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4.6 Variation of Parameters: Finding a
Particular Solution
d2 y
dy
a2(x) 2 + a1(x) + a0(x)y = f (x)
dx
dx
(1)Divide by a2(x) to put in standard form
d2 y
dy
+ q(x)y = g(x)
+
p(x)
2
dx
dx
(2)Find the general solution of the
homogeneous equation:
d2 y
dy
+ p(x) + q(x)y = 0
dx2
dx
yh(x) = c1y1(x) + c2y2(x)
where y1(x), y2(x) are two solutions of the
homogenous equation.
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(3) Look for a particular solution of
dy
d2 y
+ q(x)y = g(x)
+
p(x)
2
dx
dx
of the form
C1
C2
yp (x) = V1(x)y1(x) + V2(x)y2(x)
Our goal is to find V1(x), V2(x) so that this is
a solution of the D.E.
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(4) Compute yp′
yp = V1y1 + V2y2
yp′ = (V1y1′ + V1′y1) + (V2y2′ + V2′y2)
Regroup
yp′ = (V1y1′ + V2y2′ ) + (V1′y1 + V2′y2)
To simplify, we will require
V1′y1 + V2′y2 = 0
Equation
♯
1
(5) Compute yp′′
yp′ = (V1y1′ + V2y2′ ) +
(0)
↑
F rom Eq ♯1
yp′′ = V1y1′′ + V1′y1′ + V2y2′′ + V2′y2′
Substitute into the D.E.
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yp′′ + p(x)yp′ + q(x)yp = g(x)
(V1y1′′ + V1′y1′ + V2y2′′ + V2′y2′ ) + p(x)(V1y1′ +
V2y2′ ) + q(x)(V1y1 + V2y2) = g(x)
Regroup
0
0
↓
↓
V1(y1′′ + p(x)y1′ + q(x)y1) + V2(y2′′ + p(x)y2′ + q(x)y2)
+(V1′y1′ + V2′y2′ )
=
g(x)
y1, y2 solve the hom.eq
y1′′ + p(x)y1′ + q(x)y1
y2′′ + p(x)y2′ + q(x)y2
So
ց
= 0
= 0
V1′y1′ + V2′y2′ = g(x) Equation ♯2
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(6) Solve the two equations ♯1,♯ 2 for V1′, V2′
y1V1′ + y2V2′ =
0
y1′ V1′ + y2′ V2′ = g(x)
(7) Integrate V1′ and V2′ to find V1 and V2.
(8) The yp = V1y1 + V2y2
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Summary
d2 y
dy
+ p(x) + q(x)y = g(x) (Std Form)
dx2
dx
(1) Find homogeneous solution
yh = C1y1 + C2y2
(2)Taking y1, y2, form the equations
y1V1′ + y2V2′
=
0
y1′ V1′ + y2′ V2′
=
g(x)
↑
F romD.E. in std f orm
and solve for V1′ and V2′
(3) Integrate V1′, V2′ to find V1, V2. Then
yp = V1y1 + V2y2
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Ex
y ′′ + y = sec(x) (Std. Form)
(1) Homogenous soln.
y ′′ + y =
0
m2 + 1 =
0
m2
m
−1
√
= ± −1 = ±i
=
yh = C1 cos(x) + C2 sin(x)
(2) Look for a particular solution of the form
yp =
V1(x) cos(x) + V2(x) sin(x)
↓
Equations
♯
1, 2
↓
y1V1′
+
y2V2′
=0
y1′ V1′
+
y2′ V2′
= g(x)
cos(x)V1′
+
sin(x)V2′
=0
− sin(x)V1′ +
cos(x)V2′
= sec(x)
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Using Cramois Rule to find V1′, V2′:
1
↓
cos(x)
0
sin(x) sec(x) cos(x) 0 − sec(x) sin(x)
=
V1′ = 2(x) − (− sin2 (x))
cos(x) sin(x) cos
− sin(x) cos(x) − tan(x)
= − tan(x)
=
1
R
V1 = − tan(x)dx = − ln |sec(x)| + d1
V1′
(Pick d1 = 0; only need 1 particular soln.)
cos(x)
0 − sin(x) sec(x) cos(x) sec(x) − 0 1
=
= =1
V2′ = cos(x) sin(x) cos2(x) + sin2(x) 1
− sin(x) cos(x) R
V2 = 1dx = x + d2
(Pick d2 = 0; only need 1 particular Soln.)
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yp =
V1(x) cos(x) + V2(x) sin(x)
= (− ln | sec(x)|) cos(x) + (x) sin(x)
(3) The general solution of the D.E. is
y = C1 cos(x) + C2 sin(x) +
(− ln | sec(x)|) cos(x) + x sin(x)
Note:
V1 = − ln | sec(x)| + d1
V2 =
x + d2
If we keep the constants of integration here,the
y = (− ln |sec(x)| + d1) cos(x) + (x + d2) sin(x)
= d1 cos(x) + d2 sin(x) + (− ln | sec(x)|) cos(x)
+x sin(x)
is the general solution of the D.E.!
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Ex
x2y ′′ − 4xy ′ + 6y = 1
4 ′ 6
1
y − y + 2 y = 2 (StdF orm)
x
x
x
′′
Two solutions of the homogeneous equation
are y1 = x2 and y2 = x3.
Constrict
a particular solution to the nonhomogenous
equation.
yp = V1y1 + V2y2
yp = V1x2 + V2x3
y1V1′ + y2V2′ = 0
y1′ V1′ + y2′ V2′ =
1
x2
x2V1′ + x3V2′ = 0
2xV1′ + 3x2V2′ =
10
1
x2
x3 0
1
x3
2
0− 2
2 3x −x −1
′
x
x
=
V1 = = 4 = 3
x2 x3 3x4 − 2x4
x
x
2x 3x2 Z
1
x−2
−3
+ d1 = 2 + d1
V1 = −x dx = −
−2
2x
2
0 x
1
2x
2 1
1
′
x
=
=
V2 = x2 x3 3x4 − 2x4 x4
2x 3x2 V2 =
Z
−3
−1
x
−4
+ d2 = 3 + d2
x dx =
−3
3x
−1 3 1 1 1
1 2
So yp = ( 2 )x + ( 3 )x = − =
2x
3x
2 3 6
A particular solution of x2y ′′ − 4xy ′ + 6y = 1
is yp =
1
6
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