- Partially decoupled: dx dt = f (x), dy dt = g(x, y).
3.2 Stability
Consider a linear 2 dimensional system with two nonzero, real, distinct eigenvalues, λ 1 and λ 2.
- If both eigenvalues are positive then the origin is a source (unstable).
- If both eigenvalues are negative then the origin is a sink (stable).
- If the eigenvalues have different signs, then the origin is a saddle (unstable).
3.3 Complex Eigenvalues
- Euler’s Formula: ea+ib^ = eaeib = eacos(b) + ieasin(b).
- Given real and complex parts of a solution, the two parts can be treated as seperate independent solutions and used in the linearization theorem to determine the general solution.
- Stability: consider a linear two dimensional system with complex eigenvalues λ 1 = a+ib and λ 2 = a − ib. - If a is negative then solution spiral towards the origin (spiral sink). - If a is positive then the solutions spiral away from the origin (spiral source). - If a = 0 the solutions are periodic closed paths (neutral centers).
3.4 Repeated Eigenvalues
- Given the system, d dt X = AX with one repeated eigenvalue, λ 1.
- If V1 is an eigenvector, then X 1 (t) = eλtV 1 is a straight line solution.
- Another solution is of the form X 2 (t) = eλt(tV 1 + V 2 ).
- Where V 1 = (A − λI)V 2.
- X 1 and X 2 will be independent and the general solution is formed in the usual manner.
3.5 Zero as an Eigenvalue
- If zero is an eigenvector, nothing changes but the form of the general solution is now
X(t) = k 1 V 1 + k 2 eλ^2 tV 2.
4 Second Order Differential Equations
- Form: d^2 y dt^2 + p(t) dy dt = q(t)y = f (t).
- Homogeneous if f (t) = 0.
- given solutions y 1 and y 2 to the 2nd order differential equation, you must check the Wronskian if both solutions are from real roots of the characteristic.
- W = det
[
y 1 y 2 y′ 1 y 2 ′
]
- If W is equal to 0 anywhere on the interval of consideration, then y 1 and y 2 are not linearly independent.
- General solution given y 1 and y 2 is found as usual by the linearization theorem.
- Characteristic polynomial of a 2nd order with constant coefficients: as^2 + bs + c = 0.
- Solutions of the form y(t) = est.
- s = − b 2 a
b^2 − 4 ac 2 a
- if b^2 − 4 ac > 0, then two distinct real roots.
- if b^2 − 4 ac < 0, then complex roots.
- b^2 − 4 ac = 0, then repeated real roots.
4.1 Two real distinct Roots
- Two real roots, s 1 and s 2.
- General solution = y(t) = k 1 es^1 t^ + k 2 es−^2 t.
4.2 Complex Roots
- Complex Roots, s 1 = p + iq and s 2 = p − iq.
- General solution = y(t) = k 1 eptcos(qt) + k 2 eptsin(qt).
4.3 Repeated Roots
- Repeated Root, s 1.
- General solution = y(t) = k 1 e
− b 2 a t
− b a 2 t .
5 LaPlace Transformations
0 e −stf (t)dt = limT →∞^ ∫^ T 0 e −stf (t)dt.
- ONLY PROVIDED THAT THE INTEGRAL CONVERGES!!! MUST BE OF EX- PONENTIAL ORDER!!!
- L{f (t)} = F (s).
- L{ 1 } =
s
s^2
s − a
ω s^2 + ω^2
- Linear: L{αf (t) + βg(t)} = αF (s) + βG(s).
5.1 Inverse Laplace Transforms
- Linear: L−^1 {αF (s) + βG(s)} = αf (t) + βg(t).
5.2 Transform of a derivative
- L{f ′(t)} = sL(f (t) − f (0).
- L{f ′′(t)} = s^2 L(f (t) − sf (0) − f ′(0).