MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Wronskian, The Roots

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Published on 25 Oct 2018
School
Department
Course
Professor
Solutions of Test 2
Jordan Bell
July 9, 2013
1. The characteristic equation is r2+3r10 = 0, and solving gives r=5,2.
So solutions of the homogeneous equation are y1=e5tand y2=e2t. Calculate
the Wronskian: we get W= 7e3t.
To find u1we’ll end up doing integration by parts (going from third to fourth
line).
u1=Zy2g
Wdt
=Ze2t7tet
7e3tdt
=Zte6tdt
=te6t
6+e6t
36 .
To find u2we’ll end up doing integration by parts (going from third to fourth
line).
u2=Zy1g
Wdt
=Ze5t7tet
7e3tdt
=Ztetdt
=tetet.
Thus we can write the solution of the initial value problem as
y=c1e5t+c2e2t+e5tte6t
6+e6t
36 +e2ttetet
=c1e5t+c2e2t7
6tet35
36et.
If we want to choose c1and c2to satisfy the initial conditions, we’ll need to
evaluate y(0) and y0(0).
1
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y(0) = c1+c235
36.
y0(t) = 5c1e5t+ 2c2e2t7
6et7
6tet35
36et
y0(0) = 5c1+ 2c27
635
36
Using the initial conditions we get c1=2
7and c2=2
7. Thus the solution
of the initial value problem is
y(t) = 2
7e5t+2
7e2t7
6tet35
36et.
2. The first equation gives
u0
1=y2
y1
u0
2.
Putting this into the second equation gives
y0
1
y2
y1
u0
2+y0
2u0
2=g.
Multiplying by y1gives
y0
1y2u0
2+y1y0
2u0
2=gy1,
i.e.,
u0
2(y1y0
2y0
1y2) = gy1,
or,
u0
2W=gy1.
Thus
u2=Zgy1
Wdt.
Since u0
2W=gy1, using u0
1=y2
y1u0
2gives us
u0
1=y2
y1
gy1
W=gy2
W,
and thus
u1=Zgy2
Wdt.
3. Use the fact that y00
2=py0
2qy2and y00
1=py0
1qy1. Then
y1y00
2y00
1y2
W=y1(py0
2qy2)(py0
1qy1)y2
W
=py1y0
2qy1y2+py0
1y2+qy1y2
W=py1y0
2py0
1y2
W=p.
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