Unformatted text preview: First ﬁnd general solution of associated homogeneous equation: y0 = Ay where A = Characteristic equation is: det 5x
5x . 2
1 + 4 + 3 = 0. 1, 3
1
11
v1
0
1
2
1
Eigenvector for = 1:
=
) v2 = v1 . Choose v =
2
1
1
v2
0
1
1
v1 = 0
1
Eigenvector for = 3: 1 1
) v2 = v1 . Choose v =
2
1
2
11
v2
0
1
General solution of associated homogeneous equation:
1 +c e
2
2
1
= c1 y1 + c2 y2 .
11
22
x
x yh (x) = c1 e
h
1 1
2 .
. 1
1 3x
3x Apply Variation of Parameters to ﬁnd general solution: y(x) = u1 y1 + u2 y2 where
11
22
x
x 4xe
0 0
0
u01 y1 + u02 y2 =
11
22 0
u01 e
1
0
u01 e
1 ) x
x
x
x 0
+ u02 e
2
0
u02 e
2 3x
3x
3x
3x =
= x
x 4xe
0 . Solving gives
0
u01 = 2x
1
0
2x
u02 = 2xe2x .
2 Integrating gives
2
u1 = x2 + c1
1
1
2x
2
u2 = xe2x 1 2x
2x
2
e + c2 .
2 General solution:
2
1
y ( x ) = (x 2 + c 1 ) e = c1 e
1 1
1 x
x = 1
1 x
x 1 2x
3x
1 2x
2
+ c 2 ) e 3x
2e
2 2x
+ (xe2x + c2 e
2 1
1 3x
3x h
y h ( x) 3. x (t) = Ax + f , where A =
0 Solution: ✓ 4
2 +
+ 2
1 3
3 ◆ 1
1 1
x
2
( x2 + x 1 ) e x
2
21
2
x
1
2
(x
x + 2 )e x
2
p
y p ( x) and f (t) = ✓ 1/t
4 + 2/t ◆ We ﬁrst need to ﬁnd the solution to the corresponding homogeneous equation: 4
2
A
I=
,
2
1
so det [A
I] = ( 4
)( 1
characteristic equation are 0 and ) 4=
5. 2 + 5 , and hence the roots of the To ...
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 Fall '14
 Trigraph, general solution, BVP

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