MAT244 Review

Flashcard Icon

Flashcard

Learn Mode Icon

Learn Mode

Match Icon

Match

Coming Soon!
Library Icon

Library

View Library
Match Icon

Create

Create More Decks
Flashcard Icon Flashcards
Flashcard Icon Flashcards
Library Icon Library
Match Icon Match (Coming Soon)

Calculus - Geometry

View Results
Full Screen Icon

user_jevbwl Created by 6 mon ago

Cards in this deck(31)
Rearrange so that N(y)dy = M(x)dx - Integrate both sides and solve
Blur Image
Rearrange so that (1/y)dy = -p(t)dt - Integrate on both sides and solve
Blur Image
Find u(t) such that u'(t) = u(t)p(t) - Multiply both sides of the equation by u(t) - Simplify and integrate
Blur Image
Find the solution of y' + p(t)y = 0 - Replace the unknown constant C with the unknown equation u - Sub homogenous solution into inhomogenous equation and solve for u - Plug u value into homogenous solution to find the inhomogenous solution
Blur Image
Divide both sides by yⁿ - Sub into the equation z = y¹⁻ⁿ and z' = y'*(y¹⁻ⁿ)' - Solve as you would an inhomogenous equation - Sub y¹⁻ⁿ back into general solution
Blur Image
Sub ux into y to get (ux)' = f(u) - Solve new equation as a seperable equation for x - To find the solution for y, multiply x's solution by u
Blur Image
Rearrange so that dy = f(x)dx - Integrate both sides and solve
Blur Image
An equation M(x,y)dx + N(x,y)dy = 0 is exact if and only if Mdy = Ndx
Blur Image
Integrate M(x,y) by x and N(x,y) by y, but replace the unknown constance C1 and C2 with the unknown equations f and g - Find the values f and g such that both integrals equal eachother - The similar equation + C is the solution
Blur Image
If [Ndx - Mdy]/M is a function g(y), then the integrating factor is the function u(y) such that ln(u(y)) = ∫g(y)dy
Blur Image
If -[Ndx - Mdy]/N is a function g(x), then the integrating factor is the function u(x) such that ln(u(x)) = ∫g(x)dy
Blur Image
If [Ndx - Mdy]/[M*x - N*y] is a function g(xy), then the integrating factor is the function u(xy) such that ln(u(t)) - ∫g(t)dt where t=xy
Blur Image
y = (C₁+C₂t)eᵏᵗ
Blur Image
y = C₁eᵏ¹ᵗ+ C₂eᵏ²ᵗ
Blur Image
y=eᵃᵗ(C₁cos(Bt)+ C₂sin(Bt))
Blur Image
y = C₁y₁ + C₂y₂ for some constants C₁,C₂ and solutions y₁,y₂
Blur Image
W[y₁,y₂] = det| y₁ y₂ | | y'₁ y'₂ |
Blur Image
(d/dt)W[y₁,y₂] = (-a1/a0)W[y₁,y₂]
Blur Image
You must convert it into a Indicial Equation: ak² + (b-a)k + c = 0
Blur Image
y = (C₁+C₂ln(x))xᵏ
Blur Image
y = C₁xᵏ¹+ C₂xᵏ²
Blur Image
y=xᵃ(C₁cos(Bln(x))+ C₂sin(Bln(x)))
Blur Image
y = y???? + yₚ, with y???? being the general solution for a0(t)y''+a1(t)y'+a2(t)y=g(t) and yₚ being the particular solution of the original inhomogenous equation
Blur Image
Ax²+Bx+C
Blur Image
Aeᵈᵗ
Blur Image
Ateᵈᵗ
Blur Image
Begin with Ag(t) and change according to findings
Blur Image
Solve for the general solution of a0(t)y''+a1(t)y'+a2(t)y=0 and write it as y = u₁y₁+u₂y₂ for some unknown function u₁ and u₂, and some solutions y₁ and y₂ - Solve for the Wronskian W[y₁,y₂] - Solve for u₁ = -∫g(t)y₂(t)/W dt - Solve for u₂ = ∫g(t)y₁(t)/W dt - Sub u functions into general solution of homogenous equation to find the inhomogenous general solution
Blur Image
Blur Image
Blur Image
y₂ = y₁∫-Wy₁⁻²
Blur Image

Ask Our AI Tutor

Get Instant Help with Your Questions

Need help understanding a concept or solving a problem? Type your question below, and our AI tutor will provide a personalized answer in real-time!

How it works

  • Ask any academic question, and our AI tutor will respond instantly with explanations, solutions, or examples.
Flashcard Icon
  • Browse questions and discover topic-based flashcards
  • Practice with engaging flashcards designed for each subject
  • Strengthen memory with concise, effective learning tools