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Then a solution which is linearly independent ofϕ1is given by ϕ2(t) = ϕ1(t)
1 ϕ1(t)2 exp
−3 4
1 t2dt
dt=t
1 t2 exp
3 4 ·1
t
= −t
exp 3
4 ·1 t
d
1 t
==−4 3texp
3 4 ·1
t
. Here−4
3 can be included in the arbitrary constant, so the complete solution is x(t) =c1t+c2texp
3 4t
, t >0, wherec1 andc2are arbitrary constants.
Example 1.11 Find the complete solution of the differential equation d2x
dt2 −coshtdx
dt −(1−coth2t)x=et
in the interval ]0,∞[, given thatx= sinht is a solution of the corresponding homogeneous equation.
We give here four variants.
1) It is seen byinspectionthat the equation can be written in the following way, et = d2x
dt2 −
cotht· dx dt + d
dt{cotht} ·x
=d2x dt2 − d
dt(cotht·x) = d dt
dx
dt −cosht sinht x
= d
dt
sinht 1
sinht ·dx
dt − cosht sinh2t ·x
= d dt
sinht d
dt x
sinht
. Then by an integration,
sinht· d dt
x sinht
=et+c2, hence by a rearrangement,
d dt
x sinht
= et
sinht + c2
sinht = 2e2t
e2t−1 + 2c2et e2t−1 = d
dtln(e2t−1) +c2
1
et−1 − 1 et+ 1
et. Another integration gives
x
sinht = ln(e2t−1) +c1+c2ln
et−1 et+ 1
,
so finally we obtain (with various equivalent variants) x = sinht·ln(e2t−1) +c1sinht+c2sinht·ln
et−1 et+ 1
= sinht·ln(2etsinht) +c1sinht+c2sinht·ln
tanht 2
= sinht· {t+ ln sinht}+c1sinht+c2sinht·ln
tanh t 2
.
36 2) It we putx=ϕ1(t)·y= sinht·y, then
dx
dt = sinht· dy
dt + cosht·y, d2x
dt2 = sinht·d2y
dt2 + 2 cosht· dy
dt + sinht·y, hence by an insertion into the differential equation
et = d2x
dt2 −cotht·dx dt + 1
sinh2tx
= sinht· d2y
dt2 + 2 cosht· dy
dt + sinht·y−coshtdy
dt −cosh2t
sinht y+ 1 sinhty
= sinht· d2y
dt2 + cosht· dy dt = d
dt
sinht·dy dt
. Then by an integration,
sinht·dy
dt =et+c2, from which we e.g. get
dy
dt = 2e2t
e2t−1 + c2
2 sinhtt2cosht2 = d
dtln(e2t−1) +c2· 1
tanh2t · 1 2 cosh2 t
2 .
Another integration gives y= ln(e2t−1) +c1+c2ln
tanht
2
, so finally,
x= sinht·y= sinht·ln(e2t−1)+c1sinht+sinht·ln tanht 2. 3) The equation is already normed, sof1(t) =−cotht, thus
ϕ2(t) = ϕ1(t) 1
ϕ1(t)2 exp(−
f1(t)dt)dt
= sinht 1
sinh2t exp
cosht sinht dt
dt
= sinht 1
sinh2t exp(ln sinht)dt= sinht
sinht cosh2t−1dt
= sinht·1 2
1
cosht−1 − 1 cosht+ 1
dcosht
= sinht·1 2ln
cosht−1 cosht+ 1
== sinht·ln tanht 2.
The Wro´nski determinant is
W(t) =
ϕ1 ϕ2
ϕ1 ϕ2 =
sinht sinht·ln tanht 2 cosht cosht·ln tanht
2+ sinht· 1 2
sinht
cosht−1 − sinht cosh+1
=
sinht 0
0 sinht· sinht cosh2t−1
=
sinht 0
0 1
= sinht.
When we apply a solution formula we get the particular solution x0(t) = ϕ2(t)
ϕ1u
W dt−ϕ1(t) ϕ2u
W dt
= sinht·ln tanht 2
sinht·et
sinht dt−sinht
sinht·ln(tanh2t)et sinht dt
= etsinht·ln tanh t
2−sinht
ln
tanht 2
etdt
= etsinht·ln tanh t
2−sinht·ln(tanh t
2)·et+ sinht et
sinhtdt
= 0 + sinht·
2e2t
e2t−1dt= sinht·ln(e2t−1).
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38 Summing up the complete solution is
x= sinht·ln(e2t−1) +c1sinht+c2sinht·ln tanht 2. 4) Sinceϕ1(t) = sinht andu(t) =et, and
Ω(t) = exp(−
cotht dt) = 1 sinht, we get that
x(t) = sinht
sinht sinh2t
c1+
sinht sinht ·etdt
dt+c2
= sinht
1
sinht(c1+et)dt
+c2sinht
= sinht
2e2t
e2t−1dt+c2sinht+c1sinht
dcosht cosh2t−1
= sinht·ln(e2t−1) +c2sinht+1
2c1sinht·ln
cosht−1 cosht+ 1
.
Example 1.12 Find the complete solution of the differential equation t(t2+ 1)d2y
dt2 + (4t2+ 2)dy
dt + 2ty= 2, t∈R+,
given that the corresponding homogeneous equation has the solution y=1
t, t∈R+. Hint:
4t2+ 2
t3+t dt= ln(t4+t2), t∈R+. First method. We get by some deftness that
2 = t(t2+ 1)d2y
dt2 + (4t2+ 2)dy dt + 2ty
= (t2+ 1)
td2y
dt2 + 1· dy dt
+ (3t2+ 1)dy dt + 2ty
= (t2+1)d dt
tdy
dt + 1·y
−(t2+1)dy
dt + (3t2+1)dy dt + 2ty
= (t2+ 1)d dt
d dt(ty)
+ 2t2dy dt + 2ty
= (t2+ 1)d dt
d dt(ty)
+ 2t
tdy
dt + 1·y
= (t2+ 1)d dt
d dt(ty)
+ d
dt(t2+ 1)· d dt(ty)
= d
dt
(t2+ 1)d dt(ty)
,
thus d dt
(t2+ 1)d dt(ty)
= 2. Then by an integration,
(t2+ 1)d
dt(ty) = 2t+c2, dvs. d
dt(ty) = 2t
t2+ 1 + c2
t2+ 1. And by another integration,
ty= ln(1 +t2) +c2Arctant+c1, and the complete solution is
y=ln(1 +t2)
t +c1·1
t +c2·Arctan t
t , t∈R+, where c1 andc2 are arbitrary constants.
Second method. The standard method. Firstnorm the equation, d2y
dt2 + 4t2+ 2 t(t2+ 1)
dy dt + 2
t2+ 1y= 2 t(t2+ 1). Sinceϕ1(t) = 1
t, we have ϕ2(t) = 1
t
t2exp
−
4t2+ 2 t3+t dt
dt= 1
t
t2· 1
t4+t2dt=1 t
dt
t2+ 1 =Arctan t
t .
Then we compute the Wro´nskian,
W(t) =
1 t
Arctant t
−1
t −Arctan t
t2 + 1
t(t2+ 1)
= 1
t2(t2+ 1). A particular solution is
ϕ(t) = ϕ2(t)
ϕ1(t)u(t)
W(t) dt−ϕ1(t)
ϕ2(t)u(t) W(t) dt
= 1
t Arctant 1
t ·t2(1 +t2)· 2
t(t2+ 1)dt−1 t
Arctant
t ·t2(1 +t2)· 2 t(t2+ 1)dt
= 1
t Arctant·2t−1 t
2 Arctant dt= 2 Arctant−1 t
2tArctan t− 2t
t2+ 1dt
= 1
t ln(1 +t2). The complete solution is
y=1
t ln(1 +t2) +c1
t +c2Arctant
t , t∈R+, where c1 andc2 are arbitrary constants.
40 Example 1.13 Consider the differential equation (5) t(1 +t)d2y
dt2 + (2 + 3t)dy
dt +y= 0, t∈R+.
1) Prove thaty=t−1,t∈R+ is a solution of (5), and then find the complete solution.
Hint: Exploit that 2 + 3t t(1 +t) =2
t + 1 1 +t.
2) Find the complete solution of the differential equation t(1 +t)d2y
dt2 + (2 + 3t)dy
dt +y= 1
1 +t, t∈R+.
First variant. This problem is immediately solved by some reformulations of the linear, inhomoge- neous equation of (2),
1
1 +t = t(1 +t)d2y
dt2 + (2 + 3t)dy dt +y
=
(t+t2)d2y
dt2 + (1+2t)dy dt
+
(1+t)dy dt + 1·y
= d dt
(t+t2)dy dt
+ d
dt{(1+t)y}
= d
dt
(1 +t)
tdy dt + 1·y
= d dt
(1 +t)d dt(t·y)
. If this is integrated, we obtain with some arbitrary constantc2 that
(1 +t)d
dt(t·y) =c2+ ln(1 +t), fort >0, hence by a rearrangement
d
dt(t·y) = ln(1 +t) 1 +t + c2
1 +t, fort >0.
By another integration we obtain with another arbitrary constant c1, t·y= 1
2{ln(1 +t)}2+c1+c2ln(1 +t), fort >0, and the complete solution is er
y=1
2 ·{ln(1 +t)}2
t +c1
t +c2·ln(1 +t)
t fort >0, where c1 andc2 are arbitrary constants.
In particular,y=t−1is a solution of the homogeneous solution, and y=c1
t +c2·ln(1 +t) t
is the complete solution of the homogeneous equation.
Second variant. The standard method.
1) A check fort >0 shows that t(1 +t)d2
dt2 1
t
+ (2 + 3t)d dt
1 t
+1
t =t(1 +t)d dt
−1 t2
−(2 + 3t)1 t2 +1
t
=2t(1 +t)
t3 −2 + 3t t2 +1
t = 1
t2{2+2t−2−3t+t}= 0. It follows thatϕ1(t) =1
t is a solution of (1.13).
Then norm the equation, d2y
dt2 + 2 + 3t t(1 +t)
dy
dt + y
t(t+ 1) = 0,
where the right hand side in the inhomogeneous case is 1 t(1 +t)2.
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42 It follows from the equation above that
f1(t) = 2 + 3t t(1 +t) =2
t + 1 1 +t,
so a linearly independent solution is given by ϕ2(t) = ϕ1(t)
exp
−
f1(t)dt 1
ϕ1(t)2dt=1 t
exp
− 2 t + 1
1 +t
dt
t2dt
= 1
t
exp(−ln|t2(1 +t)|)·t2dt= 1 t
t2
t2(1 +t)dt= 1 t
dt
1 +t =ln(1 +t)
t .
The complete solution of the homogeneous equation (5) is then y=c1·1
t +c2· ln(1 +t)
t fort >0. 2) The normed, inhomogeneous equation is now
d2y dt2 +
2 t + 1
1 +t dy
dt + 1
t(t+ 1)y= 1 t(1 +t)2, thus
u(t) = 1 t(1 +t)2.
The Wro´nskian can be computed in various ways:
(a) W(t) = exp
− 2
t + 1 1 +t
dt
= 1
t2(1 +t).
(b) W(t) =
ϕ1 ϕ2
ϕ1 ϕ2 =
1 t
ln(1 +t) t
−1
t2 −ln(1 +t)
t2 + 1
t(1 +t)
= 1
t2(1 +t). We get (NB, only one of the many possible variants, Ω(t) =W(t)−1)
ϕ0(t) = ϕ2(t)
ϕ1(t)u(t)
W(t) dt−ϕ1(t)
ϕ2(t)u(t) W(t) dt
= ln(1 +t) t
1
t · 1
t(1 +t)2 ·t2(1 +t)dt−1 t
ln(1 +t)
t · 1
t(1 +t)2 ·t2(1 +t)dt
= ln(1 +t) t
dt 1 +t−1
t
ln(1 +t)
1 +t dt= {ln(1 +t)}2
t −1
2 ·{ln(1 +t)}2 t
= 1
2·{ln(1 +t)}2
t .
Due to the linearity the complete solution is y=1
2 ·{ln(1 +t)}2
t +c1·1
t +c2·ln(1 +t)
t , t >0, wherec1 andc2 are arbitrary constants.
Example 1.14 Consider the differential equation (6) t2d2y
dt2 +tdy
dt −y= 0, t∈R+.
1) Prove that (6) has the solutiony=t,t∈R+, and then find the complete solution.
2) Find the complete solution of the differential equation t2d2y
dt2 +tdy
dt −y=t, t∈R+. This example can be solved in many ways.
First variant.
1) If we puty=tinto the equation, we get t2d2y
dt2 +tdy
dt −y=t2·0 +t·1−t= 0,
proving thaty=tis a solution of the homogeneous equation, so ϕ1(t) =t. By norming, i.e. division byt2, we get fort >0 that
d2y dt2 +1
t dy dt − 1
t2y= 0,
and = 1
t i spørgsm˚al (2)
. Since
Ω(t) = exp
f1(t)dt
= exp 1
t dt
= exp(lnt) =t, we get
ϕ2(t) =ϕ1(t) 1
ϕ1(t)2 · 1
Ω(t)dt=t dt
t2·t =t
t−3dt=−1
2t·t−2=−1 2 ·1
t. The complete solution of the homogeneous equation is
y(t) =c1t+c2· 1
t, t >0, wherec1 andc2are arbitrary constants.
2) Sinceu(t) =1
t by the norming, we get the particular solution y0(t) = ϕ1(t)
1
ϕ1(t)2Ω(t){ϕ1(t)Ω(t)u(t)dt}dt=t 1
t2·t
t·t·1 t dt
dt
= t 1
t3 ·1
2 ·t2dt= 1 2tlnt.
Then the complete solution is by the linearity, y(t) = 1
2tlnt+c1t+c2·1
t, t >0, wherec1 andc2are arbitrary constants.
44 Second variant.
1) If we puty=ta into the equation, then t2d2y
dt2 +tdy
dt −y=a(a−1)ta+ata−ta= (a2−1)ta.
This expression is identical 0 for t >0, ifa=±1, so the complete solution is y(t) =c1t+c2·1
t, t >0, wherec1 andc2 are arbitrary constants.
2) Then note that the right hand side of the (non-normed) equation t2d2y
dt2 +tdy
dt −y=t,
is a solution of the homogeneous equation. We thereforeguess a particular solution of the form y0(t) =a·tlnt,
where dy0
dt =a(1 + lnt) and d2y0
dt2 =a t. Then by insertion,
t2d2y dt2 +tdy
dy −y=a{t+t+tlnt−tlnt}= 2at, which is equal to tfora= 1
2. The complete solution is
y(t) =1
2tlnt+c1t+c2· 1
t, t >0, wherec1 andc2 are arbitrary constants.
Third variant.
1) By the monotonous substitutiont=eu,u= lnt, and the chain rule we get, dy
dt = du dt · dy
du =1 t
dy
du, dvs. tdy dt =dy
du, and
t2d2y dt2 =t2d
dt 1
t dy du
=t2·
−1 t2
dy
du +t2· 1 t2
d2y du2 =d2y
du2 −dy du, hence
t2d2y dt2 +tdy
dt −y= d2y du2 −dy
du +dy
du−y= d2y
du2 −y=t=eu,
and thus d2y
du2 −y=eu, u= lnt, t >0.
The characteristic equation R2−1 = 0 has the rootsR=±1, so the homogeneous equation has the complete solution
y=c1eu+c2e−u=c1t+c21
t, t >0, wherec1 andc2are arbitrary constants.
2) Since the right hand side is of the form eu =ϕ1(u), we guess a solution of the form y = aueu where
dy
du =a(u+ 1)eu og d2y
du2 =a(u+ 2)eu. Then by insertion,
d2y
du2 −y=a(u+ 2)eu−aueu= 2aeu, which is equal toeu fora= 1
2, and the complete solution is y= 1
2ueu+c1eu+c2e−u=1
2tlnt+c1t+c21
t, t >0, wherec1 andc2are arbitrary constants.
Fourth variantWe get by some clever manipulation on the inhomogeneous equation that t = t2d2y
dt2 +tdy dt −y=
t2d2y
dt2 + 2tdy dt
−
tdy dt + 1·y
= d dt
t2dy
dt −ty
= d
dt
t3 1
t dy dt − 1
t2y
= d dt
t3d
dt y
t
, thus
d dt
t3d
dt y
t
=t.
Then by an integration, t3d
dt y t
=−2c2+t2
2, dvs. d dt
y t
=−2c2
t3 + 1 2t, and by another integration,
y
t =c1+c2
t2 +1 2lnt,
and the complete solution of the inhomogeneous equation is y=1
2tlnt+c1t+c21
t, t >0, where c1 andc2 are arbitrary constants.
46