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

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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=y02−y2+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.

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