Solutions 2.2-Page 119
Problem 9
Use the Wronskian to prove that the given functions are linearly independent on the
indicated interval.
f ( x) = e x , g ( x) = cos x, h( x) = sin x; the real line
The Wronskian will be 3 x 3 because n-1 derivatives must be taken. See pg.113. The
derivatives are as follows:
f ′( x) = e x
g ′( x) = − sin x
h ′( x) = cos x
f ′′( x) = e
g ′′( x) = − cos x
h ′′( x) = − sin x
x
ex
W (e , cos x, sin x) = e
ex
x
x
cos x
sin x
− sin x cos x = 2e x ≠ 0
− cos x − sin x
Problem 11
Use the Wronskian to prove that the given functions are linearly independent on the
indicated interval.
f ( x) = x, g ( x) = xe x , h( x) = x 2 e x ; the real line
The necessary derivatives are as follows:
f ′( x) = 1
g ′( x) = e x + xe x
f ′′( x) = 0
g ′′( x) = 2e x + xe x
x
xe x
W ( x, xe x , x 2 e x ) = 1 e x + xe x
0 2e x + xe x
h′( x) = 2 xe x + x 2 e x
h ′′( x) = 2e x + 4 xe x + x 2 e x
x 2e x
2 xe x + x 2 e x
= x 3 e 2 x ≠ 0 if x ≠ 0
2e x + 4 xe x + x 2 e x
Problem 21
A nonhomogeneous differential equation, a complementary solution y c , and a particular
solution y p are given. Find a solution satisfying the given initial conditions.
y′′ + y = 3 x; y (0) = 2, y′(0) = −2;
yc = c1 cos x + c2 sin x; y p = 3 x
The general solution is found by superposition of the particular and complementary
solutions.
Y = c1 cos x + c 2 sin x + 3 x
There are two unknowns: c1 and c 2 . The constants are solved from the initial conditions.
Differentiating the general solution.
Y ′ = −c1 sin x + c 2 cos x + 3
Now the initial conditions are used to find the constants.
Y (0) = 2 = c1
Y ′(0) = −2 = c 2 + 3
c 2 = −5
Y = 2 cos x − 5 sin x + 3 x
Problem 30
Verify that y1 = x and y 2 = x 2 are linearly independent solutions on the entire real line of
the equation
x 2 y ′′ − 2 xy ′ + 2 y = 0
but that W ( x, x 2 ) vanishes at x = 0. Why do these observations not contradict part (b) of
Theorem 3?
W ( x, x 2 ) =
x
x2
1 2x
= x 2 ≠ 0 if x ≠ 0
Theorem 3 is not contradicted because the differential equation is not written in standard
form (see eq.3 on pg.109). The equation is put into standard form by dividing through by
2
2
x 2 . The result is y ′′ − y ′ + 2 y = 0 . The coefficient functions are not continuous at
x
x
x = 0. Therefore Theorem 3 is not contradicted.
Problem 33
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
1
1
1
W = exp([r1 + r 2 + r 3]x) ⋅ r1
2
r1
r2
2
r2
r3
2
r3
is nonzero for all x.
The first part of the Wronskian, e ( r1 + r2 + r3 ) x ≠ 0 .
Now the determinant part of the Wronskian must be shown to never vanish.
1
1
r1
2
r1
r2
2
r2
1
2
2
2
2
2
2
r3 = (r2 r3 − r3 r2 ) − (r1 r3 − r3 r1 ) + (r1 r2 − r1 r2 ) = (r2 − r1 )(r3 − r2 )(r3 − r1 ) ≠ 0
2
r3
The Wronskian is the product of two nonzero numbers, and therefore it is nonzero. Since
the Wronskian is nonzero, the functions as linearly independent.