# MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Wronskian, Stationary Point

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Mat 244 Practice Test 3

Jordan Bell

July 25, 2013

You should be able to do all of the questions in this test given enough time.

This is certainly longer than the actual test will be. I wanted to give you more

worked out examples for calculus of variations.

1. Find the extremals of the functional

Zx2

x1

y2+y02−2ysin xdx.

Solution. f=y2+y02−2ysin x. So

∂f

∂y = 2y−2 sin x, ∂f

∂y0= 2y0.

Hence the Euler-Lagrange equation is

2y−2 sin x= 2y00

i.e.

y00 −y=−sin x.

This is a second order ODE. The solution of the homogeneous equation is

c1ex+c2e−x.

Now we use variation of parameters. y1=ex, y2=e−x. The Wronskian is

W=−2. g=−sin x.

u1=−Zgy2

Wdx =−1

2Zsin xe−xdx

u2=Zgy1

Wdx =1

2Zsin xexdx

To compute u1, u2we need to use integration by parts. I won’t write this out;

it is something that you should be able to skillfully do if you have suﬃcient time,

and you should have absolutely no problem remembering how to do integration

by parts. You don’t have to do it in your head; unless it’s obvious to me, I

manually set uand dv and write out the integration by parts formula.

1

u1=1

4e−x(cos x+ sin x)

u2=1

4ex(−cos x+ sin x)

So the extremals of the functional Rx2

x1y2+y02−2ysin xdx are

y=c1ex+c2e−x+1

4(cos x+ sin x) + 1

4(−cos x+ sin x)

=c1ex+c2e−x+sin x

2.

2. Find the extremals of the functional

I(y) = Zx2

x1

1+(y(x))2

(y0(x))2dx

This takes longer than I would want for a test question, but if I gave you

enough time I would expect you to be able to do something like this level of

diﬃculty.

Solution. f=1+y2

y02. Since fdoesn’t depend on x, the Euler-Lagrange

equation for this functional is

d

dx ∂f

∂y0y0−f= 0

which can also be written ∂f

∂y0y0−f=c1.

Now,

∂f

∂y0=−2(1 + y2)

y03.

Hence 2(1 + y2)

y02+1 + y2

y02=c1

so

2+2y2+1+y2=c1y02

so

1 + y2=c1y02

so (in this step I change c1, which is logically acceptable untl we have introduced

c2)

y0=c1p1 + y2

i.e. dy

c1p1 + y2=dx.

2