SOLUTION TO PROBLEM SET 7 FYS 3140
8.12.16
00
0
- Problem to solve: x2 y − 2xy + 2y = x ln x
00
0
- Rewriting the problem in standard form: y − x2 y + x22 y = lnxx (8.12.16)
- Homogeneous solutions: y1 (x) = x; y2 (x) = x2
- Particular solution: Two methods can be used: (1) Green’s function
method; and (2) variation of parameters. Note that in both cases we
solve the problem written in standard form eqn(8.12.16).
1. Green’s function Method.
Step 1: We start by assuming we have solution of the following form,
0
0
0
A(x )x x < x
G(x, x ) =
0
0
B(x )x2 x > x
and its first derivative,
0
dG(x, x )
=
dx
0
0
A(x ) x < x
0
0
2B(x )x x > x
0
0
0
0
Remember that G(x, x ) is solution for LG(x, x ) = δ(x, x ). For x 6= x
the problem reduce to homogeneous equation. Thus we have two regions
0
x < x and x > x0 . Selection of appropriate homogeneous solution in each
region depends on the given boundary conditions. Nevertheless, for computing particular solution alone we can start by arbitrarily assigning one
homogeneous solution in each region. Therefore, the above assumption is
arbitrary, you can start by interchanging the two homogeneous solutions.
step 2: apply the following two conditions: (1). The Green function must
0
be continuous at x . (2) The change of the derivative of the Green function
0
at x must have unit value. i.e.
0
0
1:
A(x )x 0 = B(x )x2 0
x=x
x=x
⇒
A(x0 )x0 − B(x0 )x02 = 0
dG(x, x0 ) 4
0 =1
dx
x=x
⇒
2B(x )x0 − A(x ) = 1
⇒
−A(x ) + 2B(x )x0 = 1
2:
0
(aa)
0
0
0
From eqn(aa) and eqn(bb) we get,
0
x −x02
A(x0 )
0
=
−1 2x0
B(x0 )
1
1
(bb)
solving for A(x0 ) and B(x0 ) using Cramer’s rule, we get
0 −x02 1 2x0 0
=1
A(x ) = 0
x −x02 −1 2x0 0
x 0
−1 1
1
0
= 0
B(x ) = 0
x −x02 x
−1 2x0 Now G(x, x0 ) can be written as
0
G(x, x ) =
0
x x<x
0
1 2
x
x>x
x0
Step 3: get the particular solution,
Z
Z
yp =
G(x, x0 )f (x0 )dx0 −
x>x0
G(x, x0 )f (x0 )dx0
x<x0
Note the negative(minus) sign in the second integral (i.e. in x < x0 region).
0
From eqn(8.12.16) f (x0 ) = lnxx0 and finally,
Z
Z
ln x0 0
1 ln x0 0
2
yp = −x
dx
+
x
dx
0
0
0
x<x0 x
x>x0 x x
Note that, x and x0 are independent variables. The integration is with respect to x0 , hence, x is taken as a constant. Note also that after the indefinite
integral is evaluated we replace x0 by x. This is not in accordance with integral rules, it is just substitution by hand inspired by ideas from Green’s
function method.
using
R
ln x0
x0
=
(ln x0 )2
2
and
R
ln x0
x02
0
x
= − 1+ln
x0 ,
∴ yp = −x ln x(ln x + 1) − x
(8.12.16sol)
2. Variation of parameters method
Again y1 and y2 are homogeneous solutions of the problem. Now we want
to write the particular solution as linear combination of the homogeneous
solution. i.e. yp = c1 y1 + c2 y2 where c1 and c2 are now parameters to be
varied till we get the right solution. When it is substituted in the differential
equation we get two conditions to be satisfied,
(1b)
c01 y1 + c02 y2 = 0
(2b)
c01 y10 + c02 y20 = f (x)
2
condition (1b) is what we wanted the parameters should satisfy, (i.e. it is
a condition imposed may be for convenience reason), but (2b) is based on
condition (1b) and the given differential equation. evaluating (1b) and (2b)
for c01 and c02
0
y2 f (x) y 0 y f (x)
2 = − 2
c01 = y1 y2 W (x)
y 0 y 0 1
2
Z
y2 f (x)
⇒ c1 = −
dx
W (x)
similarly,
Z
c2 =
y1 f (x)
dx
W (x)
the particular solution is thus,
yp = c1 y1 + c2 y2
Z
Z
y1 f (x)
y2 f (x)
dx + y2
dx
= −y1
W (x)
W (x)
(∗var∗)
which is identical with eqn(12.21) in boas(chapter 8). We can also derive
eqn(*var*) using Green’s function method.
Now we can choose y1 (x) = x; y2 (x) = x2 ; and f (x) = lnxx . Again you can
interchange y1 and y2 . Solving eqn(*var*) should then yield same result as
eqn(8.12.16sol).
8.12.18
00
0
- Problem to solve: (x2 + 1)y − 2xy + 2y = (x2 + 1)2
00
0
- Rewriting in standard form: y − x22x+1 y + x22+1 y = x2 + 1 (8.12.16)
- Homogeneous solutions: y1 (x) = x; y2 (x) = 1 − x2
- Particular solution:
1. Green’s function Method.
Step 1: We start by assuming we have solution of the following form,
0
0
0
A(x )x x < x
G(x, x ) =
0
0
B(x )(1 − x2 ) x > x
and its first derivative,
0
dG(x, x )
=
dx
0
0
A(x ) x < x
0
0
−2B(x )x x > x
3
step 2: apply the two conditions:
0
1:
A(x )x
x=x0
= B(x )(1 − x )
0
2
⇒
A(x0 )x0 − B(x0 )(1 − x02 ) = 0
dG(x, x0 ) 4
0 =1
dx
x=x
⇒
−2B(x )x0 − A(x ) = 1
⇒
−A(x ) − 2B(x )x0 = 1
2:
0
x=x0
(cc)
0
0
0
(dd)
solving for A(x0 ) and B(x0 ) from eqn(cc) and eqn(dd) using Cramer’s rule,
we obtain
0 −(1 − x02 )
1
−2x0 −(1 − x02 )
0
A(x ) = 0
=
x −(1 − x02 )
x02 + 1
−1
−2x0 0
x 0
−1 1
−x0
0
= 02
B(x ) = 0
02
x −(1 − x )
x +1
0
−1
−2x
Now G(x, x0 ) can be written as
(
0
G(x, x ) =
−(1−x02 )
x
x02 +1
−x0
2)
(1
−
x
x02 +1
0
x<x
0
x>x
Step 3: get the particular solution:
Now f (x0 ) = (x02 + 1)
Z
Z
−(1 − x02 ) 02
−x0
0
2
(x
+
1)dx
+
(1
−
x
)
(x02 + 1)dx0
yp = −x
02
x02 + 1
x>x0 x + 1
x<x0
=
x2 x4
+
2
6
(8.12.18sol)
2. Variation of parameters method
Let y1 (x) = x; y2 (x) = 1 − x2 ; and f (x) = x2 + 1, solving eqn(*var*)
should then yield same result as eqn(8.12.18sol).
8.13.7
- problem to solve: 3x2 y 2 y 0 − x2 y 3 = 1
1. Bernoulli equation:
4
Bernoulli equantion is of the form y 0 + P y = Qy n , rewriting the problem in
this form we get
1
y −2
y0 −
(8.13.7a)
y= 3
3x
3x
First multiply eqn(8.13.7a) by (1−n)y −n = 3y 2 . and then make substitution
z = y 1−n = y 1−(−2) = y 3 , and z 0 = 3y 2 y 0 . i.e.
3y 2 y 0 −
1 3
1
y = 3
x
x
⇒ z0 −
1
1
z= 3
x
x
solving for z using integrating factor,
eI
R
= e
−1
dx
x
−I
z = e
=
⇒ z = y3 =
Z
=
1
x
I
e Qdx + C
−1
+ Cx
x2
−1
+ Cx
x2
2. some other trick
The basic idea is just to try to rewrite the problem in simpler form. To
that end, observe the two terms 3x2 y 2 y 0 and x2 y 3 . If we start by assuming
z = x3 y 3 then differentiating z once will generate the first and the second
terms form. Thus, if we dare to write the problem in terms of variable z,
we get
4
z0 − z = 1
x
Solving this equation will give identical solution as Bernoulli’s method.
12.1.8
- problem to solve: (x2 + 2x)y 00 − 2(x + 1)y 0 + 2y = 0 (12.1.8a)
1. elementary method:
get one solution first by guessing u(x) = x + 1, then the second solution
would be v(x)u(x) = v(x)(x + 1), and find v(x) by substituting back in the
eqn(12.1.8a). i.e.
(x2 + 2x)(v(x)(x + 1))00 − 2(x + 1)(v(x)(x + 1))0 + 2v(x)(x + 1) = 0
⇒ (x2 + 2x)(x + 1)v 00 − 2v 0 = 0
Let z = v 0 the above equation then becomes
(x2 + 2x)(x + 1)z 0 − 2z = 0
5
it is separable equation, thus
dz
2
=
dx
z
x(x + 2)(x + 1)
1
2
dz
1
dx
⇒
=
+
−
z
x x+2 x+1
⇒ ln z = ln x + ln(x + 2) − 2 ln(x + 1)
x(x + 2)
⇒z =
(x + 1)2
1
x(x + 2)
=1−
and v 0 = z =
2
(x + 1)
(x + 1)2
1
⇒v = x+
x+1
⇒
The second solution is therefore,
v(x)(x + 1) = x2 + x + 1
and the general solution is the linear combination of the two
y(x) = A(x + 1) + B(x2 + x + 1)
⇒ y(x) = A(x + 1) + Bx2
(1.8sol)
where A and B are arbitrary constants.
2. Method of Frobenius:
Assume the solution to the problem is written in power series of the form
y=x
s
∞
X
n=0
n
an x =
∞
X
an xn+s
(3)
n=0
then its first and second derivatives are
0
y =
∞
X
an (n + s)xn−1+s
(4)
n=0
00
y =
∞
X
an (n + s)(n − 1 + s)xn−2+s
(5)
n=0
Now substitute eqn(3-5) in eqn(12.1.8a)
(x2 +2x)
∞
X
an (n+s)(n−1+s)xn−2+s −2(x+1)
n=0
∞
X
n=0
6
an (n+s)xn−1+s +2
∞
X
n=0
an xn+s = 0
⇒
∞
X
an (n + s)(n − 1 + s)x
n+s
+ 2
n=0
− 2
∞
X
n=0
∞
X
an (n + s)(n − 1 + s)x
n−1+s
−2
∞
X
an (n + s)xn+s
n=0
an (n + s)xn−1+s + 2
n=0
∞
X
an xn+s = 0
n=0
collecting sums of identical exponents together, we get
⇒
∞
X
an {(n + s)(n + s − 3) + 2}xn+s +
|n=0
{z
A
}
∞
X
an 2(n + s)(n + s − 2)xn−1+s = 0
|n=0
{z
B
}
Now the sum of coefficients of each term must be zero. Thus we set up the
following table
A
B
xs−1
0
a0 2s(s − 2)
xs
a0 (s(s − 3) + 2)
a1 2(s + 1)(s − 1)
xs+1
a1 ((1 + s)(s − 2) + 2)
a2 2(s + 2)(s)
···
xs+n
an ((n + s)(n + s − 3) + 2)
an+1 2(n + 1 + s)(n + s − 1)
From the first term (xs−1 ) we get the indicial equation. i.e.
2s(s − 2) = 0
⇒ s = 0;
s=2
Now we solve the problem for s = 0 and s = 2, separately.
case 1, s=0:
set coefficients of xn+s to zero
an+1 2(n + 1 + s)(n + s − 1) + an ((n + s)(n + s − 3) + 2) = 0
⇒ an+1 = −
(n)(n − 3) + 2
an
2(n + 1)(n − 1)
this implies that
a1 = −
2
a0 = a0
−2
a2 = 0
a3 = 0
..
.
an = 0
7
case 2, s = 2:
set coefficients of xn+s to zero, and changing the variable from an → bn
bn+1 2(n + 1 + s)(n + s − 1) + bn ((n + s)(n + s − 3) + 2) = 0
⇒ bn+1 = −
(n + 2)(n − 1) + 2
bn
2(n + 3)(n + 1)
this implies that
0
b1 = − b0 = 0
6
b2 = 0
b3 = 0
..
.
bn = 0
from eqn (3), the General solution is thus ,
y(x) = xs=0 a0 x + a0 + xs=2 b0
⇒ a0 (x + 1) + b0 x2
which is same result as eqn(12.1.8sol).
12.11.2
- problem to solve: x2 y 00 + xy 0 − 9y = 0 (12.11.2a)
1. elementary method
using Euler-Cauchy method you can find that the solution for eqn(12.11.2.a).
i.e.
y(x) = Ax3 + Bx−3
Method of Frobenius:
substituting eqn(3),(4), and (5) in eqn (12.11.2a) we get
∞
X
an (n + s)(n − 1 + s)x
n=0
n+s
+(
∞
X
an (n + s)x
n=0
⇒
∞
X
n+s
−9
∞
X
an xn+s = 0
n=0
an {(n + s)(n − 1 + s) + (n + s) − 9}xn+s = 0
n=0
coefficient of each term must be zero, hence
an {(n + s)(n − 1 + s) + (n + s) − 9} = 0
8
⇒ an {(n + s)(n + s) − 9} = 0
from the first term we get the indicial equation
s2 − 9 = 0
⇒ s = 3;
s = −3
case 1, s=3
all terms except a0 is zero
case 2, s=-3
all terms except b0 is zero.
The general solution is thus
y(x) = a0 x3 + b0 x−3
12.11.6
- Problem to solve: 3xy 00 + (3x + 1)y 0 + y = 0
Method of Frobenius:
(12.11.6)
substituting eqn(3),(4), and (5) in eqn (12.11.6) we get
(3x)
∞
X
n−2+s
an (n+s)(n−1+s)x
+(3x+1)
n=0
⇒
∞
X
∞
X
an (n+s)x
n−1+s
n=0
3an (n+s)(n−1+s)x
n−1+s
+
n=0
∞
X
+
∞
X
an xn+s = 0
n=0
3an (n+s)x
n=0
n+s
+
∞
X
an (n+s)x
n−1+s
n=0
+
∞
X
an xn+s = 0
n=0
grouping sums of identical exponents together, we obtain
∞
X
an (n + s)(3n + 3s − 2)x
n−1+s
n=0
|
+
∞
X
an (3n + 3s + 1)xn+s = 0
n=0
{z
A
}
|
{z
B
}
Now the sum of coefficients of each term must be zero. Thus we set up the
following table
A
B
xs−1
a0 s(3s − 2)
0
xs
a1 (s + 1)(3s + 1)
a0 (3s + 1)
xs+1
a2 (s + 2)(3s + 2)
a1 (3s + 4)
···
From the first term (xs−1 ) we get the indicial equation. i.e.
s(3s − 2) = 0
⇒ s = 0;
9
s=
2
3
xs+n
an+1 (n + s + 1)(3n + 3s + 1)
an (3n + 3s + 1)
Now we solve the problem for s = 0 and s = 32 .
case 1, s=0:
set coefficients of xn+s to zero
an+1 (n + 1)(3n + 1) + an (3n + 1) = 0
⇒ an+1 = −
1
an
n+1
this implies that
1
a1 = − a0
1
1
1
a2 = −
a1 = a0
1+1
2
1
1
a3 = −
a2 = −
a0
2+1
3·2
1
1
a4 = −
a3 =
a0
3+1
4·3·2
..
.
1
(11.6a)
an = (−1)n a0
n!
case 2, s = 32 :
set coefficients of xn+s to zero, and changing the variable from an → bn
bn+1 (n + 5/3)(3n + 3) + bn (3n + 3) = 0
⇒ bn+1 = −
3
1
bn = −
bn
n + 5/3
3n + 5
this implies that
3
b1 = − b0
5
3
32
b2 = −
b1 =
b0
3+5
8·5
3
33
b3 = −
b2 = −
b0
6+5
11 · 8 · 5
3
34
b4 = −
b3 =
b0
9+5
14 · 11 · 8 · 5
..
.
bn = no simple compact expression.
10
(11.6b)
The General solution is thus from eqn (3), (11.6a) and (11.6b)
∞
X
2
3x
(3x)2
(3x)3
(3x)4
n
n 1
3
a0 x +x
b0 −
b0 +
b0 −
b0 +
b0
y(x) =
(−1)
n!
5
8·5
11 · 8 · 5
14 · 11 · 8 · 5
n=0
−x
⇒ a0 e
+ b0 x
2
3
3x (3x)2
(3x)3
(3x)4
1−
+
−
+
5
8·5
11 · 8 · 5 14 · 11 · 8 · 5
12.11.11
- problem to solve: 36x2 y 00 + (5 − 9x2 )y = 0
Method of Frobenius:
(12.11.11a)
substituting eqn(3),(4), and (5) in eqn (12.11.11a) we get
36
⇒ 36
∞
X
n=0
∞
X
an (n + s)(n − 1 + s)xn+s + (5 − 9x2 )
⇒
an xn+s = 0
n=0
an (n + s)(n − 1 + s)xn+s + 5
n=0
∞
X
∞
X
an {36(n + s)(n − 1 + s) + 5}xn+s −
∞
X
an xn+s − 9
n=0
∞
X
∞
X
an xn+2+s = 0
n=0
9an xn+2+s = 0
n=0
n=0
{z
A
|
}
|
{z
B
}
Now the sum of coefficients of each term must be zero. Thus we set up the
following table
A
B
xs
a0 36s(s − 1) + 5
0
xs+1
a1 36(s + 1)s + 5
0
xs+2
a2 36(s + 2)(s + 1) + 5
-9a0
From the first term (xs ) we get the indicial equation. i.e.
s2 − s +
5
=0
36
1
5
⇒s= ;
s=
6
6
1
Now we solve the problem for s = 6 and s = 56 .
case 1, s = 16 :
setting coefficients of xn+s to zero, we get
an 36(n + s)(n + s − 1) + 5 + −9an−2 = 0
an =
36(n +
9
+
1
6 )(n
1
6
− 1) + 5
11
an−2 =
3
an−2
4n(3n − 2)
···
xs+n
an 36(n + s)(n + s − 1) + 5
-9an−2
this implies that
a1 = 0
3
3
a0 = 5 a0
a2 =
32
2
a3 = 0
3
32
a4 =
a
=
a0
2
25 · 5
210 · 5
..
.
Here odd exponents coefficients are zero
case 2, s = 56 :
set coefficients of xn+s to zero, and changing the variable from an → bn
bn =
36(n +
9
+
5
6 )(n
5
6
− 1) + 5
an−2 =
3
an−2
4n(3n − 2)
this implies that
b1 = 0
3
3
b2 =
b0 = 5 b0
32
2
b3 = 0
3
32
b4 =
b
=
b0
2
25 · 5
210 · 5
..
.
Again odd exponents coefficients are zero.
The General solution is thus,
y(x) = xs=1/6 {a0 +
32 x4
3x2
32 x4
3x2
s=5/6
a
+
a
·
·
·
}+x
{b
+
b
+
b0 · · · }
0
0
0
0
25
210 · 5
25
210 · 5
⇒ y(x) = a0 x1/6{1 +
−1
3x2
3x3
32 x4
32 x5
6 {x + +
+
·
·
·
}
+
b
x
+
···}
0
25
210 · 5
25
210 · 5
12