MATH2270

FW00

Solutions to Assignment 5

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

(i) Let y = xr. Substituting in the DE yields r(r-1) + 3r - 8 = 0 with solutions r = 2 and -4 giving the general solution y = C1x2 + C2x-4.

(ii) Let y = xr. Substituting in the DE yields r(r-1) + 2r - 6 = 0 with solutions r = 2 and -3 giving the general solution y = C1x2 + C2x-3.

(b) Solve the boundary-value problem

Let y = xr. Substituting in the DE yields 2r(r-1) + r - 1 = 0 with solutions r = 1 and -1/2 giving the general solution y = C1x + C2x-1/2. Applying the initial conditions we get

C1 + C2 = 5 and 4C1 + C2/2 = 13 with solution C1 = 3 and C1 = 2.

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

The Wronskian of y1 and y2 is definned as W(y1, y2) = y1 y2´ - y2 y1´

(a)(i) We have y1 = x2 and y2 = x-4. Then W = x2(-4x-5) - x-4(2x) = -6x-3.

(a)(ii) We have y1 = x2 and y2 = x-3. Then W = x2(-3x-4) - x-3(2x) = -5x-2.

(b) We have y1 = x and y2 = 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 - x2)( 2u´ + xu´´) - 2x( u + xu´) + 2xu = x(1 - x2)u´´+ (2 - 4x2)u´ = 0. If we let v = u´ then we have a separable first-order DE for v which yields

which yields v = 1/[x2(1-x2)]. 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 y1 satisfies the DE. Similarly if

showing that y2 is also a solution of the DE.

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

The Wronskian is y1 y2´ - y2 y1´ which from above is

which reduces to reduces to sin2(x) + cos2(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 y1 and y2 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 = z1y1 + z2y2. Then in the notation of section 8.3 of the text we have a2 = 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 x0 = 0.

(i) y´´ + 2xy´ - 4y = 0

(ii) (1 - x2) y´´ - xy´ + r2y = 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 xn is (n+2)(n+1)cn+2 + 2ncn - 4cn giving the recurrence relation

cn+2 = 2(2-n)cn/[(n+2)(n+1)] Thus c2 = 2c0, c4 = 0 which implies c2n = 0 for n > 1. Also c3 = c1/3,

c5 = -c3/10 = -c1/30, c7 = -6c5/42 = c1/210 and c2n+1 = (-1)n+1c1/[(2n+1)(2n-1)n!]

Thus y = c0(1 + 2x2) + c1(x + x3/3 - x5/30 + x7/210 ... + (-1)n=1x2n+1/[(2n+1)(2n-1)n!]+ ...)

In part (ii) the coefficient of xn is (n+2)(n+1)cn+2 - n(n-1)cn - ncn + r2cn giving the recurrence relation

cn+2 = (n-r)(n+r)cn/[(n+2)(n+1)] giving c2 = -r2c0/2 and c3 = (1-r2)c1/6 giving

y = c0(1 - r2x2/2 + ...) + c1(x + (1-r2)x3/6 + ...).

The general coefficient is c2n = -[(2n-2)2-r2][(2n-4)2-r2]...[4-r2]r2c0/(2n)! and

c2n+1 = [(2n-1)2-r2][(2n-3)2-r2]...[9-r2][1-r2]c1/(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 cr+2 = 0 so that cr+4 = cr+6 = ... = 0 giving one solution as a polynomial of degree r.

If y(0) = 1 then c0 = 1 and if y´(0) = 1 then c1 = 1. Thus the polynomial solutions are y = 1 if r = 0, y = x if r = 1 and y = 1 - 2x2 if r = 2.

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

(i) x2y´´ + x(x-1)y´ + y = 0

(ii) x2y´´ + (2x + 3x2) y´ - 2y = 0

The power series in both cases is of the form

In part (i) the lowest power of x is xs with coefficient s(s-1)c0 - sc0 + c0 giving the indicial equation

s2 - 2s + 1 = 0 which has roots s = 1 (repeated).

In part (ii) the lowest power of x is xs with coefficient s(s-1)c0 + 2sc0 - 2c0 giving the indicial equation

s2 + 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 xn+s is (n+s)(n+s-1)cn + (n+s-1)cn-1 - (n+s)cn + cn giving the recurrence relation cn = -(n+s-1)cn-1/(n+s-1)2 = -cn-1/(n+s-1). With s = 1 this becomes cn = -cn-1/n. Thus c1 = -c0, c2 = -c1/2 = c0/2, c3 = -c2/3 = -c0/6 and in general cn = (-1)nc0/n!

The series solution is y = c0x(1 - x + x2/2 - x3/6 + ...) = c0xe-x

In part (ii) the coefficient of xn+s is (n+s)(n+s-1)cn + 2(n+s)cn + 3(n+s-1)cn-1 - 2cn giving the recurrence relation cn = -3(n+s-1)cn-1/(n+s-1)(n+s+2) = -3cn-1/(n+s+2). Taking the larger exponent

s = 1 this becomes cn = -3cn-1/(n+3). Thus c1 = -3c0 /4, c2 = -3c1/5 = 9c0/20, c3 = -3c2/6 = -9c0/40 and in general cn = 6(-3)nc0/(n+3)!

The series solution is y = c0x(1 - 3x/4 + 9x2/20 - 9x3/40 + ...)

(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.