Math 20D Final Exam
Practice Problems
1. Be able to define/explain all of the following terms/ideas, and
understand what they mean and are used for:
(a) Wronskian determinant for a second order differential equation
(b) Fundamental set of solutions
(c) General solution to a differential equation
(d) Ordinary point vs. Singular point
(e) Wronskian of a set of solutions of a first order system in n unknown
functions
(f) Laplace transform
(g) Linear Independence
(h) Eigenvalue/eigenvector
(i) Linear differential equation
(j) Homogeneous linear differential equation/nonhomogeneous linear
differential equation
(k) Exact equation
(l) Direction field/slope field
(m) Autonomous equation
2. Consider the differential equation
dy
= y(2 + y).
dt
(a) What are the equilibrium solution(s) of the equation?
(b) Classify the equilibrium solution(s) as stable, unstable, or
semistable.
(a) The equilibrium solutions are where dy/dt = 0, so y = 0 and y = −2 are the
equilibrium solutions.
(b) y = −2 is stable, y = 0 is unstable.
1
3. Find the general solution to each of the following differential equations
or systems of differential equations:
(a) y 0 =
3x2 −1
3+2y
dy
= 3x2 − 1. Separating variables we
We multiply by 3 + 2y to obtain (3 + 2y) dx
2
obtain (3 + 2y)dy = (3x − 1)dx. Integrating both sides yields 3y + y 2 = x3 − x + C.
We can use the quadratic formula to solve for y, obtaining
p
−3 ± 9 + 4(x3 − x + C)
y=
.
2
(b) y 0 =
y(3x2 −ex )
2y−5
2
x
Again we separate variables to obtain 2y−5
y dy = (3x − e )dx. Integrating both
sides yields 2y − 5 ln y = x3 − ex + C. We leave the solution in implicit form.
(c)
dy
dx
=
x+3y
x−y
We solve this by rewriting in terms of v = xy . We have
1+3v
1−v , so we can rewrite the differential equation as
v+x
dy
dx
dv
= v + x dx
, and
x+3y
x−y
=
1 + 3v
dv
=
.
dx
1−v
1−v
Using algebraic manipulation to separate variables, we have v2 +2v+1
dv = x1 dx.
1
2
Integrating both sides of the equation, we have ln( v+1
) − v+1
= ln x + C. Finally,
substituting v = y/x, we obtain the final (implicit) solution
ln(
2x
x
)−
= ln x + C.
y+x
y+x
(d) 2y 0 + y = 2t
We use the method of integrating factors here. We first multiply throughR by 12 to put
1
the equation in standard form y 0 + 12 y = t. The integrating factor is e 2 dt = et/2 .
Multiplying through by et/2 , we obtain et/2 y 0 + 12 et/2 y = tet/2 . Integrating yields
et/2 y = 2tet/2 − 4et/2 + C. Therefore,
y = 2t − 4 + Ce−t/2 .
2
(e) y 0 + 2ty = 2te−t
2
R
2
We use the method of integrating factors. The integrating factor is e 2tdt = et .
2
2
2
2
Multiplying through by et we obtain et y 0 + 2tet y = 2t. Integrating yields et y =
2
t + C. Therefore,
2
2
y = t2 e−t + Ce−t .
(f)
x
x2 +y 2
+
y
0
x2 +y 2 y
=0
x
We first verify that the differential equation is exact. Let M (x, y) = x2 +y
2 and
−2xy
−2xy
y
N (x, y) = x2 +y2 . Then My = x2 +y2 and Nx = x2 +y2 , so the equation is exact.
Therefore,
the solution is ψ(x, y) = C, where ψx = M and ψy = N . Thus, ψ =
R x
dx
= 12 ln(x2 + y 2 ) + h(y). Differentiating with respect to y gives ψy =
x2 +y 2
y
0
0
x2 +y 2 + h (y) = N , so h (y) = 0 and thus h(y) = 0. Therefore, we obtain the
solution
1
ln(x2 + y 2 ) = C.
2
(g) y 00 + y 0 − 2y = 2t
We first solve the corresponding homogeneous equation y 00 + y 0 − 2y = 0. Using
the characteristic equation, we obtain the solution y = c1 e−2t + c2 et . To solve the
nonhomogeneous equation, we can use the method of undetermined coefficients. We
assume a particular solution of the form Y = At + B, so Y 0 = A and Y 00 = 0. Thus,
A − 2At − 2B = 2t, so A = −1 and B = − 12 . Therefore, we obtain the general
solution
1
y = c1 e−2t + c2 et − t − .
2
(h) y 00 + 4y = t2 + 3et
We first solve the corresponding homogeneous equation y 00 + 4y = 0. Using the
characteristic equation, we obtain the solution y = c1 sin(2t) + c2 cos(2t). To solve
the nonhomogeneous equation, we use the method of undetermined coefficients.
We assume a particular solution of the form Y = At2 + Bt + C + Det , so Y 0 =
2At+B+Det and Y 00 = 2A+Det . Thus, 2A+Det +4(At2 +Bt+C +Det ) = t2 +3et .
Therefore, 4A = 1, 4B = 0, 4C + 2A = 0, and 5D = 3. Therefore, A = 41 , B = 0,
C = − 18 , and D = 35 . Therefore, we obtain the general solution
1
1 3
y = c1 sin(2t) + c2 cos(2t) + t2 − + et .
4
8 5
3
(i) y 00 + 2y 0 + y = e−t
We first solve the corresponding homogeneous equation y 00 + 2y 0 + y = 0. Using
the characteristic equation, we obtain the solution y = c1 e−t + c2 te−t . Using the
method of undetermined coefficients for the nonhomogeneous equation, we assume
a particular solution of the form Y = At2 e−t . Then Y 0 = 2Ate−t − At2 e−t and
Y 00 = 2Ae−t − 2Ate−t + At2 e−t . Therefore 2Ae−t − 2Ate−t + At2 e−t + 2(2Ate−t −
At2 e−t ) + At2 e−t = e−t , so A = 12 . Thus, we obtain the general solution
1
y = c1 e−t + c2 te−t + t2 e−t .
2
(j) y 00 + 4y = 3sin(2t)
We first solve the corresponding homogeneous equation y 00 + 4y 0 = 0. Using the
characteristic equation, we obtain the solution y = c1 sin(2t) + c2 cos(2t). To solve
the nonhomogeneous equation, we use the method of undetermined coefficients.
We assume a particular solution of the form Y = At sin(2t) + Bt cos(2t). Then
Y 0 = A sin(2t) + 2At cos(2t) + B cos(2t) − 2Bt sin(2t), and Y 00 = 4A cos(2t) −
4At sin(2t) − 4B sin(2t) − 4Bt cos(2t). Plugging into the differential equation, we
have
4A cos(2t)−4At sin(2t)−4B sin(2t)−4Bt cos(2t)+4(At sin(2t)+Bt cos(2t)) = 3 sin(2t).
Therefore, A = 0 and B = −3/4. Thus, we have the general solution
y = c1 sin(2t) + c2 cos(2t) − 3/4t cos(2t).
(k) y 00 − 2y 0 + y =
et
1+t2
We first solve the corresponding homogeneous equation y 00 − 2y 0 + y = 0 to obtain
the solution y = c1 et + c2 tet . We use the method of variation of parameters to solve
the nonhomogeneous equation. Let y1 = et and y2 = tet . Then W (y1 , y2 )(t) =
et (et + tet ) − et (tet ) = e2t . Then
t
Z
u1
and
4
= −
e
tet 1+t
2
dt
e2t
Z
t
= −
dt
1 + t2
1
= − ln(1 + t2 )
2
t
Z
u2
=
e
et 1+t
2
dt
e2t
1
dt
1 + t2
tan−1 (t).
Z
=
=
Therefore, we have a particular solution Y = u1 y1 + u2 y2 = − 12 ln(1 + t2 )et +
tan−1 (t)tet , and we obtain the general solution
y = c1 et + c2 tet −
1
ln(1 + t2 )et + tan−1 (t)tet .
2
(l) y 00 + y = tan t
We solve the corresponding homogeneous equation y 00 + y = 0 to obtain the solution
y = c1 sin(t) + c2 cos(t). We use the method of variation of parameters to solve the
nonhomogeneous equation. Let y1 = sin(t) and y2 = cos(t). Then W (y1 , y2 )(t) =
sin(t)(− sin(t)) − cos(t)(cos(t)) = −1. Therefore,
Z
cos(t) tan(t)
u1 = −
dt
−1
Z
=
sin(t)dt
= − cos(t)
and
Z
u2
sin(t) tan(t)
dt
−1
Z
− sin2 (t)
=
dt
cos(t)
Z
=
(cos(t) − sec(t))dt
=
=
sin(t) − ln(sec(t) + tan(t)).
Therefore, we obtain the particular solution Y = u1 y1 + u2 y2 = − cos(t) sin(t) +
(sin(t) − ln(sec(t) + tan(t))) cos(t) and the general solution
y = c1 sin(t) + c2 cos(t) − cos(t) sin(t) + (sin(t) − ln(sec(t) + tan(t))) cos(t).
5
(m) x0 =
1
4
1
−2
x
We first seek eigenvalues for the matrix. Since
1−r
1
det
= (1 − r)(−2 − r) − 4 = r2 + r − 6 = (r + 3)(r − 2),
4
−2 − r
the eigenvalues are −3 and 2. Taking r = −3 and solving
4 1
1 − (−3)
1
0
=
4
−2 − (−3) 0
4 1
yields an eigenvector ξ1 =
1 − (2)
4
yields an eigenvector ξ2 =
1
−5
2
−1
=
−1
4
1
−4
0
0
(n) x0 =
. Taking r = 2 and solving
1
0
−2 − (2) 0
x = c1
1
−4
1
1
0
0
. Therefore, we obtain the general solution
1
−4
e
−3t
+ c2
1
1
e2t .
x
We first seek eigenvalues for the matrix. Since
1−r
2
det
= (1 − r)(−1 − r) + 10 = r2 + 9,
−5 −1 − r
the eigenvalues are ±3i. Taking r = 3i and solving
1 − 3i
2
0
−5
−1 − 3i 0
−2
−2
yields an eigenvector ξ =
. Therefore we have a solution
e3it .
1 − 3i
1 − 3i
Since e3it = cos(3t) + i sin(3t), we can write this solution as
−2
−2
e3it =
(cos(3t) + i sin(3t))
1 − 3i
1 − 3i
−2 cos(3t)
−2 sin(3t)
=
+i
.
cos(3t) + 3 sin(3t)
−3 cos(3t) + sin(3t)
6
Therefore, our final solution is
−2 cos(3t)
−2 sin(3t)
x = c1
+ c2
.
cos(3t) + 3 sin(3t)
−3 cos(3t) + sin(3t)
(o) x0 =
2
− 32
3
2
−1
x
We first seek eigenvalues for the matrix. Since
det
2−r
− 32
3
2
−1 − r
= (2 − r)(−1 − r) +
9
1
= r2 − r + =
4
4
we have the eigenvalue r = 12 . Solving
3
3
2 − 21
0
2
2
=
− 32
−1 − 21 0
− 32
we obtain an eigenvector ξ =
1
−1
0
(p) x =
2
3
−1
−2
x+
1
−1
et
t
2
,
3
2
− 23
3
2
− 32
et/2 + c2
1
−1
2
3
0
1
−1
. Therefore, our final solution is
tet/2 +
2
3
0
et/2 .
We first solve the corresponding homogeneous system, that is, we solve
2 −1
0
x =
x.
3 −2
Finding eigenvalues for the matrix, we have that
2−r
−1
det
= (2 − r)(−2 − r) + 3 = r2 − 1,
3
−2 − r
so the eigenvalues are r = 1 and r = −1.
7
1
2
. Solving
we obtain a generalized eigenvector η =
0
0
r−
x = c1
3
2
− 23
Taking r1 = 1 and solving
2−1
3
0
0
1
1
2 − (−1)
−1
0
3
−2 − (−1) 0
−1
−2 − 1
=
yields a corresponding eigenvector ξ1 =
1
3
−1
−3
0
0
. Taking r2 = −1 and solving
=
3
3
−1
−1
0
0
1
yields a corresponding eigenvector ξ2 =
. Therefore, the solution to the
3
1
1
homogeneous system is xh = c1
et + c2
e−t .
1
3
To solve the nonhomogeneous system, we apply the method of undetermined coefficients. Since ξ1 et is a solution to the corresponding homogeneous system, we
assume a particular solution of the form X = Atet + Bet + Ct + D. Then
X0 = Atet + (A + B)et + C, and thus
2 −1
1
0
t
t
t
t
t
Ate + (A + B)e + C =
(Ate + Be + Ct + D) +
e +
t.
3 −2
0
1
Equating coefficients on the left and right hand sides, we have the following four
equations:
2 −1
i. A =
A
3 −2
2 −1
1
ii. A + B =
B+
3 −2
0
2 −1
0
iii. 0 =
C+
3 −2
1
2 −1
iv. C =
D
3 −2
Equation
(i)tellsus that
A is an eigenvector corresponding to the eigenvalue 1, so
1
α
A=α
=
. Taking this together with equation (ii) yields
1
α
α
α
B1
B2
α + B1
α + B2
+
8
B1
1
=
+
B2
0
2B1 − B2 + 1
=
3B1 − 2B2
2
3
−1
−2
Therefore, we have the equations α + B1 = 2B1 − B2 + 1 and α + B2 = 3B1 − 2B2 .
α
Rewriting these two equations yields B1 − B2 = α−1
2 and B1 − B2 = 3 . Therefore,
α
α−1
2 = 3 , and thus α = 3. This yields that B1 − B2 = 1, so we may choose B1 = 1
and B2 = 0.
2 −1
0
Equation (iii) corresponds to solving the system
C=
, that is,
3 −2
−1 1
2 −1 0
. This yields C =
.
reducing the augmented matrix
3 −2 −1
2
Finally,
we solve
equation (iv)
for D by reducing the augmented matrix
2 −1 1
0
to obtain D =
. Therefore, the final solution to the system
3 −2 2
−1
is
1
1
1
1
1
0
x = c1
et + c2
e−t +
tet +
et +
t+
.
1
3
1
0
2
−1
4. A scientist has two tanks being used in an experiment. Tank one
contains fifty gallons of fresh water, and tank two contains fifty gallons
of water with a concentration of one pound of salt per gallon at the start
of the experiment. When the experiment begins, water with a
concentration of one pound of salt per gallon is pumped into tank one at
a rate of two gallons per minute, and fresh water is pumped into tank
two at a rate of two gallons per minute. In addition, water flows from
tank two to tank one at a rate of one gallon per minute. There are two
drains, one from tank one at a rate of three gallons per minute, and one
from tank two at a rate of one gallon per minute.
(a) Write an initial value problem that models the amount of salt in
each tank in this system.
Let Q1 (t) denote the number of pounds of salt in tank one at time t and Q2 (t)
denote the number of pounds of salt in tank two at time t. Then we have the initial
value problem
 0
3
1
Q1 + 50
Q2
 Q1 = 2 − 50
2
0
Q2 = − 50 Q2

Q1 (0) = 0, Q2 (0) = 50
9
(b) Solve the initial value problem from part (a).
We rewrite the two equations as Q0 =
3
− 50
0
1
50
2
− 50
Q+
2
0
. Focusing first
1
3
− 50
50
Q, we obtain
on the corresponding homogeneous equation Q0 =
0
−2
50
1
1
3
2
eigenvalues − 50 and − 50 , with corresponding eigenvectors
and
, re0
1
spectively. Therefore, the homogeneous equation has solution
1
1
−3t/50
c1
e
+ c2
e−2t/50 .
0
1
Using the method of undetermined coefficients to solve the nonhomogeneous equation, we assume
ofthe form
Q0 = 0,so we need only solve
Q=
A. Then a solution
1
3
100
−2
− 50
50
3
A=
, so A =
. Therefore, we have
the equation
2
0
0
0
− 50
the solution
100 1
1
−3t/50
−2t/50
3
Q = c1
.
e
+ c2
e
+
0
1
0
To solve the initial value problem, we then use the initial condition Q(0) =
0
50
to obtain c1 = − 250
3 and c2 = 50, so we have
100 250
1
1
−3t/50
−2t/50
3
Q=−
e
+ 50
e
+
,
0
1
0
3
and thus Q1 (t) = 13 (−250e−3t/50 + 150e−2t/50 + 100) and Q2 (t) = 50e−2t/50 .
(c) After how long do the two tanks contain the same amount of salt?
We wish to find t so that 31 (−250e−3t/50 + 150e−2t/50 + 100) = 50e−2t/50 . We
2
can simplify this to 100 = 250e−3t/50 , so e−3t/50 = 25 , so t = − 50
3 ln( 5 ) ≈ 15.27
minutes.
(d) Discuss the terminal behavior of this system. What does this mean
in terms of the salt concentration?
As t → ∞, the system approaches 0 along the line Q1 = Q2 . In terms of salt
concentration, this tells us that as t → ∞, both tanks have no salt, and that
moreover the amount of salt in them is approximately equal after a substantial
amount of time.
10
5. Consider the differential equation
(1 − x)y 00 + xy 0 − y = 1 − x.
(a) Verify that y1 (x) = ex and y2 (x) = x are solutions to the
corresponding homogenous equation.
To verify that these are solutions, we simply take derivatives and plug them in
to the corresponding homogeneous equation. For y1 , we have y10 = y100 = ex , and
plugging this into the equation yields (1 − x)ex + xex − ex = 0, so this solves the
homogeneous equation.
For y2 , we have y20 = 1 and y200 = 0, so plugging this into the equation yields
(1 − x)0 + x − x = 0, so this solves the homogeneous equation.
(b) Use the method of variation of parameters to find a general solution
to the nonhomogeneous equation.
Now, W (y1 , y2 ) = ex − xex = ex (1 − x). To use variation of parameters, we must
have an equation of the form y 00 + p(t)y 0 + q(t)y = g(t), so we divide by 1 − x to
x
1
obtain y 00 + 1−x
y 0 − 1−x
y = 1. Variation of parameters tells us that we have a
particular solution of the form Y = u1 y1 + u2 y2 , where
Z
Z
Z
x
e−x
e−x
−x
−x
u1 = −
dx
=
−
(−e
+
)dx
=
e
+
dx
ex (1 − x)
1−x
x−1
(this integral does not have a closed form) and
Z
Z
ex
1
u2 =
dx
=
dx = − ln(1 − x).
ex (1 − x)
1−x
Z
e−x
Therefore, we have a particular solution Y = (e−x +
dx)ex − x ln(1 − x).
x−1
Thus, the general solution to the equation is
Z
e−x
y = c1 ex + c2 x + 1 + ex
dx − x ln(1 − x).
x−1
6. Consider the differential equation
(4 − x2 )y 00 + 2y = 0.
(a) Is x0 = 0 an ordinary point or a singular point?
Since 4 − x20 = 4 6= 0, 4 is an ordinary point.
11
(b) Seek power series solutions of the differential equation about x0 ;
find the first four terms in each of two solutions y1 and y2 .
Assume y =
∞
X
an xn , so y 00 =
n=0
∞
X
n(n − 1)an xn−2 . Then, plugging this in to the
n=2
equation, we obtain
0
=
(4 − x2 )
∞
X
n(n − 1)an xn−2 + 2
n=2
=
=
=
∞
X
an xn
n=0
4n(n − 1)an x
n=2
∞
X
∞
X
n−2
−
∞
X
n
n(n − 1)an x +
n=2
4(n + 2)(n + 1)an+2 xn −
n=0
∞
X
∞
X
2an xn
n=0
∞
X
n(n − 1)an xn +
n=0
∞
X
2an xn
n=0
(4(n + 2)(n + 1)an+2 − n(n − 1)an + 2an )xn .
n=0
Therefore, for each n, we have 4(n + 2)(n + 1)an+2 − n(n − 1)an + 2an = 0, i.e.
(n(n − 1) − 2)an
. If a0 = c1 and a1 = c2 , we have
an+2 =
4(n + 2)(n + 1)
a2
=
a3
=
a4
=
a5
=
(−2)a0
c1
=−
4(2)(1)
4
c2
(−2)a1
=−
4(3)(2)
12
(2 − 2)a2
=0
4(4)(3)
a3
c2
((3)(2) − 2)a3
=
=−
4(5)(4)
20
240
and we obtain
y
c1 2 c2 3
c2 5
x − x −
x + ...
4
12
240
1
1
1 5
c1 (1 − x2 − . . . ) + c2 (x − x3 −
x − . . . ).
4
12
240
= c1 + c2 x −
=
Therefore we have y1 = 1 − 41 x2 − . . . and y2 = x −
12
1 3
12 x
−
1
5
240 x
− ....
(c) Show that the solutions y1 and y2 form a fundamental set of
solutions.
To verify these are a fundamental set of solutions, we use the Wronskian determinant. We have that W (y1 , y2 )(0) = 1 6= 0, so y1 and y2 are a fundamental set of
solutions to the system.
7. Answer the same three questions as in the previous problem for the
equation
(1 + x2 )y 00 − 4xy 0 + 6y = 0.
Evaluating 1 + x2 at x = 0 yields 1 6= 0, so x0 = 0 is an ordinary point of the
∞
X
an xn , we obtain
differential equation. Assuming that y takes the form y =
y0 =
∞
X
nan x
n−1
and y 00 =
n=0
n(n − 1)an x
n−2
. Therefore
n=2
n=1
0
∞
X
(1 + x2 )y 00 − 4xy 0 + 6y
∞
∞
∞
X
X
X
an xn
nan xn−1 + 6
n(n − 1)an xn−2 − 4x
= (1 + x2 )
=
=
=
=
n(n − 1)an x
n=0
n=1
n=2
∞
X
n−2
+
∞
X
n
n(n − 1)an x −
(n + 2)(n + 1)an+2 xn +
n=0
∞
X
n
4nan x +
∞
X
n(n − 1)an xn −
n=0
∞
X
∞
X
4nan xn +
n=0
((n + 2)(n + 1)an+2 + n(n − 1)an − 4nan + 6an )xn
n=0
Therefore we must have that for all n ≥ 0,
(n + 2)(n + 1)an+2 + n(n − 1)an − 4nan + 6an = 0,
i.e., an+2 =
(−n2 +5n−6)an
(n+2)(n+1) .
Taking a0 = c1 and a1 = c2 , we have
13
6an xn
n=0
n=1
n=2
n=2
∞
X
∞
X
∞
X
n=0
6an xn
a2
=
a3
=
a4
=
a5
=
−6a0
= −3c1
2
(−1 + 5 − 6)a1
c2
=−
3(2)
3
(−4 + 10 − 6)a2
=0
4(3)
3c2
c2
(−25 + 25 − 6)a3
=−
=−
5(4)
30
10
and thus we obtain
y
c2 3 c2 5
x − x + ...
3
10
1 3 c2 5
2
= c1 (1 − 3x + . . . ) + c2 (x − x − x + . . . ).
3
10
= c1 + c2 x − 3c1 x2 −
c2 5
Thus, we have solutions y1 = 1 − 3x2 + . . . and y2 = x − 31 x3 − 10
x + . . . . To verify
that these are a fundamental set of solutions, we use the Wronskian determinant.
Since W (y1 , y2 )(0) = 1 6= 0, this is a fundamental set of solutions.
8. Use the Laplace Transform to solve the following initial value problems:
(a) y 00 + 2y 0 + 5y = 0,
y(0) = 2,
y 0 (0) = −1
Applying the Laplace Transform to both sides of the equation and setting Y = L(y),
we obtain
L(y 00 + 2y 0 + 5y)
=
0
L(y ) + 2L(y ) + 5L(y)
=
0
s Y − sy(0) − y (0) + 2sY − 2y(0) + 5Y
=
0
=
0
00
0
0
2
2
(s + 2s + 5)Y − 2s + 1 + 2
Therefore, Y =
2s−3
s2 +2s+5 .
Completing the square, we obtain
s2 + 2s + 5 = s2 + 2s + 1 + 4 = (s + 1)2 + 4.
Therefore,
14
2s − 3
s2 + 2s + 5
2s − 3
=
(s + 1)2 + 4
2(s + 1) − 5
=
(s + 1)2 + 4
5
2
s+1
−
= 2
(s + 1)2 + 4
2 (s + 1)2 + 4
5
= 2L(e−t cos(2t)) − L(e−t sin(2t))
2
Y
=
Thus, y = e−t cos(2t) − 25 e−t sin(2t).
(b) y 00 + 2y 0 + y = 4e−t ,
y(0) = 0,
y 0 (0) = 1
Applying the Laplace Transform to both sides of the equation and setting Y = L(y),
we obtain
L(y 00 + 2y 0 + y)
00
0
= L(4e−t )
L(y ) + 2L(y ) + L(y)
=
s2 Y − sy(0) − y 0 (0) + 2sY − 2y(0) + Y
=
(s2 + 2s + 1)Y − 1
=
4L(e−t )
1
4
s+1
4
s+1
Therefore,
Y
4
1
+
(s + 1)(s2 + 2s + 1) (s2 + 2s + 1
1
4
+
=
3
(s + 1)
(s + 1)2
2!
1!
= 2
+
(s + 1)3
(s + 1)2
=
=
2L(t2 e−t ) + L(te−t ).
Therefore, y = 2t2 e−t + te−t .
15
(c) y 00 + 9y = δ(t − 3),
y(0) = 0, y 0 (0) = 3
Applying the Laplace Transform to both sides of the equation and setting Y = L(y),
we obtain
L(y 00 ) + 9L(y)
0
2
s Y − sy(0) − y (0) + 9Y
2
(s + 9)Y − 3
=
L(δ(t − 3))
=
e−3s
=
e−3s .
Therefore,
3
1
+
s2 + 9 s2 + 9
3
1 −3s 3
e
+
3
s2 + 9 s2 + 9
1 −3s
e L(sin(3t)) + L(sin(3t)).
3
= e−3s
Y
=
=
Therefore, y = 13 u3 (t) sin(3(t − 3)) + sin(3t).
(d) y 00 + 4y = sin t − uπ (t) sin(t − π),
y(0) = 0,
y 0 (0) = 0
Applying the Laplace Transform to both sides of the equation and setting Y = L(y),
we obtain
L(y 00 ) + 4L(y)
s2 Y − sy(0) − y 0 (0) + 4Y
(s2 + 4)Y
L(sin t) − L(uπ (t) sin(t − π))
1
1
=
− e−πt 2
s2 + 1
s +1
1
1
− e−πt 2
.
=
s2 + 1
s +1
=
Therefore,
Y =
(s2
1
1
− e−πt 2
2
+ 1)(s + 4)
(s + 1)(s2 + 4)
Using partial fraction decomposition, we have that
Thus,
Y
=
1/3
s2 +1
1/3
−1/3
1/3
−1/3
+
− e−πt ( 2
+
)
s2 + 1 s2 + 4
s + 1 s2 + 4
1/3
1
2
1/3
1
2
−πt
−
−
e
−
s2 + 1 6 s2 + 4
s2 + 1 6 s2 + 4
1
1
1
1
L(sin t) − L(sin(2t)) − e−πt
L(sin t) − L(sin(2t))
3
6
3
6
=
=
=
Therefore, y =
1
(s2 +1)(s2 +4)
1
3
sin t −
1
6
sin(2t) − uπ (t)( 13 sin(t − π) −
16
1
6
sin(2(t − π))).
+
−1/3
s2 +4 .
17