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2326 Solutions to Problem Sheet 31
Second Order Differential Equations II
Question 1: Find the general solution of the following second order ode using
the method of variation of parameters.
(a) y 00 + y = sec x,
yp (x) = x sin x + cos x ln(cos x)
(b) y 00 + y = tan x ,
yp (x) = − cos x log(sec x + tan x)
(c) y 00 + 2y 0 + y = 4e−x log x,
yp (x) = x2 e−x (2 log x − 3)
(d) y 00 + y = 4x cos x,
yp (x) = x cos x + (x2 − 1) sin x
(e) y 00 + 2y 0 + 2y = e−x sec x,
yp (x) = e−x (x sin x + cos x ln(cos x))
(f) y 00 + y = sec x tan x,
yp (x) = x cos x − sin x − sin x ln(cos x)
1
, (F)
(g) y 00 − 3y 0 + 2y =
1 + e−x
yp (x) = (ex + e2x ) ln(1 + e−x ) − ex
2
(h) y 00 − y =
.
1 + ex
yp (x) = (ex − e−x ) ln(1 + ex ) − xex − 1
Sample method.
(g)
Independent solutions of the homogeneuos equation are:
y1 (x) = ex
1
y2 (x) = e2x
Esther Vergara Diaz, evd@maths.tcd.ie, see also http://www.maths.tcd.ie/~evd
1
whose Wronskian is:
W = y1 y20 − y10 y2 = 2ex e2x − ex e2x = e3x .
By Lagrange’s ‘variation of parameters’ a particular integral is u1 y1 + u2 y2 where:
2r
u1 = − yW
=
−e2x
e3x (1+e−x )
=
−e−x
1+e−x
so take
u1 = ln(1 + e−x ),
u02 =
=
y1 r
W
=
ex
e3x (1+e−x )
e−2x
1+e−x
−2x
R
e
dx
1 + e−x
R u
du,
substituting u = e−x ,
= −
1+u
R 1−v
dv,
substituting v = 1 + u,
=
v
= ln |v| − v + C = ln(1 + e−x ) − e−x ,
setting C = 1.
u2 =
So a particular integral is:
ex log(1 + e−x ) + e2x log(1 + e−x ) − e−x = (ex + e2x ) log(1 + e−x ) − ex .
Question 2: (F) The following is a special case of ‘Bessel’s equation’:
1
2 00
0
2
x y + xy + x −
y = 0.
4
√ x is a solution over any interval with x > 0. Find a linear
Verify that y1 (x) = sin
x
independent solution y2 using the reduction of order method.
sin x
cos x
y = c1 √ + c2 √
x
x
Question 3: (F) Find the general solution of the equation
xy 00 − 2(x + 1)y 0 + (x + 2)y = x3 e2x
for x > 0, given that the reduced equation has a solution of the form y1 = emx .
2
y = (c1 + c2 x3 )ex + (x2 − 2x + 2)e2x
Question 4: Find the general solution of each of the following equations.
(a) x2 y 00 − 2xy 0 + 2y = x3 ex
y = c1 x + c2 x2 + xex .
(b) (1 − x)y 00 + xy 0 − y = (1 − x)2
y = c1 x + c2 x2 + xex .
(c) xy 00 − (1 + x)y 0 + y = x2 e2x ,
y = c1 (1 + x) + c2 ex + 12 e2x (x − 1).
(d) (x2 − 1)y 00 − 2xy 0 + 2y = (x2 − 1)2
x2
x4
−
.
y = c1 x + c2 (x2 + 1) +
6
2
Sample method.
(c) y1 = 1 + x is an obvious solution of the reduced equation. To find a
second independent solution y2 first write the equation in standard form:
1
y + p(x)y + q(x)y = y − 1 +
x
00
0
00
y0 +
1
y = xe2x = r(x).
x
Then
Z
−
p(x)dx = x + ln |x|
and
e−
R
p(x)dx
= |x|ex
and Liouville’s method says that y2 = vy1 where
R
R e− p
v = K
y12
R |x|ex
= K
dx
(1 + x)2
x
R xe
=
dx,
(1 + x)2
choosing K = 1 if x > 0 K = −1 if x < 0
R (1 + x)ex
x
= −xe
+
dx
1+x
1+x
ex
= 1+x
3
Therefore y2 = ex . Now apply Lagrange’s ‘variation of parameters’. The Wronskian of y1 and y2 is:
W (y1 , y2 ) = y1 y20 − y10 y2 = (1 + x)ex − ex = xex
and a particular integral is u1 y1 + u2 y2 where:
u1 = − yW2 r
x
2x
xe
= − e xe
= −e2x ,
x
so take
u1 = −e2x /2,
u02 , =
=
y1 r
W
(1+x)xe2x
xex
= (1 + x)ex ,
so take
u2 = xex .
Thus a particular integral is:
−
x − 1 2x
1 + x 2x
e + xe2x =
e
2
2
and the general solution is:
y = c1 (1 + x) + c2 ex +
4
x − 1 2x
e .
2