midterm2 solution

AMS 361 Midterm Exam 2 Solutions
April 17, 2014
1)
y 00 + 5y 0 + 4y = 0
y(0) = 2
y 0 (0) = 2
Characteristic equation is
r2 + 5r + 4 = 0
So r =
4, 1.
4x
y(x) = c1 e
+ c2 e
x
Now appliy the initial conditions:
y(0) = c1 + c2 = 2
y 0 (0) =
Which implies c1 =
4
3 , c2
=
4c1
c2 = 2
10
3 .
y(x) =
4
e
3
4x
+
10
e
3
2)
y (4)
2y 00 + y = 0
Characteristic equation is
r4
(r2
2r2 + 1 = 0
1)2 = 0
So r = ±1 each with multiplicity 2.
1
x
y(x) = c1 ex + c2 xex + c3 e
x
+ c4 xe
x
3)
These functions are linearly dependent. To see this by definition, recall
2sin2 x
cos2x = 1
So that
1
y3 (x) + 2y2 (x) = 0
2
y1 (x)
4)
x2 y 00
2xy 0
10y = 0
Substitute y = xr and we want to find r.
x2 r(r
1)xr
r(r
2
2xrxr
1)
r2
(r
2r
3r
1
10xr = 0
10 = 0
10 = 0
5)(r + 2) = 0
So r = 5, 2 and
y(x) = c1 x5 + c2 x
2
5)
y 00
3y 0 + 2y = 2x + cos2x
To get yc :
r2
(r
3r + 2 = 0
2)(r
2
1) = 0
So r = 2, 1 and the complementary solution is
yc = c1 e2x + c2 ex
Now guess
yp = Ax + B + Ccos2x + Dsin2x
Plugging this in results in
3A + 2B + 2Ax
2(C + 3D)cos2x + (6C
2D)sin2x = 2x + cos2x
Equating coeﬃcients results in
A=1
B=
3
2
C=
1
20
D=
3
20
So
yp = x +
3
2
1
cos2x
20
6)
y 000
y 0 = ex
y(0) = y 0 (0) = 0
y 00 (0) = 1
To get yc :
r3
r(r2
r=0
1) = 0
3
3
sin2x
20
r(r
1)(r + 1) = 0
yc = c 1 + c 2 e x + c 3 e
x
Now guess (due to the duplication in yc ):
yp = Bxex
Plugging this in gives
2Bex = ex
So
B=
yp =
1
2
1 x
xe
2
The general solution is
y(x) = c1 + c2 ex + c3 e
x
1
+ xex
2
The initial conditions are
y(0) = c1 + c2 + c3 = 0
y 0 (0) = c2
c3 +
1
=0
2
y 00 (0) = c2 + c3 + 1 = 1
Which gives c1 = 0, c2 =
1
4,
y(x) =
c3 = 14 .
1 x 1
e + e
4
4
4
x
1
+ xex
2
7)
y 00 + y = tanx
The right hand side is not easily handled with undermined coeﬃcients. Variation of parameters is more appropriate.
yc = c1 sinx + c2 cosx
The Wronskian is
W (x) = det

sinx
cosx
cosx
sinx
=
sin2 x
cos2 x =
1
Applying variation of parameters,
ˆ
ˆ
cos(x)tan(x)
sin(x)tan(x)
yp (x) = sinx
dx + cosx
dx
1
1
= sinx
=
=
ˆ
sinxdx
sin(x)cos(x)
sin(x)cos(x)
=
cosx
cos(x)
ˆ
ˆ
sin2 x
dx
cosx
[sec(x)
cos(x)]dx
cos(x)[ln|secx + tanx|
cos(x)ln|sec(x) + tan(x)|
5
sinx]

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