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Summary Calculus I
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This document summarizes everything (including examples) for the exam Calculus-IB for the study Aerospace Engineering.
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WI1421LR Calculus ICalculus I
WI1421LR
Delft University of Technology
1
, WI1421LR Calculus I
Table of Contents
Second-order differential equations……………………………………………………………………………………………………3
- Homogeneous equations……………………………………………………………………………………………………….3
- Nonhomogeneous equations…………………………………………………………………………………………………3
Series………………………………………………………………………...……………………………………………………………………….4
The integral test…………………….……………………………………………………….…………………………………………………..5
A p-series……………………………………………………………………………………………………………………………………………5
The comparison tests………………………………………………………………………………………………………………………….5
- The limit comparison test……………………………………………………………………………………………………….5
Alternating series………………………………………………………………………………………………………………………………..5
Absolute convergence………………………………………………………………………………………………………………………..6
The ratio test………………………………………………………………………………………………………………………………………6
Power series and radius of convergence………………………………………………………………………………………………6
Differentiation and integration of power series…………………………………………………………………………………..6
Taylor and Maclaurin series………………………………………………………………………………………………………………..7
The Binomial series……………………………………………………………………………………………………..………………………7
Series solutions………………………………………………………………………………………………………………………….……….8
Taylor series in limits…………………………………………………………………………………………………………………..………8
Arc lengths………………………………………………………………………………………………………………………………………….9
Partial derivatives……………………………………………………………………………………………………………….………………9
Tangent planes…………………………………………………………………………………………………………………….…………..10
Linear approximations…………………………………………………………………………………………………………….………..10
Limits………………………………………………………………………………………………………………………………………………..10
2
, WI1421LR Calculus I
Second-order differential equations
A second-order differential equation has the form
𝑃(𝑥)𝑦 ′′ (𝑥) + 𝑄(𝑥)𝑦 ′ (𝑥) + 𝑅(𝑥)𝑦(𝑥) = 𝐺(𝑥)
If 𝐺(𝑥) = 0 it is homogeneous, otherwise it is non-homogeneous.
Homogeneous equations
Second-order homogeneous equations have the form
𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0
We use the general solution
𝑦 = 𝑒 𝑟𝑥
This gives:
𝑦 ′ = 𝑟𝑒 𝑟𝑥 ; 𝑦 ′′ (𝑥)𝑟 2 𝑒 𝑟𝑥
We can find 𝑟 by substituting these into the formula:
𝑎𝑟 2 𝑒 𝑟𝑥 + 𝑏𝑟𝑒 𝑟𝑥 + 𝑐𝑒 𝑟𝑥 = 0 → 𝑎𝑟 2 + 𝑏𝑟 + 𝑐 = 0
Solving this equation gives a value for 𝑟 (or two). The solutions are thus:
𝑦 = 𝑒 𝑟1 𝑥 ; 𝑦 = 𝑒 𝑟2 𝑥
However, multiples of these are also solutions, giving:
𝑦 = 𝑐1 𝑒 𝑟1 𝑥 ; 𝑦 = 𝑐2 𝑒 𝑟2 𝑥
This is the case for 𝑟1 ≠ 𝑟2 . We distinguish between 3 cases:
- 𝑟1 and 𝑟2 are real and 𝑟1 ≠ 𝑟2 → 𝑦(𝑥) = 𝑐1 𝑒 𝑟1 𝑥 + 𝑐2 𝑒 𝑟2 𝑥
- 𝑟1 and 𝑟2 are real and 𝑟1 = 𝑟2 → 𝑦(𝑥) = 𝑐1 𝑒 𝑟𝑥 + 𝑐2 𝑥𝑒 𝑟𝑥
- 𝑟1 and 𝑟2 are nonreal and 𝑟1,2 = 𝛼 ∓ 𝑖𝛽 → 𝑦(𝑥) = 𝑐1 𝑒 𝛼𝑥 cos(𝛽𝑥) + 𝑐2 𝑒 𝛼𝑥 sin(𝛽𝑥)
Example:
Find (if possible) the solution of the boundary-value problem
𝑦 ′′ (𝑥) + 4𝑦 ′ (𝑥) + 4𝑦(𝑥) = 0, 𝑦(0) = 2, 𝑦(1) = 0
Auxiliary equation: 𝑟 + 4𝑟 + 4 = 0 → (𝑟 + 2)2 = 0 → 𝑟 = −2
2
This gives a general solution: 𝑦(𝑥) = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑥𝑒 −2𝑥
The condition 𝑦(0) = 2 → 𝑐1 = 2
Then the condition 𝑦(1) = 0 → 𝑐2 = −𝑐1 = −2
Thus the (unique) solution of the boundary-value problem is 𝑦(𝑥) = 2𝑒 −2𝑥 − 2𝑥𝑒 −2𝑥
Nonhomogeneous equations
A nonhomogeneous differential equation has the form
𝑃(𝑥)𝑦 ′′ (𝑥) + 𝑄(𝑥)𝑦 ′ (𝑥) + 𝑅(𝑥)𝑦(𝑥) = 𝐺(𝑥)
The general solution can be written as
𝑦(𝑥) = 𝑦𝑝 (𝑥) + 𝑦𝑐 (𝑥)
- 𝑦𝑝 : particular solution
- 𝑦𝑐 : solution of the complementary problem:
𝑃(𝑥)𝑦 ′′ (𝑥) + 𝑄(𝑥)𝑦 ′ (𝑥) + 𝑅(𝑥)𝑦(𝑥) = 0
We use the method of undetermined coefficients to find the particular solution:
- 𝐺(𝑥) = polynomial, use 𝑦𝑝 (𝑥) = a polynomial of the same degree as 𝐺
- 𝐺(𝑥) = the form of 𝑐𝑒 𝑘𝑥 , use 𝑦𝑝 (𝑥) = 𝐴𝑒 𝑘𝑥
- 𝐺(𝑥) = 𝐶 cos(𝑘𝑥) or 𝐶 sin(𝑘𝑥), use 𝑦𝑝 (𝑥) = 𝐴 cos(𝑘𝑥) + 𝐵 sin(𝑘𝑥)
Example:
Find the general solution of the differential equation
3
, WI1421LR Calculus I
𝑦 ′′ (𝑥) + 𝑦 ′ (𝑥) − 2𝑦(𝑥) = 9𝑒 𝑥 − 4𝑒 2𝑥
2
Auxiliary equation: 𝑟 + 𝑟 − 2 = 0 → (𝑟 + 2)(𝑟 − 1) = 0 → 𝑟 = −2; 𝑟 = 1
𝑦𝑐 (𝑥) = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥
For the particular solution we take
𝑦𝑝 (𝑥) = 𝐴𝑥𝑒 𝑥 + 𝐵𝑒 2𝑥
This gives:
𝑦𝑝′ (𝑥) = 𝐴(𝑥 + 1)𝑒 𝑥 + (4 + 2 − 2)𝐵𝑒 2𝑥 = 9𝑒 𝑥 − 4𝑒 2𝑥 → 3𝐴𝑒 𝑥 + 4𝐵𝑒 2𝑥 = 9𝑒 𝑥 − 4𝑒 2𝑥
This gives 𝐴 = 3 and 𝐵 = −1:
𝑦𝑝 (𝑥) = 3𝑥𝑒 𝑥 − 𝑒 2𝑥
Therefore the general solution is:
𝑦(𝑥) = 3𝑥𝑒 𝑥 − 𝑒 2𝑥 + 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥
Series
A sequence is an ordered list of numbers:
𝑎1 , 𝑎2 , 𝑎3 , … = {𝑎𝑛 }∞
𝑛=1
The series is then defined as:
∞
𝑎1 + 𝑎2 + 𝑎3 + … = ∑ 𝑎𝑛
𝑛=1
The partial sum is given by:
𝑛
𝑆𝑛 = ∑ 𝑎𝑘 = 𝑎1 + 𝑎2 + 𝑎3 +. . . +𝑎𝑛
𝑘=1
If:
- lim 𝑆𝑛 = 𝑆 exists as a real number and sequence {𝑆𝑛 }∞
𝑛=1 is convergent, then the series ∑ 𝑎𝑛
𝑛→∞
is convergent: ∑∞
𝑛=1 𝑎𝑛 = 𝑆
- Sequence {𝑆𝑛 }∞𝑛=1 is divergent the series is divergent
The geometric series ∑∞
𝑛=1 𝑎𝑟
𝑛−1
= 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 +. ..:
∞ 𝑎
- For |𝑟| < 1 → ∑𝑛=1 𝑎𝑟 𝑛−1 = 1−𝑟
- For |𝑟| ≥ 1 the series is divergent
A telescopic series has the form:
∞
1
∑
𝑛(𝑛 + 1)
𝑛=1
The nth partial sum is then given by:
𝑛 𝑛
1 1 1 1 1 1 1 1 1 1 1
𝑆𝑛 = ∑ = ∑( − ) = − + − + − +⋯+ −
𝑘(𝑘 + 1) 𝑘 𝑘+1 1 2 2 3 3 4 𝑛 𝑛+1
𝑘=1 𝑘=1
This simplifies to:
1
𝑆𝑛 = 1 −
𝑛+1
This gives:
lim 𝑆𝑛 = 1
𝑛→∞
1
Thus the series is convergent and ∑∞
𝑛=1 𝑛(𝑛+1) = 1
4
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