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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (2)

0 answers

Post by Cho HyuJang on March 12, 2013

Besides....

for the 5th example,

c=2/3p^3-2pi^4+(-1),

the cos part. that one should be -(-1).

0 answers

Post by James Xie on August 21, 2012

for the 5th example, (2/3)pi^2 should read (2/3)pi^3

Differential Equations Eulers Method

  • Approximation method
    • Pinpoint initial conditions and
    • Find , and
    • Find , and
    • Repeat until approximation target value reached
  • Tip: be very organized and systematic with your work flow

Differential Equations Eulers Method

Solve for y given [dy/dx] + xy = 0 and y(0) = 5
  • Separate x and y
  • [dy/dx] + xy = 0
  • [dy/dx] = −xy
  • [dy/y] = −x  dx
  • ∫[1/y] dy = ∫−x  dx
  • lny + c1 = −[(x2)/2] + c2
  • lny = −[(x2)/2] + c
  • elny = e−[(x2)/2] + c
  • y = ec e−[(x2)/2]
  • 5 = ec e0
  • 5 = ec
y = 5 e−[(x2)/2]
Solve for y given [dy/dx] − 2x = 0 and y(1) = 3
  • [dy/dx] − 2x = 0
  • [dy/dx] = 2x
  • dy = 2x  dx
  • ∫dy = ∫2x  dx
  • y = x2 + c
  • 3 = 12 + c
  • 2 = c
y = x2 + 2
Solve for y given y[dy/dx] − cosx = 0 and y(0) = 2
  • y[dy/dx] − cosx = 0
  • y[dy/dx] = cosx
  • y  dy = cosx  dx
  • ∫y  dy = ∫cosx  dx
  • [(y2)/2] = sinx + c1
  • y2 = 2 sinx + c
  • y = ±√{2 sinx + c}
  • 2 = ±√{2 sin0 + c}
  • 2 = ±√c
  • c = 4
y = √{2 sinx + c}
Solve for y given ex [dy/dx] + e4x = 0 and y(0) = 1
  • ex [dy/dx] + e4x = 0
  • ex [dy/dx] = − e4x
  • [dy/dx] = −e3x
  • ∫dy = − ∫e3x dx
  • y = −[1/3] e3x + c
  • 1 = −[1/3] e0 + c
  • 1 = −[1/3] + c
  • c = [4/3]
y = −[1/3] e3x + [4/3]
Approximate y(.01) if y′ = 2y and y(0) = 2, using Euler's method with h = .01
  • y0′ = 2(2) = 4
  • x1 = x0 + h
  • x1 = 0 + .01
  • x1 = .01
  • y1 = y0 + h  y0
  • y1 = 2 + .01(4)
y(.01) ≈ 2.04
Approximate y(1.2) if y′ = 4y3 − 1 and y(1) = 2, using Euler's method with h = .1
  • y0′ = 4(2)3 − 1 = 31
  • x1 = 1 + .1 = 1.1
  • y1 = 2 + .1(31) = 5.1
  • y1′ = 4 y13 − 1 = 4 (5.1)3 − 1 = 529.604
  • x2 = 1.1 + .1 = 1.2
  • y2 = y1 + h  y1′ = 5.1 + .1 (529.604)
y(1.2) ≈ 58.06
Approximate y(0.5) if y′ = 2y + 3x and y(1) = 2, using Euler's method with h = .5
  • y0′ = 2(2) + 3(1) = 7
  • x1 = 0 + .5 = .5
  • y1 = 2 + .5 (7)
y(.5) ≈ 5.5
Approximate y(2.02) if y′ = y2 + t and y(2) = 1, using Euler's method with h = .01
  • y0′ = 12 + 2 = 3
  • t1 = 2 + .01 = 2.01
  • y1 = 1 + .01(3) = 1.03
  • y1′ = y12 + t1 = 1.032 + 2.01 = 3.102727
  • t2 = 2.01 + .01 = 2.02
  • y2 = y1 + h  y1
  • y2 = 1.03 + .01(3.102727)
y(2.02) ≈ 1.06
Solve for y given [dy/dx] + 2xy − yex = 0 and y(0) = 1
  • [dy/dx] = −2xy + yex
  • [dy/dx] = y(−2x + ex)
  • [dy/y] = −2x + ex dx
  • ∫[1/y] = ∫−2x + ex dx
  • lny = −x2 + ex + c
  • y = e(−x2 + ex + c)
  • y = ec e(−x2 + ex)
  • 1 = ec e0 + 1
  • 1 = ec e1
  • ec = [1/e]
y = [1/e] e(ex − x2)
Solve for y given [dy/dx] − √{1 − y2} = 0 and y(0) = 0
  • [dy/dx] − √{1 − y2} = 0
  • dy = √{1 − y2}
  • [dy/(√{1− y2})] = 1  dx
  • ∫[1/(√{1 − y2})] dy = ∫dx
  • sin−1 y = x + c
  • sin(sin−1 y) = sin(x + c)
  • 0 = sin(0 + c)
y = sin(x)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Differential Equations Eulers Method

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Differential Equations 0:08
    • Example 1
  • Differential Equations: Euler's Method 2:33
    • Rules
    • Example 2
  • Example 3 5:42
  • Example 4 9:44
  • Example 5 14:14