AMS 361- Applied Calculus IV
Homework 6 - Solution
Due Date: March 26th
1. Chapter 3.2 Problem 12, Page 167
Use the Wronskian to prove that the given functions are linearly independent on the
indicated interval.
f (x) = x,
g(x) = cos(lnx),
h(x) = sin(lnx); x > 0
Solution
x
cos(lnx)
sin(lnx)
−sin(lnx)
cos(lnx)
W = 1
x
x
0 −cos(lnx) + sin(lnx) −sin(lnx) − cos(lnx) x2
x2
x2
x2
−sin(lnx) −sin(lnx) cos(lnx)
cos(lnx) −cos(lnx) sin(lnx)
= x
−
−
+
x
x2
x2
x
x2
x2
−cos(lnx) sin(lnx)
−sin(lnx) cos(lnx)
− sin(lnx)
−
+
−
cos(lnx)
x2
x2
x2
x2
−2
2
2
= x
2sin (lnx) + 2sin(lnx)cos(lnx) + 2cos (lnx) − 2sin(lnx)cos(lnx)
= 2x−2 6= 0
f or x > 0.
2. Chapter 3.2 Problem 16, Page 168
A third-order homogeneous linear equation and three linearly independent solutions are
given. Find a particular solution satisfying the given initial conditions.
y (3) − 5y ′′ + 8y ′ − 4y = 0; y(0) = 1, y ′ (0) = 4, y ′′ (0) = 0;
y1 = ex ,
y2 = e2x ,
y3 = xe2x
Solution: If the third-order homogeneous linear equation’s linearly independent solutions
are given, we can write the General Solution as follows.
y = c1 ex + c2 c2x + c3 xe2x
Using the initial conditions
c1 + c2 + c3
c1 + 2c2 + c3
c1 + 4c2 + 4c3
= 1
= =4
= 0
We can find c1 = −12 c2 = 12 c3 = −10 solving the linear system. Hence the particular
solution satisfying the given initial conditions is
y(x) = −12ex + 12e2x − 10xe2x
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3. Chapter 3.2 Problem 22, Page 168
A nonhomogeneous differential equation, a complementary solution yc and particular
solution yp are given. Find the solution satisfying the given initial conditions.
y ′′ − 4y = 12; y(0) = 0, y ′ (0) = 10; yc = c1 e2x + c2 e−2x , yp = −3
Solution: Genereal solution of the nonhonogeneous DE is y(x) = yc + yp .
y(x) = c1 e2x + c2 e−2x − 3
The general solution should satisfy the initial conditions
y(0) = 0 = c1 + c2 − 3
y (0) = 10 = 2c1 − 2c2
′
We can find c1 = 4, c2 = −1 solving the linear system. Hence the solution satisfying the
given initial conditions is
y(x) = 4e2x − e−2x − 3
4. Chapter 3.2 Problem 26, Page 168
(a) Find by inspection particular solutions of the two nonhomogeneous equations
y ′′ + 2y = 4 and y ′′ + 2y = 6x
(b) Use the method of problem 25 to find a particular solution of the differential equation
y ′′ + 2y = 6x + 4.
Solution:
(a) Let y1 = c1 is the solution of y ′′ + 2y = 4 and y2 = c2 x + c3 is the solution of
y ′′ + 2y = 6x
y1′ = y1′′ = 0 ⇒ 2c1 = 4 ⇒ y1 = 2
y2′ = c2 , y2′′ = 0 ⇒ 2y2 = 6x ⇒ c2 = 3; c3 = 0 ⇒ y2 = 3x
(b) y = y1 + y2 = 3x + 2 is the particular solution of the differential equation y ′′ + 2y =
6x + 4 where y ′′ = 0.
5. Chapter 3.2 Problem 33, Page 168 Suppose that the three numbers r1 , r2 and r3
are distinct. Show that the three functions exp(r1 x), exp(r2 x) and exp(r3 x) are linearly
independent by showing that their Wronskian is nonzero for all x.
Solution: The Wronskian of three functions is
rx
e1
e r2 x
er3 x rx
r2 er2 x r3 er3 x W = r1 e 1
r12 er1 x r22 er2 x r32 er3 x 1 1 1
= er1 x er2 x er3 x r1 r2 r3 r12 r22 r32 = er1 x er2 x er3 x r2 r32 − r3 r22 − (r1 r32 − r3 r12 ) + r1 r22 − r2 r12
= er1 x er2 x er3 x [r2 r3 (r3 − r2 ) − r1 r3 (r3 − r1 ) + r1 r2 (r2 − r1 )]
6= 0 f or r1 6= r2 6= r3
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6. Chapter 3.2 Problem 38, Page 168 Use the method of reduction of order as in Problem
37 to find a second linearly independent solution y2 .
x2 y ′′ + xy ′ − 9y = 0
(x > 0); y1 (x) = x3
Solution: The method of resuction of order
y(x)
y ′ (x)
= v(x)x3
= v ′ (x)x3 + 3v(x)x2
y ′′ (x)
= v ′′ (x)x3 + 6v ′ (x)x2 + 6v(x)x
The given differential equation transforms into
x2 v ′′ (x)x3 + 6v ′ (x)x2 + 6v(x)x + x v ′ (x)x3 + 3v(x)x2 − 9v(x)x3 = 0
After simplification
v ′′ x5 + 7v ′ x4 = 0
Reducible second order differential equation (chapter1.6). Let v ′ = p
xp′ + 7p = 0 ⇒
dp
−7
=
p
x
Integrating the both sides gives us
lnp = −7lnx + lnc1 ⇒ p = c1 x−7
p = v ′ = c1 x−7 ⇒ v(x) = −
c1
+ c2
6x6
With c1 = −6 and c2 we get v(x) = 1/x6 , the second linearly independent solution
y2 (x) = 1/x3 .
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