1. Solve the differrential equation:
2t2
d2 y
dy
+ 3t − 12y = t3
2
dt
dt
(1)
Solution:
First we perform the transformation to the z variable, defined by t = ez .
For a general differential equation, this transforms the equation from
the form:
d2 y
dy
(2)
αt2 2 + βt + γy = tn
dt
dt
to
dy
d2 y
(3)
α 2 + (β − α) + γy = enz
dz
dz
Applying this to Eq.1 we get:
d2 y dy
−
− 12y = e3z
(4)
dz 2 dz
To solve the equation we must obtain the complimentary solution and
the particular solution. After we have obtained both the total solution
is just their sum. The final step is then to convert z back to t.
2
(a) First the complimentary solution. We get this by solving the homogenous version of the differential equation, i.e. setting the right
hand side to zero.
d2 y dy
− 12y = 0
(5)
2 2−
dz
dz
This is just a simple second order O.D.E., which can be solved via
the usual substitution of y = eλt . This produces the characteristic
polynomial:
2λ2 − λ − 12 = 0
(6)
Which can be factorised as:
(2λ + 3)(λ − 4)
(7)
Hence the roots are λ1 = − 32 and λ2 = 4. So we have the complimentary solution:
3
yc (z) = C1 e− 2 z + C2 e4z
1
(8)
(b) To obtain the particular solution, we restore the right hand side:
2
d2 y dy
− 12y = e3z
−
dz 2 dz
(9)
The ansatz we use is y = Ce3z , i.e. the right hand side with a
constant. Each of the derivatives is then:
dy
dy
= 3Ce3z 2 = 9Ce3z
dz
dz
(10)
If we place these into Eq.9 we obtain:
18Ce3z − 3Ce3z − 12Ce3z = e3z
(11)
Dividing by e3z we get:
18C − 3C − 12C = 1
3C = 1
1
C=
3
(12)
(13)
(14)
Hence the particular solution is yp (z) = 31 e3z
(c) Hence the entire solution is:
3
1
y(z) = yp (z) + yc (z) = e3z + C1 e− 2 z + C2 e4z
3
(15)
The final step is to convert this back to a function of t, for which
we use z = ln(t). So,
3
1
y(t) = e3ln(t) + C1 e− 2 ln(t) + C2 e4ln(t)
3
3
1 3
= t + C1 t− 2 + C2 t4
3
(16)
(17)
Which is the solution.
2. Solve the differential equation:
1 + t2 dy
+ y = t2 + 2t4 + t6
2t dt
2
(18)
Solution:
This is a first order differential equation. Usually, these are written in
the form:
dy
+ p(t)y = f (t)
dt
So we divide by
(19)
1+t2
2t
to cast it in this form:
dy
2t
2t
+
t2 + 2t4 + t6
y=
dt 1 + t2
1 + t2
(20)
We can simplfy the right hand side:
=
2t
1 + t2
2t
1 + t2
t2 + 2t4 + t6
(21)
(1 + t2 )(t2 + t4 ) = (2t)(t2 + t4 ) = 2t3 + 2t5
(22)
And so we have:
2t
dy
+
y = 2t3 + 2t5
2
dt 1 + t
(23)
2t
3
5
This is now in the form of Eq.23, with p(t) = 1+t
2 and f (t) = 2t + 2t .
To solve a first-order differential equation such as this we construct the
function I(t):
Z t
Z t
2s
p(s)ds =
I(t) =
ds
(24)
2
c 1+s
c
To perform this integral will require a substitution. The only possibly
useful ones are u = s2 and u = 1 + s2 . Let’s try the second one first:
u = s2 + 1
du
= 2s
ds
With this substitution the integral is:
Z t2 +1
2
1
I(t) =
du = [ln(u)]tc2+1
= ln(t2 + 1) − ln(c2 + 1)
+1
c2 +1 u
3
(25)
(26)
(27)
We choose c = 0 for simplicity, in which case ln(c2 + 1) = 0.
We then multiply the differential equation Eq.23 by eI(t) to obtain:
2 +1)
eln(t
dy
2t
2
2
+ eln(t +1)
y = eln(t +1) (2t3 + 2t5 )
2
dt
1+t
As always this can be written as:
d ln(t2 +1) 2
e
y = eln(t +1) (2t3 + 2t5 )
dt
2 +1)
However eln(t
(28)
(29)
= t2 + 1, so this is just:
d
(t2 + 1)y = (t2 + 1)(2t3 + 2t5 )
dt
d
(t2 + 1)y = 2t7 + 4t5 + 2t3
dt
To solve we integrate both sides:
Z
Z
d
2
(t + 1)y dt = 2t7 + 4t5 + 2t3 dt
dt
1
2
1
(t2 + 1)y = t8 + t6 + t4 + C
4
3
2
We just divide by (t2 + 1) to obtain the solution:
Z
Z
d
2
(t + 1)y dt = 2t7 + 4t5 + 2t3 dt
dt
1 8
t + 23 t6 + 12 t4
C
4
+ 2
y=
2
t +1
t +1
(30)
(31)
(32)
(33)
(34)
(35)
3. Solve the differential equation:
(t2 + 1)
d2 y
dy
− 8t + 12y = 0
2
dt
dt
(36)
Solution:
This cannot be solved by using t = ez , since that only works in the
case where t2 multiplies the second derivative, where as here we have
t2 + 1. So we use a series solution:
y(t) =
∞
X
n=0
4
an tn
(37)
First we work out the expression for each term in the equation. First,
the first derivative:
∞
dy X
=
an (n)tn−1
dt
n=0
(38)
∞
dy X
t =
an (n)tn
dt
n=0
(39)
and also the second derivative:
∞
d2 y X
an (n)(n − 1)tn−2
=
dt2
n=0
(40)
∞
t2
d2 y X
=
an (n)(n − 1)tn
dt2
n=0
(41)
Putting all of these into the original equation we have:
∞
X
n
an (n)(n − 1)t +
n=0
∞
X
n−2
an (n)(n − 1)t
n=0
−
∞
X
n
8an (n)t +
n=0
∞
X
12an tn = 0
n=0
(42)
However the second term does not have the correct power of t, so to
correct this will shift the sum variable. We use instead m = n − 2 in
which case
∞
X
an (n)(n − 1)tn−2
(43)
n=0
becomes:
∞
X
am + 2(m + 2)(m + 1)tm
(44)
m=−2
however m is just an name for the variable, which we can replace by n,
and thus we have:
∞
X
an + 2(n + 2)(n + 1)tn
n=−2
5
(45)
However this series still does not begin at n = 0. However if you work
out the first two terms they are zero, both because one of the (n + 2) or
(n+1) terms will be zero, or because a−1 and a−2 are zero by definition.
So we have:
∞
X
an + 2(n + 2)(n + 1)tn
(46)
n=0
Hence the expression Eq.(42), becomes:
∞
X
n
an (n)(n − 1)t +
∞
X
n
an + 2(n + 2)(n + 1)t −
n
8an (n)t +
n=0
n=0
n=0
∞
X
∞
X
12an tn = 0
n=0
(47)
Since each sum starts with n = 0 and involves the same power of t,
they can be combined into one sum:
∞
X
(an (n)(n − 1) + an + 2(n + 2)(n + 1) − 8an (n) + 12an )tn = 0
n=0
(48)
∞
X
(an (n)(n − 1) + an + 2(n + 2)(n + 1) − 8an (n) + 12an )tn = 0
n=0
(49)
Which means the expression involving the coeffecients is zero:
an (n)(n − 1) + an + 2(n + 2)(n + 1) − 8an (n) + 12an = 0
(50)
Which we can manipulate to obtain the recursion relation:
an ((n − 1)(n) − 8(n) + 12)
(n + 2)(n + 1)
an (n2 − 9n + 12)
an + 2 = −
(n + 2)(n + 1)
an + 2 = −
(51)
(52)
4. Solve the differential equation:
d2 y
dy
−
4
− 12y = f (t)
dt2
dt
6
(53)
Where f (t) is the function:
f (t) = cosh(t − 1)
(54)
which has Fourier coefficients:
cn =
sinh(1)
1 + n2 π 2
(55)
Solution:
We will get the particular solution first.
To solve this differential equation we must express f (t) as a Fourier
series. The expression for f(t) is:
f (t) =
∞
X
cn eiπnt
(56)
sinh(1)
1 + n2 π 2
(57)
n=0
cn =
Hence we are trying to solve:
∞
X
dy
d2 y
cn eiπnt
−
4
−
12y
=
dt2
dt
n=0
(58)
Since the equation is linear, we can just solve the equation for a general
exponential function and then sum the answers. So we solve:
d2 y
dy
− 4 − 12y = eiπnt
2
dt
dt
(59)
We make the ansatz yp (t) = Ceiπnt . So the derivatives are:
d2 y
= C(−π 2 n2 )eiπnt
dt2
dy
= C(iπn)eiπnt
dt
(60)
(61)
Placing these derivatives into Eq.(59) we get:
C(−π 2 n2 )eiπnt − 4C(iπn)eiπnt − 12Ceiπnt = eiπnt
7
(62)
If we divide by eiπnt we get:
C(−π 2 n2 ) − 4C(iπn) − 12C = 1
C(−π 2 n2 − 4iπn − 12) = 1
1
C=
2
2
−π n − 4iπn − 12
(63)
(64)
(65)
This can be placed directly into the original Fourier expansion formula
of Eq.(56) to obtain yp (t), since the equation is linear:
yp (t) =
∞
X
n=0
cn
−π 2 n2
1
eiπnt
− 4iπn − 12
(66)
sinh(1)
1 + n2 π 2
(67)
cn =
Now we only need to get the complimentary solution. The differential
equation with the right hand side set to zero is:
d2 y
dy
− 4 − 12y = 0
2
dt
dt
(68)
From which we get the characteristic polynomial:
λ2 − 4λ − 12 = 0
(λ − 6)(λ + 2) = 0
(69)
(70)
So the roots are λ = 6, −2. Hence the complimentary solution is:
yc (t) = C1 e6t + C2 e−2t
(71)
So the full solution is:
∞
X
n=0
y(t) = yp (t) + yc (t) =
(72)
sinh(1)
−1
eiπnt + C1 e6t + C2 e−2t
2
2
2
2
1 + n π π n + 4iπn + 12
(73)
8