Download Solutions - Math Berkeley

Name: Solutions
Math 1B: Quiz 5
April 8, 2015
1. Find the forms of the particular solutions for the following problems. Do not solve for
the unknown coefficients.
(a) 2y ′′ + 7y ′ + 6y = e4x sin(x).
The corresponding homogeneous equation is 2y ′′ + 7y ′ + 6y = 0 which has the
auxiliary equation 2r2 + 7r + 6 = 0 ⇒ (2r + 3)(r + 2) = 0 ⇒ r = −3
, −2. The
2
−3x
−2x
complementary solution therefore is yc (x) = c1 e 2 + c2 e .
For the trial particular solution, try yp (x) = e4x (A cos(x) + B sin(x)). Since there
is no overlap between the trial yp and the yc found above, this form of yp is good
enough.
(b) y ′′ − 4y ′ + 4y = xe2x .
The corresponding homogeneous equation is y ′′ − 4y ′ + 4y = 0 which has the
auxiliary equation r2 − 4r + 4 = 0 ⇒ (r − 2)2 = 0 ⇒ r = 2. The complementary
solution therefore is yc (x) = c1 e2x + c2 xe2x .
For the trial particular solution, try yp (x) = e2x (A + Bx). Since both e2x and
xe2x appear in yc , we need to consider instead the form yp (x) = e2x (Ax2 + Bx3 )
to avoid any overlap between the particular and complementary solutions.
1
2. Find the general solution of
y ′′ + y = csc(x).
The corresponding homogeneous equation is y ′′ +y = 0 which has the auxiliary equation
r2 + 1 = 0 ⇒ r = ±i. The complementary solution therefore is yc (x) = c1 cos(x) +
c2 sin(x) with basic solutions y1 (x) = cos(x) and y2 (x) = sin(x).
We assume the particular solution is of the form
yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x)
where u1 and u2 are unknown functions. By the method of variation of parameters,
we know then that u′1 and u′2 satisfy
u′1 (x)y1 (x) + u′2 (x)y2 (x) = 0
u′1 (x)y1′ (x) + u′2 (x)y2′ (x) = csc(x).
(1)
(2)
(1) : cos(x)u′1 (x) + sin(x)u′2 (x) = 0
(2) : − sin(x)u′1 (x) + cos(x)u′2 (x) = csc(x).
(3)
(4)
We have
Note that sin(x)(3) + cos(x)(4) gives
u′2 (x)(sin2 (x) + cos2 (x)) = csc(x) cos(x) ⇒ u′2 (x) = cot(x).
∫
Use (3) to get u′1 (x) = − tan(x) cot(x) = −1. Thus, u1 (x) = −1 dx = −x and
∫
∫
cos(x)
u2 (x) = cot(x) dx =
dx = ln | sin(x)|.
sin(x)
We therefore have yp (x) = −x cos(x) + sin(x) ln | sin(x)| so the general solution to the
problem is
y(x) = yc (x) + yp (x) = c1 cos(x) + c2 sin(x) − x cos(x) + sin(x) ln | sin(x)|.
2