**MATH2270**

**FW00**

**Solutions to
Assignment 5 **

1. (a) Find the general solutions to the following Cauchy-Euler differential equations:

(i) Let y = x^{r}. Substituting in the DE
yields r(r-1) + 3r - 8 = 0 with solutions r = 2 and -4 giving the
general solution y = C_{1}x^{2} + C_{2}x^{-4}.

(ii) Let y = x^{r}. Substituting in the DE
yields r(r-1) + 2r - 6 = 0 with solutions r = 2 and -3 giving the
general solution y = C_{1}x^{2} + C_{2}x^{-3}.

(b) Solve the boundary-value problem

Let
y = x^{r}. Substituting in the DE yields 2r(r-1) + r - 1 = 0
with solutions r = 1 and -1/2 giving the general solution y = C_{1}x
+ C_{2}x^{-1/2}. Applying the initial conditions we
get

C_{1
}+ C_{2} = 5 and 4C_{1} + C_{2}/2
= 13 with solution C_{1} = 3 and C_{1} = 2.

(c) Find the Wronskian of the pair of linearly independent solutions of the differential equations in parts (a) and (b).

The Wronskian of y_{1} and y_{2} is
definned as W(y_{1}, y_{2}) = y_{1 }y_{2}´
- y_{2 }y_{1}´

(a)(i) We have y_{1} = x^{2} and y_{2}
= x^{-4}. Then W = x^{2}(-4x^{-5}) - x^{-4}(2x)
= -6x^{-3}.

(a)(ii) We have y_{1} = x^{2} and y_{2}
= x^{-3}. Then W = x^{2}(-3x^{-4}) - x^{-3}(2x)
= -5x^{-2}.

(b) We have y_{1} = x and y_{2} =
x^{-1/2}. Then W = x(-1/2)x^{-3/2} - x^{-1/2}
= -(3/2)x^{-1/2}.

2. Given that y = x is a solution of the differential equation

find a second linearly independent solution.

Let y = xu. Then y´ = u + xu´ and y´´ = 2u´ + xu´´. Substituting this into the DE yields

(1 - x^{2})( 2u´ + xu´´) - 2x(
u + xu´) + 2xu = x(1 - x^{2})u´´+ (2 -
4x^{2})u´ = 0. If we let v = u´ then we have a
separable first-order DE for v which yields

which yields v = 1/[x^{2}(1-x^{2})].
Thus

Thus the second solution is

The DE is Legendre's equation of order 1.

3. (a) Verify that

and

are solutions of the second-order differential equation

showing that y_{1 }satisfies the DE. Similarly
if

showing that y_{2} is also a solution of the DE.

(b) Find the Wronskian of this pair of solutions.

The Wronskian is y_{1 }y_{2}´ -
y_{2 }y_{1}´ which from above is

which reduces to reduces to sin^{2}(x) + cos^{2}(x)
= 1

(c) Use the Oscillation Theorem to show that these solutions have an infinite number of zeros for x > 2.

Writing the DE as y´´ + Q(x) y = 0 we have
Q(x) = 1 - 2x^{-2} > 1/2 for x > 2. Thus from the
Oscillation Theorem y_{1} and y_{2} have an infinite
number of zeros and only one zero of one solution falls between
consecutive zeros of the other.

(d) Use MAPLE to plot these two solutions to show that the zeros of one solution fall between consecutive zeros of the other solution.

See MAPLE solution.

(e) Use variation of parameters to find the general solution to the differential equation

Write y = z_{1}y_{1} + z_{2}y_{2}.
Then in the notation of section 8.3 of the text we have a_{2}
= 1 and f = 2x. In addition we have shown that W = 1 so that

and

4. (a) Find the general solution of the following
differential equations as a power series about x_{0} = 0.

(i) y´´ + 2xy´ - 4y = 0

(ii)
(1 - x^{2}) y´´ - xy´ + r^{2}y =
0.

Find the recurrence relation for the coefficients, write out the first four terms of each series and, if possible, find a general expression for the coefficients.

The power series in both cases is of the form

In part (i) the coefficient of x^{n} is
(n+2)(n+1)c_{n+2} + 2nc_{n } - 4c_{n} giving
the recurrence relation

c_{n+2} = 2(2-n)c_{n}/[(n+2)(n+1)] Thus
c_{2} = 2c_{0}, c_{4} = 0 which implies c_{2n}
= 0 for n > 1. Also c_{3} = c_{1}/3,

c_{5} = -c_{3}/10 = -c_{1}/30,
c_{7} = -6c_{5}/42 = c_{1}/210 and c_{2n+1}
= (-1)^{n+1}c_{1}/[(2n+1)(2n-1)n!]

Thus y = c_{0}(1 + 2x^{2}) + c_{1}(x
+ x^{3}/3 - x^{5}/30 + x^{7}/210 ... +
(-1)^{n=1}x^{2n+1}/[(2n+1)(2n-1)n!]+ ...)

In part (ii) the coefficient of x^{n} is
(n+2)(n+1)c_{n+2} - n(n-1)c_{n} - nc_{n } +
r^{2}c_{n} giving the recurrence relation

c_{n+2} = (n-r)(n+r)c_{n}/[(n+2)(n+1)]
giving c_{2} = -r^{2}c_{0}/2 and c_{3}
= (1-r^{2})c_{1}/6 giving

y = c_{0}(1 - r^{2}x^{2}/2 +
...) + c_{1}(x + (1-r^{2})x^{3}/6 + ...).

The general coefficient is c_{2n} =
-[(2n-2)^{2}-r^{2}][(2n-4)^{2}-r^{2}]...[4-r^{2}]r^{2}c_{0}/(2n)!
and

c_{2n+1} =
[(2n-1)^{2}-r^{2}][(2n-3)^{2}-r^{2}]...[9-r^{2}][1-r^{2}]c_{1}/(2n+1)!

(b) In part (a) (ii) show that one solution is a polynomial of degree r if r is an integer. Write out the specific polynomials for r = 0, 1 and 2 if y(0) = 1 and y´(0) = 1.

From the recurrence relation we see that if r is an
integer the c_{r+2} = 0 so that c_{r+4} = c_{r+6}
= ... = 0 giving one solution as a polynomial of degree r.

If y(0) = 1 then c_{0} = 1 and if y´(0)
= 1 then c_{1} = 1. Thus the polynomial solutions are y = 1
if r = 0, y = x if r = 1 and y = 1 - 2x^{2} if r = 2.

5. (a) Use the method of Frobenius to find the indicial equation for the following differential equations

(i) x^{2}y´´ + x(x-1)y´ + y
= 0

(ii) x^{2}y´´ + (2x + 3x^{2})
y´ - 2y = 0

The power series in both cases is of the form

In part (i) the lowest power of x is x^{s} with
coefficient s(s-1)c_{0} - sc_{0} + c_{0}
giving the indicial equation

s^{2} - 2s + 1 = 0 which has roots s = 1
(repeated).

In part (ii) the lowest power of x is x^{s} with
coefficient s(s-1)c_{0} + 2sc_{0} - 2c_{0}
giving the indicial equation

s^{2} + s - 2 = 0 which has roots s = 1 and -2.
Note these roots differ by an integer.

(b) Find one solution to these differential equations, i.e. find the recurrence relation for the coefficients, write out the first four terms of each series and, if possible, find a general expression for the coefficients.

In part (i) the coefficient of x^{n+s}_{ }is
(n+s)(n+s-1)c_{n} + (n+s-1)c_{n-1} - (n+s)c_{n}
+ c_{n} giving the recurrence relation c_{n} =
-(n+s-1)c_{n-1}/(n+s-1)^{2 }= -c_{n-1}/(n+s-1).
With s = 1 this becomes c_{n} = -c_{n-1}/n. Thus c_{1}
= -c_{0}, c_{2 }= -c_{1}/2_{ } =
c_{0}/2, c_{3} = -c_{2}/3 = -c_{0}/6
and in general c_{n} = (-1)^{n}c_{0}/n!

The series solution is y = c_{0}x(1 - x + x^{2}/2
- x^{3}/6 + ...) = c_{0}xe^{-x}

In part (ii) the coefficient of x^{n+s}_{ }is
(n+s)(n+s-1)c_{n} + 2(n+s)c_{n }+ 3(n+s-1)c_{n-1}
- 2c_{n} giving the recurrence relation c_{n} =
-3(n+s-1)c_{n-1}/(n+s-1)(n+s+2) ^{ }= -3c_{n-1}/(n+s+2).
Taking the larger exponent

s = 1 this becomes c_{n} = -3c_{n-1}/(n+3).
Thus c_{1} = -3c_{0 }/4, c_{2 }= -3c_{1}/5
= 9c_{0}/20, c_{3} = -3c_{2}/6 = -9c_{0}/40
and in general c_{n} = 6(-3)^{n}c_{0}/(n+3)!

The series solution is y = c_{0}x(1 - 3x/4 +
9x^{2}/20 - 9x^{3}/40 + ...)

(c) Write down the form of the second solution in both cases.

In part (i) since the roots of the indicial equation are equal the second solution is of the form

In part (ii) the roots of the indicial equation differ by an integer so the second solution is of the form

It turns out that for this DE the coefficient C = 0 and the series terminates after 3 terms.