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