M2Z03: SAMPLE FINAL EXAM C SOLUTIONS
1. The general solution of the differential equationy0= 6x
2
2y+ cosy is →(A) y2+ siny= 2x3+c
(B) 2y2+ cosy= 6x3+c
(C) 2y2+ siny= 6x3+c
(D) y2+ cosy=x3+c
(E) y2+ siny=x2+c
Solution. This is a separable equation.
(2y+ cosy)y0= 6x2.
Hence,
Z
2y+ cosy dy=
Z
6x2dx or
y2+ siny= 2x3+c.
2. The solution of the initial value problem (1 +x2)y0+ 2x y= cosx,y(0) = 1 is
(A) y=1 +x
2
sin(x)
(B) y= x
2+ 2
2 + sin(x)
(C) y= x
2+ 1
1 + cos(x)
→(D) y= 1 + sin(x) x2+ 1
(E) y= 2−cos(x) x2+ 1
3. Which of the following homogeneous linear differential equations hasy(x) =xsin(√2x) as a solution?
(A) y00+ 2y= 0
(B) y00+ 2y0+ 2y= 0 →(C) y(4)+ 4y00+ 4y= 0
(D) y(4)+ 2y00+ 2y= 0
(E) None of the above
Solution. y00+ 2y= 0 iffy=c
1 sin(
√
2x) +c2cos(
√ 2x). y00+ 2y0+ 2y= 0 iffy=c
1e−xsinx+c2e−xcosx
y(4)+ 4y00+ 4y iffy=c
1sin(
√
2x) +c2 cos(
√
2x) +c1xsin(
√
2x) +c2xcos(
√ 2x) y(4)+ 2y00+ 2y= 0 has an auxiliary equation with complex, but no multiple roots.
4. Which of the following is a fundamental set of solutions for the homogeneous differential equationy00−6y0+ 9y= 0 ?
(A) {e3x, e3x}
(B) {e−3x, e3x}
→(C) {e3x, x e3x}
(D) {e−3x, x e−3x}
(E) {ex/3, e−x/3}
Solution. The auxiliary equation is m2−6m+ 9 = 0 or (m−3)2 = 0. The general solution is thus
5. The Wronskian determinant of the set of functions{x, ex, x+ex}is
(A) ex
(B) e2x
(C) xe2x
→(D) 0
(E) 1
Solution. W(x) = 0 since the functions are linearly dependent.
6. The homogeneous equationy00+ 16y= 0 has for general solution
y(x) =c1cos(4x) +c2 sin(4x)
(you need not check that). The boundary–value problemy00+16y= 0,y(0) = 0,y(π) = 1 has:
(A) the unique solutiony= 1−cos(4x) →(B) no solution
(C) the unique solutiony= sin(4x)
(D) two solutions: y = cos(4x) andy= sin(4x)
(E) infinitely many solutions: y = Csin(4x), whereC is any number
Solution. The condition y(0) = 0 implies that c1 = 0 and thusy(x) =c2sin(4x). In
that case, the conditiony(π) = 1 cannot be satisfied since sin(4π) = 0.
7. The homogeneous differential equationy(4)+y(3)−2y00= 0 has for general solution
y(x) =c1+c2x+c3e−2x+c4ex
(you need not check that). For this equation, the initial value problem y(0) = 1, y0(0) =−1,y00(0) = 4,y(3)(0) =−8 has solution
(A) y(x) = 0
(B) y(x) =x+e−x
(C) y(x) = 1 +e−2x
→(D) y(x) =x+e−2x
(E) y(x) = 1 +e−x
Solution. We have
y0(x) =c2−2c3e−2x+c4ex, y00(x) = 4c3e−2x+c4ex, y000(x) =−8c3e−2x+c4ex.
The initial condition give the equations
c1+c3+c4= 1, c2−2c3+c4=−1, 4c3+c4= 4, −8c3+c4=−8
8. The solution of the initial value problem 4y00−y = x ex2 subject to initial condition
y(0) = 1,y0(0) = 0 is
→(A) y(x) =3 4e
x 2 +1
4e −x
2 +1
8x
2ex 2 −1
4xe
x 2
(B) y(x) =ex2 +e−x2 +x2ex2 −xex2
(C) y(x) =3 4e
−x 2 +1
4e
x 2 +1
8x
2e−x 2 −1
4xe −x
2
(D) y(x) =7 8
(E) None of the above
Solution. The auxiliary equation is 4m2−1 = 0 with roots m=±1
2. The
complemen-tary solution is thus
yc(x) =c1ex/2+c2e−x/2.
A particular solution has the formyp(x) = (A x2+B x)ex/2. We have
y0p(x) = (2A x+B)ex/2+
A
2 x
2+B
2 x
ex/2=
A
2 x
2+
2A+B 2
x+B
ex/2
and
yp00(x) =
A x+
2A+B 2
ex/2+
A
4 x
2+
A+B 4
x+B 2
ex/2
=
A
4 x
2+
2A+B 4
x+ (2A+B)
ex/2.
Thus
4yp00(x)−yp(x) = [(8A)x+ (8A+ 4B)]ex/2=x ex/2
which yieldsA= 18 andB =−1
4. The general solution is thus
y(x) =c1ex/2+c2e−x/2+
1 8x
2ex/2
−1 4x e
x/2.
and
y0(x) = c1 2 e
x/2−c2
2 e −x/2+
1 4x+
1 16x
2
ex/2−
1 4+ 1 8x
ex/2.
The conditiony(0) = 1 yieldsc1+c2= 1 and the conditiony0(0) = 0 that c21−c22−14 = 0.
This impliesc1= 34 andc2=14. The solution is thus
y(x) =3 4e
x/2+1
4e
−x/2+1
8x
2ex/2−1
4x e
9. If the method of undetermined coefficients is used to solve the linear differential equation y00+ 9y= sin(3x), then the form of a particular solution is
(A) yp(x) =Asin(3x) +Bcos(3x)
→(B) yp(x) =A xsin(3x) +B xcos(3x)
(C) yp(x) =Asin(3x)
(D) yp(x) =A x2sin(3x) +B x2cos(3x)
(E) yp(x) =A x3sin(3x) +B x3cos(3x)
Solution. The complementary solution is yc(x) = c1sin(3x) +c2 cos(3x). Hence,
yp(x) =A xsin(3x) +B xcos(3x).
10. The general solution of the linear differential equationx2y00−6x y0−8y=x2, for which
the homogeneous part has the solutionsy1(x) =x8 andy2(x) = 1x, is given by
(A) y(x) =c1x8+
c2
x − 1 48x6
(B) y(x) =c1x8+
c2
x − 1 27x3
(C) y(x) =c1x8+
c2
x − 1 48x2
(D) y(x) =c1x8+
c2
x − 1 12x
4
→(E) y(x) =c1x8+
c2
x − 1 18x
2
Solution. The Wronskian associated with y1(x) =x8 andy2(x) = 1x is
W(x) =
y1 y2
y01 y02
=
x8 x−1
8x7 −x−2
=−9x6.
When written in standard form, the right-hand side of the DE isf(x) = 1. Hence
yp(x) =−
Z x−1
−9x6dx
x8+
Z x8
−9x6dx
x−1
=
Z x−7
9 dx
x8−
Z x2
9 dx
x−1
=
−x −6
54
x8−
Z x3
27dx
x−1=−1 18x
2
The general solution is thusy(x) =c1x8+
c2
x − 1 18x
11. Consider the boundary–value problemy00+λ y= 0,y0(0) = 0,y0(π) = 0. The smallest eigenvalueλfor this problem is
(A) 1
(B) 12
→(C) 0
(D) 1 4
(E) π
Solution. Whenλ <0, we haveλ=−ω2, whereω >0. The auxiliary equation becomes
m2−ω2= 0 with rootsm=±ω. The general solution is theny(x) =c1eωx+c2e−ωx.
Hence, y0(x) = ω(c1eωx−c2e−ωx). The condition y0(0) = 0 implies c1 = c2, so that
y0(x) =ω c1(eωx−e−ωx). The condition y0(π) = 0 then implies ω c1(eωπ−e−ωπ) = 0
which yieldsc1= 0 and no non-trivial solution exists in that case. Ifλ= 0, the general
solution isy(x) =c1+c2x. The condition y0(0) = 0 impliesc2= 0 and thus y(x) =c1
and the other conditiony0(π) = 0 is then also satisfied. It follows that 0 is an eigenvalue with corresponding eigenvectorφ0(x) = 1. Clearly, 0 is then the smallest eigenvalue.
12. Suppose thatA is a 2×2 real matrix and that one of its eigenvalue isλ= 2 +i with corresponding eigenvector v =
2 i
. Then, the general solution of the 1st order system
X0=AXis given by
(A) X(t) =c1
etcos(2t)
−2etsin(2t)
+c2
etsin(2t)
2etcos(2t)
→ (B) X(t) =c1
2e2tcost −e2tsint
+c2
2e2tsint e2tcost
(C) X(t) =c1
2e2tsint
−e2tcost
+c2
−e2tsint
2e2tcost
(D) X(t) =c1
2etcost
−etsint
+c2
e2tsint
2e2tcost
(E) X(t) =c1
2e2tsint −e2tsint
+c2
2e2tcost e2tcost
Solution. A complex-valued solution is
Z(t) =e(2+i)t
2 i
=e2t (cost+isint)
2 0
+i
0 1
=
2e2tcost
−e2tsint
+i
2e2tsint
e2tcost
.
Two linearly independent real solutions are thus
2e2tcost
−e2tsint
and
2e2tsint
e2tcost
,
and the general solution is thus
X(t) =c1
2e2tcost
−e2tsint
+c2
2e2tsint
e2tcost
13. Suppose thatAis a 3×3 real matrix with characteristic polynomial
P(λ) = det(A−λI) =−λ3+ 2λ+ 4.
FindA if
A3=
8 4 −2
0 2 2
0 −2 2
.
(A) A=
1 1 −2
2 1 −2
−2 −2 2
(B) A=
2 1 −2
1 2 −1
0 −2 1
(C) A=
0 1 −2
2 1 1
1 −2 4
(D) A=
0 2 −1
2 1 2
4 −1 2
→(E) A=
2 2 −1
0 −1 1 0 −1 −1
Solution. By the Cayley-Hamilton theorem, we have P(A) = −A3+ 2A+ 4I = 0.
Thus,
A= 1/2A3−2I=
4 2 −1
0 1 1
0 −1 1
−
2 0 0 0 2 0 0 0 2
=
2 2 −1
0 −1 1 0 −1 −1
14. The symmetric matrix
A=
8 −2 −2
−2 11 −1 −2 −1 11
.
has characteristic polynomial P(λ) = −(λ−6) (λ−12)2 and v 1 =
1 −3
1
is an
eigen-vector associated with the eigenvalueλ= 12. Another eigenvector associated with the eigenvalueλ= 12 and orthogonal tov1 is given by
→(A) v2=
−4 1 7
(B) v2=
−1 0 1
(C) v2=
2 −1 −1
(D) v2=
3 1 0
(E) v2=
1 2 −4
Solution.Ifv=
x1
x2
x3
is an eigenvector associated with the eigenvalueλ= 2, we have
−4 −2 −2 −2 −1 −1 −2 −1 −1
x1
x2
x3
=
0 0 0
or 2x1+x2+x3 = 0. Furthermore, the fact that v is orthogonal to v1 means that
x1−3x2+x3= 0. The set of solutions to both equations has the form
v=t
−4 1 7
, t∈R.
15. Let
A=
2 a 3 −1
Find the value ofafor whichλ= 5 is an eigenvalue ofA.
(A) a= 2
(B) a=−5
(C) a=−3
→(D) a= 6
(E) a= 7
Solution. We have
det(A−5I) =
−3 a 3 −6
16. Find the value ofafor which the vectorv=
1 2 2
is an eigenvector of the matrix
A=
−1 0 2
0 1 a
a 2 0
.
(A) a= 1
(B) a=−1
(C) a=−3
→(D) a= 2
(E) a= 0
Solution. Ifλis the corresponding eigenvalue, we have
A=
−1 0 2
0 1 a
a 2 0
1 2 2
=λ
1 2 2
or
3 2 + 2a
a+ 4
=
λ 2λ 2λ
,
which implies thatλ= 3 anda= 2.
17. The eigenvaluesλof the matrix
A=
2 0 1 0 2 2 13 1 0
are given by:
(A) λ= 0,2,3
(B) λ=−2,1,4 →(C) λ=−3,2,5
(D) λ=−1,1,2
(E) λ=−13,2,3
Solution. We have
det(A−λ I) =
2−λ 0 1
0 2−λ 2
13 1 −λ
= (2−λ) (λ2−2λ−15) = (2−λ) (λ−5) (λ+ 3) = 0
18. A 3×3 matrix A has eigenvalues λ1 = −2, λ2 = 1 and λ3 = 4 with corresponding
eigenvectors
v1= −1 0 2
, v2= 1 1 1
, v3= 0 1 2 ,
respectively. Then, a matrixP satisfying
P−1AP =
4 0 0
0 −2 0
0 0 1
is given by:
(A) P=
−1 0 2
1 1 1
0 1 2
(B) P=
−1 1 0
0 1 1
2 1 2
(C) P=
1 −1 0
1 0 1
1 2 2
(D) P =
1 0 −1 1 1 0 1 2 2
→(E) P =
0 −1 1
1 0 1
2 2 1
Solution. P= [v3, v1, v2].
19. The 3rd order linear differential equation x000(t)−2x00(t) + 3x(t) = 0 can be written
as a 1st order linear system of the form X0(t) = AX(t) where X(t) =
x(t) y(t) z(t)
, with
y(t) =x0(t),z(t) =x00(t) andA being the matrix
(A)
1 0 0 0 1 0 0 3 −2
(B)
0 1 0 1 0 0 2 0 3
→(C)
0 1 0
0 0 1
−3 0 2
(D)
0 1 0
1 0 1
3 −2 1
(E)
1 1 1
1 1 0
−3 2 0
20. The solution of the initial value problem
(
x0(t) =−x(t),
y0(t) =−4x(t) + 3y(t),
with initial conditionsx(0) = 2,y(0) = 3 is given by
(A) x(t) =e−t+e3t, y(t) =e−t+ 2e3t
(B) x(t) = 3e−t−e3t, y(t) = 2e−t+e3t
→(C) x(t) = 2e−t, y(t) = 2e−t+e3t
(D) x(t) =−e−t+ 2e3t, y(t) =−e−t+ 3e3t
(E) x(t) = 2e−t, y(t) =−e−t+ 4e3t
Solution. LettingX(t) =
x(t) y(t)
, the system can be written asX0(t) =AX(t) in matrix
form, where A =
−1 0 −4 3
. The eigenvalues of A are λ = −1,3 with corresponding
eigenvectors
1 1
and
0 1
, respectively. The general solution is thus
c1e−t
1 1
+c2e3t
0 1
.
Lettingt= 0 yieldsc1= 2 and c1+c2= 3, i.e. c2= 1. Hence,
X(t) = 2e−t
1 1
+e3t
0 1
=
2e−t
2e−t+e3t
21. Suppose thatAis a real, 3×3 matrix with eigenvalues has eigenvaluesλ1=−1,λ2= 0
andλ3= 2 with corresponding eigenvectors
v1=
1 0 2
, v2=
1 1 1
, v3=
0 1 2
,
respectively. Let X(t) =
x(t) y(t) z(t)
be the unique solution of the initial value problem
X0(t) =AX(t) with initial conditionX(0) =
3 3 9
. Then, the value of the 1st component
ofX(t) att= 1 is
(A) x(1) = 2 +e
(B) x(1) =e+ 2e−1 →(C) x(1) = 1 + 2e−1
(D) x(1) =−1 +e−1+ 2e
(E) x(1) = 1−2e−1−e
Solution. The general solution is thus
c1e−t
1 0 2
+c2
1 1 1
+c3e2t
0 1 2
.
Lettingt= 0 yields
c1+c2= 3, c2+c3= 3, 2c1+c2+ 2c3= 9,
orc1= 2, c2= 1 andc3= 2. Hence,x(1) =c1e−1+c2= 2e−1+ 1.
22. Let A be an n×n real, symmetric matrix. Which of the following statements is not alwayscorrect?
→(A) A B=B Afor anyn×nsymmetric matrixB
(B) A is diagonalizable
(C) A2−A is symmetric (D) A2 is symmetric
(E) A2 is diagonalizable
Solution. (A) is false since, for example A =
0 0 0 1
and B =
0 1 1 0
are both
symmetric, but
A B=
0 0 0 1
0 1 1 0
=
0 0 1 0
6=
0 1 0 0
=
0 1 1 0
0 0 0 1
23. LetAbe an n×nreal matrix. Which of the following does notnecessarily imply that Ais diagonalizable:
(A) Ahasndistinct eigenvalues.
(B) Ais symmetric.
(C) A=B2, whereB is diagonalizable. (D) Ais invertible and A−1 is diagonalizable.
→(E) The 1st order systemX0(t) =AX(t), whereX(t) is a column vector withn com-ponents, admitsnlinearly independent solutions for−∞< t <∞.
Solution. (E)is true for any real matrix A, even ifAis not diagonalizable.
24. The Laplace transformL
(3t−1)2 is
→(A) 18
s3 −
6 s2 +
1 s
(B) 9
s2 −
6 s + 1
(C) 9s e−s/3
(D) (3s−1) 2
s3
(E) 1
9s e
s/3
Solution. We have (3t−1)2= 9t2−6t+ 1. Hence,
L
(3t−1)2 =18 s3 −
6 s2 +
1 s.
25. The Laplace transformL
e3t−2 is
(A) 1
3s−5
(B) 3
s−2 3
(C) e
2
s+ 3
→(D) e
−2
s−3
(E) 1
(3s−2)3
Solution. We havee3t−2=e−2e3t. Hence,
L
26. The Laplace transformL {t sin(2t)}is
(A) 4
(s2+ 4)2
→(B) 4s
(s2+ 4)2
(C) − 4s (s2+ 4)2
(D) 2s
(s2+ 4)2
(E) − 2s
2
s2+ 4
Solution. SinceL {sin(2t)}= s22+4, we have
L {tsin(2t)}=−d ds
2 s2+ 4
= 4s (s2+ 4)2.
27. The inverse Laplace transformL−1
1
s(s2+ 4)
is
(A) 1−cos(2t)
(B) 4 (1−cos(2t))
→(C) 1
4(1−cos(2t))
(D) 1
2t sin(2t)
(E) 1
4(1 + cos(2t))
Solution. Using partial fractions,
1 s(s2+ 4) =
1 4
1
s− s s2+ 4
.
Hence,
L−1
1
s(s2+ 4)
=1
4{1−cos(2t)}.
28. Suppose thatL {f(t)}=s e −πs
s2+ 4. Then, the valuef(3π) is (A) 0
→(B) 1
(C) −1
(D) 1
2
(E) 3π e
−3π2
9π2+ 4
Solution. SinceL {cos(2t)}= s
s2+ 4, we have
f(t) = cos(t−π)U(t−π).
29. The convolution (f ∗g) (t) off(t) =e−2tandg(t) =e2t is equal to
(A) 1
(B) t
(C) 1
2 e
2t−e−2t
(D) 1
4 e
2t+e−2t
→(E) 1
4 e
2t−e−2t
Solution. SinceL
e−2t = 1
s+ 2 andL
e2t = 1
s−2, we have
L {f∗g}= 1 s+ 2
1 s−2 =
1 4
1 s−2 −
1 s+ 2
.
It follows that
(f∗g) (t) = 1 4 e
2t−e−2t
.
30. Consider the differential equation
y0+ 2y=
(
1, 0≤t <3, 0, t≥3,
with initial conditiony(0) = 0. The Laplace transformL {y(t)} of the solution is
(A) e
−3s
s+ 2
→(B) 1−e
−3s
s(s+ 2)
(C) 1−e
−2s
s(s+ 3)
(D) e
−3s−1
s(s+ 2)
(E) 1 +e
−3s
s(s+ 2)
Solution. We havey0+ 2y=g(t) whereg(t) = 1− U(t−3). LettingY(s) =L {y(t)}, we have thus
s Y(s)−y(0) + 2Y(s) =s Y(s) + 2Y(s) = (s+ 2)Y(s) = 1 s−
e−3s
s and
Y(s) = 1−e −3s
31. Ify(t) is the solution of the differential equationy00+1
4y=δ(t−π) with initial conditions y(0) = 0,y0(0) = 1, theny(2π) is equal to
→(A) 2
(B) −2
(C) 0
(D) 4
(E) −4
Solution. LettingY(s) =L {y(t)}, we have
s2Y(s)−s y(0)−y0(0) +1
4Y(s) =s
2Y(s)
−1 +1
4Y(s) =e −πs
or
s2+1 4
Y(s) =e−πs+ 1.
Hence,
Y(s) = 2e −πs1
2
s2+1 4
+ 2
1 2
s2+1 4
and y(t) = 2 sin
t−π
2
U(t−π) + 2 sin
t
2
.
Thus,y(2π) = 2 sin π2+ 2 sin(π) = 2.
32. The system of differential equations
x0= 2x−y y0 = 4x−2y
with initial conditionsx(0) = 1 andy(0) = 1 has for solution
(A) x(t) = 2e2t−et
(B) x(t) = 1 + 2t
(C) x(t) = 1−t
→(D) x(t) = 1 +t
(E) x(t) =e2t
Solution. LettingX(s) =L {x(t)} andY(s) =L {y(t)}, we have
s X(s)−1 = 2X(s)−Y(s) s Y(s)−1 = 4X(s)−2Y(s)
or, in matrix form,
s−2 1 −4 s+ 2
X(s) Y(s)
=
1 1
Using Cramer’s rule,
X(s) =
1 1
1 s+ 2
s−2 1 −4 s+ 2
= s+ 1 s2 =
1 s +
1 s2.
33. The differential equation (4−x2)x3y00(x) +x y0(x)−2y(x) = 0 has all of its singular points classified as
→(A) x= 2 regular,x=−2 regular,x= 0 irregular
(B) x= 4 regular,x= 0 regular
(C) x= 4 regular,x=−4 regular,x= 0 regular
(D) x= 2 irregular,x=−2 irregular,x= 0 regular
(E) x= 2iregular,x=−2iregular,x= 0 irregular
Solution. In standard form, the DE is written as
y00(x)− 1
(x−2) (x+ 2)x2y
0(x) + 2
(x−2) (x+ 2)x3y(x) = 0.
We deduce that−2,0,2 are the singular points. Since,
− 1
(x+ 2)x2 and
2 (x+ 2)x3
are both analytic atx= 2, 2 is regular. Similarly, Since,
− 1
(x−2)x2 and
2 (x−2)x3
are both analytic atx=−2,−2 is regular. 0 is irregular since− 1
(x−2) (x+ 2)x is not analytic atx= 0.
34. The differential equation 9x y00+ 3y0+x y= 0 has a general solution of the form
→(A) y= ∞
X
n=0
anxn+
∞
X
n=0
bnxn+
2 3
(B) y= ∞
X
n=0
anxn+
∞
X
n=0
bnxn−
2 3
(C) y= ∞
X
n=0
anxn+
1
3 + (lnx)
∞
X
n=0
bnxn+
1 3
(D) y= ∞
X
n=0
anxn−
1
3 + (lnx)
∞
X
n=0
bnxn−
1 3
(E) y= ∞
X
n=0
anxn+
∞
X
n=0
bnxn−
1 3
Solution. In standard form, the DE is written as
y00(x) + 1 3xy
0(x) +1
9y(x) = 0.
We haveP(x) = 31x andQ(x) = 19 Thus x P(x) = 13 =a0+a1x+a2x2+. . .. Hence,
a0= 13. Also, x2Q(x) = x
2
9 =b0+b1x+b2x
2+. . . which yieldsb
0= 0. The indicial
equationr(r−1) +a0r+b0 reduces thus tor(r−1) +13r= 0 orr(r−23) and its roots
arer= 0 andr=23. The general solution is thus of the form
y= ∞
X
n=0
anxn+
∞
X
n=0
bnxn+
35. The differential equation y00+x y0+y = 0 has a power series solution y = ∞
X
n=0
anxn
where the recursion formula for the coefficientsan is
(A) an+2=
n−1 (n+ 2) (n+ 1)an
(B) an+1=−
1 n+ 2an
→(C) an+2=−
1 n+ 2an
(D) an+2=−
n−1 (n+ 2) (n+ 1)an
(E) an+2=−
1
n+ 1an+1− 1 n+ 2an
Solution. We have
y00+x y0+y= ∞
X
n=2
ann(n−1)xn−2+x
∞
X
n=0
ann xn−1+
∞
X
n=0
anxn
= ∞
X
n=0
an+2(n+ 2) (n+ 1)xn+
∞
X
n=0
ann xn+
∞
X
n=0
anxn
= ∞
X
n=0
{an+2(n+ 2) (n+ 1) +an(n+ 1)} xn = 0
It follows that
an+2(n+ 2) (n+ 1) +an(n+ 1) = 0, n≥0,
or
an+2=−
an(n+ 1)
(n+ 2) (n+ 1) =− an