Worksheet 16
Math 1B, GSI: Andrew Hanlon
1. Find the general solution of the equation y 00 − y = t2 and the particular solution when y(0) = 0
and y 0 (0) = 0.
Solution:
We first solve y 00 − y = 0. This was done on Worksheet 15 and the general solution is y =
Aet + Be−t . Now, we need to find one solution to y 00 − y = t2 . We guess that this solution is
of the form yo = c1 t2 + c2 t + c3 . Plugging this in gives
2c1 − c1 t2 − c2 t − c3 = t2
or
(1 + c1 )t2 + c2 t + (c3 − 2c1 ) = 0.
In order for this to be true for every t, all the coefficients must be zero. Thus, we must have
1 + c1 = 0, c2 = 0, and c3 − 2c1 = 0. From that, we get c1 = −1, c2 = 0, and c3 = −2. Thus,
we have that a solution to the nonhomogeneous equation is yo = −t2 − 2. Then, the general
solution to the nonhomogeneous equation is
y = Aet + Be−t − t2 − 2
and we see y 0 = Aet − Be−t − 2t. The initial conditions give
0=A+B−2
and
0 = A − B.
We solve these and conclude A = B = 1. Thus, the particular solution with these initial
conditions is
y = et + e−t − t2 − 2.
2. Find the general solution of the equation y 00 − 2y 0 − 3y = e2t + cos(t). Find the particular
solution with y(0) = 1 and y 0 (0) = 0.
Solution:
We first solve the homogeneous equation y 00 − 2y 0 − 3y = 0. For this, we find the roots of the
polynomial x2 − 2x − 3 = (x − 3)(x + 1) so the roots are 3, −1. Thus, the general solution to the
homogeneous equation is Ae3t +Be−t . Now, we need to find one solution to the nonhomogeneous
equation. To do this, we guess the solution has the form yo = c1 e2t + c2 cos t + c3 sin t. Plugging
in gives
4c1 e2t − c2 cos t − c3 sin t − 2(2c1 e2t − c2 sin t + c3 cos t) − 3(c1 e2t + c2 cos t + c3 sin t) = e2t + cos t
or
(1 + 3c1 )e2t + (1 + 4c2 + 2c3 ) cos t + (4c3 − 2c2 ) sin t.
This implies 1 + 3c1 = 0, 1 + 4c2 + 2c3 = 0, and 4c3 − 2c2 = 0 since it must be true for all t. We
solve this to obtain c1 = −1/3, c2 = −1/5, and c3 = −1/10 so yo = −(e2t /3 + cos t/5 + sin t/10).
Thus, the general solution to the nonhomogeneous equation is
y = Ae3t + Be−t −
e2t cos t sin t
−
−
3
5
10
and we see y 0 = 3Ae3t − Be−t − 2e2t /3 + sin t/5 − cos t/10. Thus, the initial conditions give us
1=A+B−
1 1
−
3 5
and
1
2
−
3 10
which we solve to get A = 23/40 and B = 23/24. Thus, the particular solutions for these initial
conditions is
23
23
e2t cos t sin t
y = e3t + e−t −
−
−
.
40
24
3
5
10
0 = 3A − B −
3. Find the general solution of the equation y 00 + y = sin t.
Solution:
We first solve the homogeneous solution y 00 + y = 0. This was done is Worksheet 15 to
obtain the general solution A cos t + B sin t. Now, we would normally guess something of
this form for the solution of the nonhomogeneous solution, but since they agree we multiply
by t. Thus, we guess a solution of the form yo = t(c1 cos t + c2 sin t). We calculate yo0 =
c1 cos t + c2 sin t + t(−c1 sin t + c2 cos t) and yo00 = −2c1 sin t + 2c2 cos t − t(c1 cos t + c2 sin t).
Plugging in, we get
−2c1 sin t + 2c2 cos t − t(c1 cos t + c2 sin t) + t(c1 cos t + c2 sin t) = sin t
or
(1 + 2c1 ) sin t − 2c2 cos t = 0.
Thus, we conclude 1 + 2c1 = 0 and 2c2 = 0 since this must be true for all t. Solving, we get
c1 = −1/2 and c2 = 0. Thus, yo = −t cos t/2 and the general solution to the nonhomogeneous
equation is
t cos t
.
y = A cos t + B sin t −
2
t
4. Find the general solution of the equation y 00 −2y 0 +y = t2e+1 . (Hint: use variation of parameters)
Solution: We first solve the homogeneous equation y 00 − 2y 0 + y = 0. We did this in Worksheet
15 to obtain the general solution Aet + Btet . Then, for variation of parameters we look for
a solution to the homogeneous equation of the form yo = A(t)et + B(t)tet . Then, we get the
variation of parameters equations
A0 (t)et + B 0 (t)tet = 0
2
and
et
.
t2 + 1
Combining these, we get the equation B 0 (t) = 1/(t2 + 1) so B(t) = arctan t (we ignore the
integrating factor since we look for just one solution). We also get A0 (t) = −tB 0 (t) = −t/(t2 +1)
so A(t) = − 12 ln(1 + t2 ). Thus, we get
A0 (t)et + B 0 (t)(et + tet ) =
yo = −
et ln(1 + t2 )
+ arctan(t)tet
2
and the general solution of the nonhomogeneous equation is
y = Aet + Btet −
et ln(1 + t2 )
+ arctan(t)tet .
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5. Use Euler’s formula (eix = cos x + i sin x) to derive the formulas for cos(x + y) and sin(x + y).
Solution: On one hand we have
ei(x+y) = cos(x + y) + i sin(x + y)
and on the other
ei(x+y) = eix eiy = (cos x+i sin x)(cos y+i sin y) = cos x cos y−sin x sin y+i(sin x cos y+sin y cos x).
Thus,
cos(x + y) + i sin(x + y) = cos x cos y − sin x sin y + i(sin x cos y + sin y cos x).
This means that the real and imaginary parts must be equal so we get
cos(x + y) = cos x cos y − sin x sin y
and
sin(x + y) = sin x cos y + sin y cos x.
6. Show that if f is a solution of y 00 +q(t)y 0 +p(t)y = 0 and g is a solution of y 00 +q(t)y 0 +p(t)y = h(t)
then f + g is a solution of y 00 + q(t)y 0 + p(t)y = h(t).
Deduce using the fact from the previous worksheet that the general solution of y 00 + q(t)y 0 +
p(t)y = h(t) is fgen + g where fgen is the general solution of y 00 + q(t)y 0 + p(t)y = 0.
Proof. We are given that
f 00 + q(t)f 0 + p(t)f = 0
and
g 00 + q(t)g 0 + p(t)g = h(t).
Now, we check
(f + g)00 + q(t)(f + g)0 + p(t)(f + g) = (f 00 + q(t)f 0 + p(t)f )+ (g 00 + q(t)g 0 + p(t)g) = 0 + h(t) = h(t)
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as desired.
Then, we know that fgen is a two parameter family so by what we have just done fgen + g is a
two parameter family of solutions to y 00 + q(t)y 0 + p(t)y = h(t). Then, by the fact from the fact
on Worksheet 15, we get the fgen + g is the general solution to y 00 + q(t)y 0 + p(t)y = h(t).
7. Show that if f1 is a solution of y 00 +q(t)y 0 +p(t)y = h1 (t) and f2 is a solution of y 00 +q(t)y 0 +p(t)y =
h2 (t) then f1 + f2 is a solution of y 00 + q(t)y 0 + p(t)y = h1 (t) + h2 (t) Thus, when you have a sum
of terms on the right hand side, you treat them separately.
Proof. We are given that
f100 + q(t)f10 + p(t)f1 = h1 (t)
and
f200 + q(t)f20 + p(t)f2 = h2 (t).
Now, we check
(f1 +f2 )00 +q(t)(f1 +f2 )0 +p(t)(f1 +f2 ) = (f100 +q(t)f10 +p(t)f1 )+(f200 +q(t)f20 +p(t)f2 ) = h1 (t)+h2 (t)
as desired.
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