MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Phase Portrait

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Published on 25 Oct 2018
School
Department
Course
Professor
Solutions of Test 3
Jordan Bell
August 9, 2013
1. f=y02
x3. Then
f
y = 0,f
y0=2y0
x3.
Hence the Euler-Lagrange equation is
0 = d
dx 2y0
x3,
so 2y0
x3=c1.
This is a separable equation:
2dy =c1x3dx.
Integrating both sides gives
2y=c1x4
4+c2,
or
y=c1x4+c2.
Using y(0) = 1 and y(1) = 1 gives us c1= 1 and c2=2. Hence
y=2x4+ 1.
2. F=y02y2+λy.
F
y =2y+λ,F
y0= 2y0
2y+λ= 2y00 so 2y00 + 2y=λ
If you are lucky you can solve this by looking at it (if you remember that cos
and sin solve y00 +y= 0). But you can solve this by finding the homogeneous
solutions and then using variation of parameters also.
y=c1cos x+c2sin x+λ
2
Since y(0) = 0 and y(π) = 1, we get c1+λ
2= 0 and c1+λ
2= 1. This gives
us λ= 1 and so c1=1
2.
So y=cos x
2+c2sin x+1
2. Now we use the final condition to figure out c2.
1
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