Summary
Summary Differential Equations Cheat Sheet
- Module
- Institution
- Book
A one page summary for Math244 Analysis
[Show more]
Connected book
Book Title:
Author(s):
- Edition:
- ISBN:
- Edition:
More summaries for
-
Multivariable Calculus Notes: Partial Derivatives
-
MAT1512 ASSIGNMENT 2 2023 SOLUTIONS
-
AP Calculus Review: U1 Limits and Continuity
Seller
Follow
FreshStick6969
Reviews received
Content preview
ODE Cheat Sheet Nonhomogeneous Problems Series SolutionsMethod of Undetermined Coefficients Taylor Method
First Order Equations f (x) yp (x) P∞
f (x) ∼ n c x , cn =
f (n) (0)
an xn + · · · + a1 x + a0 A n xn + · · · + A 1 x + A 0 n=0 n n!
Separable ae bx Aebx 1. Differentiate DE repeatedly.
y 0 a cos ωx + b sin ωx A cos ωx + B sin ωx
R (x)dy
= f (x)g(y)
R 2. Apply initial conditions.
g(y)
= f (x) dx + C Modified Method of Undetermined Coefficients: if any
term in the guess yp (x) is a solution of the homogeneous 3. Find Taylor coefficients.
Linear First Order equation, then multiply the guess by xk , where k is the 4. Insert coefficients into series form for y(x).
y 0 (x) + p(x)y(x) smallest positive integer such that no term in xk yp (x) is a
R x = f (x) solution of the homogeneous problem. Power Series Solution
µ(x) = exp p(ξ) dξ Integrating factor. P∞
(µy)0 = f µ Exact Derivative. 1. Let y(x) = c (x
n=0 n
− a)n .
R Reduction of Order
1
Solution: y(x) = µ(x) f (ξ)µ(ξ) dξ + C 2. Find y 0 (x), y 00 (x).
Homogeneous Case
Exact 3. Insert expansions in DE.
Given y1 (x) satisfies L[y] = 0, find second linearly independent
0 = M (x, y) dx + N (x, y) dy solution as v(x) = v(x)y1 (x). z = v 0 satisfies a separable ODE. 4. Collect like terms using reindexing.
Solution: u(x, y) = const where
Condition: My = Nx Nonhomogeneous Case 5. Find recurrence relation.
du = ∂u
∂x
dx + ∂u∂y
dy
∂u
= M (x, y), ∂u
= N (x, y) 6. Solve for coefficients and insert in y(x) series.
∂x ∂y 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.
Non-Exact Form Ordinary and Singular Points
µ(x, y) (M (x, y) dx + N (x, y) dy) = du(x, y) Method of Variation of Parameters y 00 + a(x)y 0 + b(x)y = 0. x0 is a
My = Nx yp (x) = c1 (x)y1 (x) + c2 (x)y2 (x) Ordinary point: a(x), b(x) real analytic in |x − x0 | < R
c01 (x)y1 (x) + c02 (x)y2 (x) = 0 Regular singular point: (x − x0 )a(x), (x − x0 )2 b(x) have
N ∂µ
∂x
− M ∂µ
∂y
= µ ∂M ∂y
− ∂N∂x
.
f (x)
c01 (x)y10 (x) + c02 (x)y20 (x) = a(x) convergent Taylor series about x = x0 .
Special cases
M −N R Irregular singular point: Not ordinary or regular singular
If yM x = h(y), then µ(y) = exp h(y) dy point.
My −Nx R Applications
If N
= −h(x), then µ(y) = exp h(x) dx
Free Fall Frobenius Method
P∞
Second Order Equations x00 (t) = −g 1. Let y(x) = c (x
n=0 n
− x0 )n+r .
v 0 (t) = −g + f (v)
Linear 2. Obtain indicial equation r(r − 1) + a0 r + b0 .
a(x)y 00 (x) + b(x)y 0 (x) + c(x)y(x) = f (x) Population Dynamics 3. Find recurrence relation based on types of roots of
y(x) = yh (x) + yp (x) P 0 (t) = kP (t) indicial equation.
yh (x) = c1 y1 (x) + c2 y2 (x) P 0 (t) = kP (t) − bP 2 (t) 4. Solve for coefficients and insert in y(x) series.
Constant Coefficients Newton’s Law of Cooling
ay 00 (x) + by 0 (x) + cy(x) = f (x) Laplace Transforms
T 0 (t) = −k(T (t) − Ta )
y(x) = erx ⇒ ar2 + br + c = 0 Transform Pairs
Cases Oscillations c
c
Distinct, real roots: r = r1,2 , yh (x) = c1 er1 x + c2 er2 x mx00 (t) + kx(t) = 0 s
1
One real root: yh (x) = (c1 + c2 x)erx mx00 (t) + bx0 (t) + kx(t) = 0 eat , s>a
s−a
Complex roots: r = α ± iβ, yh (x) = (c1 cos βx + c2 sin βx)eαx mx00 (t) + bx0 (t) + kx(t) = F (t) n!
tn , s>0
Types of Damped Oscillation sn+1
ω
Cauchy-Euler Equations Overdamped, b2 > 4mk sin ωt
s2 + ω 2
ax2 y 00 (x) + bxy 0 (x) + cy(x) = f (x) Critically Damped, b2 = 4mk s
cos ωt
y(x) = xr ⇒ ar(r − 1) + br + c = 0 Underdamped, b2 < 4mk s2 + ω 2
a
Cases sinh at
s2 − a2
Distinct, real roots: r = r1,2 , yh (x) = c1 xr1 + c2 xr2 Numerical Methods cosh at
s
One real root: yh (x) = (c1 + c2 ln |x|)xr s2 − a2
Euler’s Method e−as
Complex roots: r = α ± iβ, y0 = y(x0 ), H(t − a) , s>0
s
yh (x) = (c1 cos(β ln |x|) + c2 sin(β ln |x|))xα yn = yn−1 + ∆xf (xn−1 , yn−1 ), n = 1, . . . , N. δ(t − a) e−as , a ≥ 0, s > 0
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller FreshStick6969. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for £3.38. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
76669 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy revision notes and other study material for 14 years now