2E2 Tutorial Sheet 13 Second Term1
has eigenvalue λ1 = 5 with eigenvector
x1 =
1 February 2004
1. (2) Find the general solution to
y 0 − 2y = −t
(1)
Solution: This follows from the general solution to
0
y + ry = f (t)
which is
y = Ce−rt + e−rt
Z
ert f dt
so here r = −2 and f (t) = −t so, using integration by parts
Z
y = Ce2t − e2t te−2t dt
Z
1
1
= Ce2t − e2t − te−2t +
e−2t dt
2
2
1 −2t 1 −2t
2t
2t
= Ce − e
− te − (e )
2
4
t 1
= Ce2t + +
2 4
= 5y2 − 23
= 5y1 + 15.
(2)
(3)
A=
1
0 5
5 0
(8
y = f 1 x1 + f 2 x2
(10
and subsituting this into the differential equation gives
−23
.
(f10 − 5f1 )x1 + (f20 + 5f2 )x2 =
15
(11
or, writing it out,
(4)
−23
15
=
g1 + g 2
g1 − g 2
(13
and, hence, g1 = −4 and g2 = −19. Putting this back into the equation leads to
(f10 − 5f1 )x1 + (f20 + 5f2 )x2 = −4x1 − 19x2
(14
f10 − 5f1 = −4.
(15
Hence
(5)
Thus, this is of the form y 0 + ry = f with r = −5, f (t) = −4 and so
Z
f1 = C1 e5t − 4e5t e−5t dt
(6)
and so
f1 = C1 e5t +
(7)
Conor Houghton, houghton@maths.tcd.ie and http://www.maths.tcd.ie/~houghton/ 2E2.html
4
5
(16
(17
Similarly,
2(f20 + 5f2 ) =
(1 − 1)
−23
15
= −38
(18
or
f20 + 5f2 = −19.
1
(9
Now to seperate the equation lets decompose the inhomeneous part, sometimes called
the forcing term, over the two eigenvectors:
−23
= g 1 x1 + g 2 x2
(12
15
Now, the matrix
so if we write
with y1 (0) = −3 and y2 (0) = 5.
Solution: First of all rewrite the equation in matrix form
−23
0 5
.
y+
y0 =
15
5 0
1
1
and eigenvalue λ1 = −5 with eigenvector
1
x2 =
−1
2. (3) Find the general solution to
y10
y20
2
(19
Thus, r = −5, f (t) = 19 and using integrating gives above
f2 = C2 e−5t −
19
.
5
The general solution is therefore
4
19
1
1
+ C2 e−5t −
.
y = C1 e5t +
1
−1
5
5
If y1 (0) = −3 and y2 (0) = 5 then we get
4
19
−3
1
1
= C1 +
+ C2 −
.
5
1
−1
5
5
(20)
Once again, we split the forcing term over the two eigenvectors:
t et
et
e
= x1 + x2
0
2
2
We get
1
f1 − 3f1 = et
2
Z
1
f1 = C1 e3t + e3t e−2t dt.
2
(21)
so
and so,
1
f1 = C1 e3t − et
4
(22)
In the same way
and hence
−3 = C1 + C2 − 3
23
5 = C 1 − C2 +
5
so C1 = −C2 = 1/5 and
1 5t 4
1
19
1
1
y=
e +
+ − e−5t −
.
1
−1
5
5
5
5
(23)
and so
Integrating gives
(24)
y=
Solution: Here we have
where
A=
et
0
(26)
1 2
2 1
,
(27)
this has eigenvalue λ1 = 3 with eigenvector
1
x1 =
1
and eigenvalue λ1 = −1 with eigenvector
1
x2 =
.
−1
3
(32
(33
(34
(35
1
f2 = C2 e−t + et
4
(36
1
C1 e3t − et
4
(37
1
1
(25)
y = Ay +
(31
This means
3. (3) Find the solution to
y10 = y1 + 2y2 + et
y20 = 2y1 + y2
1
f2 + f 2 = et
2
Z
1
f2 = C2 e−t + e−t e2t dt.
2
(30
(28)
(29)
4
1
1
+ C2 e−t + et
−1
4