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MATH3705 Tutorial 3

1 Find the two roots of the indicial equation to the following differential equation

x(1 minus x)yprimeprime + (3minus 5x)yprime minus 4y = 0

Solution We rewrite the equation as

yprimeprime + 3minus 5x

x(1 minus x)yprime +

minus4x

x2(1 minus x)y = 0

We have

xp(x) = 3 minus 5x1 minus x

= (3 minus 5x)(1 + x + x2 + x3 + ) = 3 minus 2xminus rArr p0 = 3

x2q (x) = minus4x

(1 minus x) = minus4x(1 + x + x2 + x3 + ) = minus4xminus 4x2 minus rArr q 0 = 0

Then we get the indicial equation r2 + 2r = 0 Then r1 = 0 r2 = minus2

2 Find the two roots of the indicial equation to the following differential equation

x2(1 minus x)yprimeprime + (3 minus 5x)xyprime + y = 0

Solution We rewrite the equation as

yprimeprime + 3minus 5x

x(1 minus x)yprime +

1

x2(1 minus x)y = 0

We have

xp(x) = 3 minus 5x

1 minus x = (3 minus 5x)(1 + x + x2 + x3 + ) = 3 minus 2xminus rArr p0 = 3

x2

q (x) =

1

(1 minus x) = 1(1 + x + x2

+ x3

+ ) = 1 + x + x2

+ x3

+ rArr q 0 = 1

Then we get the indicial equation r2 + 2r + 1 = 0 Then r1 = r2 = minus1

3 Find the general solution of

2x2yprimeprime + 7xyprime + (3 + 2x)y = 0 (1)

for x gt 0 near x0 = 0

1

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Solution Step 1 Determine whether x0 = 0 is an ordinary point or a regular singular

point Write the equation as

yprimeprime + 7

2x

yprime + 3 + 2x

2x2

y = 0 (2)

We have

xp(x) = 35 x2q (x) = 15 + x

So x0 = 0 is a regular singular point

Step 2 Find and solve the indicial equation Note that p0 = 35 q 0 = 15 The indicial

equation is

r2 + ( p0 minus 1)r + q 0 = 0 rArr r2 + 25r + 15 = 0 rArr r1 = minus1 r2 = minus15

Note that r1 minus r2 = 05 so we have Case (i)

Step 3 Find the recursive relation about cn

(r) Let

y =infinsumn=0

cn

(r)xn+r c0(r) = 1 rArr

yprime =infinsumn=0

(n + r)cn(r)xn+rminus1 yprimeprime =infinsumn=0

(n + r)(n + r minus 1)cn(r)xn+rminus2

Substitute them into (1) we have

2x2infinsumn=0

(n+r)(n+rminus1)cn

(r)xn+rminus2+7x

infinsumn=0

(n+r)cn

(r)xn+rminus1+(3+2x)infinsumn=0

cn

(r)xn+r = 0

infinsumn=0

2(n + r)(n + r minus 1)cn

(r)xn+r +infinsumn=0

7(n + r)cn

(r)xn+r

+infin

sumn=03c

n(r)xn+r +

infin

sumn=02c

n(r)xn+r+1 = 0

infinsumn=0

[2(n + r)(n + r minus 1) + 7(n + r) + 3]cn

(r)xn+r +infinsumn=0

2cn

(r)xn+r+1 = 0

infinsumn=0

[(2n + 2r + 3)(n + r + 1)]cn

(r)xn+r +infinsumn=1

2cnminus1(r)xn+r = 0

(2r + 3)(r + 1)]c0(r)xr +infinsumn=1

[(2n + 2r + 3)(n + r + 1)]cn

(r) + 2cnminus1(r)xn+r = 0

2

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(2r + 3)(r + 1)]c0(r) = 0 [(2n + 2r + 3)(n + r + 1)]cn

(r) + 2cnminus1 = 0 n ge 1

Since c0(r) = 1 = 0 the above first equation results in our indicial equation (2 r +3)(r +

1) = 0 The second equation gives

cn

(r) = minus2(2n + 2r + 3)(n + r + 1)

cnminus1(r) n ge 1 (3)

Step 4 Find y1 Take r = r1 = minus1 by (3)

cn(minus1) = minus2

n(2n + 1)cnminus1(minus1) n ge 1 rArr

cn

(1) = (minus2)n

n(2n + 1) n ge 1

Therefore

y1 = xminus1 +infinsumn=1

(minus2)n

n(2n + 1)xnminus1 = xr1

9830801 +

infinsumn=1

(minus2)n

n(2n + 1)xn

983081

Step 5 Find y2 Take r = r2 = minus15 by (3)

cn

(minus15) = minus2

n(2n minus 1)cnminus1(minus15) rArr

cn

(minus15) = (minus2)n

n(2n minus 1)

Therefore

y2 = xminus15 +infinsumn=1

(minus2)n

n(2n minus 1)xnminus15

Step 6 The general solution is y(x) = c1y1 + c2y2 where c1 and c2 are constants

4 Find two linearly independent solutions x2yprimeprime(x) + xyprime(x) + (3x2 minus 4)y(x) = 0 valid for

x gt 0

Solution Note that λ2 = 3 and ν 2 = 4 rArr λ =radic

3 and ν = 2 Hence

y1(x) = J 2(radic 3x) y2(x) = Y 2(radic 3x)

5 Find two linearly independent solutions x2yprimeprime(x) + xyprime(x) + (4x2 minus 3)y(x) = 0 valid for

x gt 0

Solution Note that λ2 = 4 and ν 2 = 3 rArr λ = 2 and ν =radic

3 Hence

y1(x) = J radic 3(2x) y2(x) = J minus

radic 3(2x)

3

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Solution Step 1 Determine whether x0 = 0 is an ordinary point or a regular singular

point Write the equation as

yprimeprime + 7

2x

yprime + 3 + 2x

2x2

y = 0 (2)

We have

xp(x) = 35 x2q (x) = 15 + x

So x0 = 0 is a regular singular point

Step 2 Find and solve the indicial equation Note that p0 = 35 q 0 = 15 The indicial

equation is

r2 + ( p0 minus 1)r + q 0 = 0 rArr r2 + 25r + 15 = 0 rArr r1 = minus1 r2 = minus15

Note that r1 minus r2 = 05 so we have Case (i)

Step 3 Find the recursive relation about cn

(r) Let

y =infinsumn=0

cn

(r)xn+r c0(r) = 1 rArr

yprime =infinsumn=0

(n + r)cn(r)xn+rminus1 yprimeprime =infinsumn=0

(n + r)(n + r minus 1)cn(r)xn+rminus2

Substitute them into (1) we have

2x2infinsumn=0

(n+r)(n+rminus1)cn

(r)xn+rminus2+7x

infinsumn=0

(n+r)cn

(r)xn+rminus1+(3+2x)infinsumn=0

cn

(r)xn+r = 0

infinsumn=0

2(n + r)(n + r minus 1)cn

(r)xn+r +infinsumn=0

7(n + r)cn

(r)xn+r

+infin

sumn=03c

n(r)xn+r +

infin

sumn=02c

n(r)xn+r+1 = 0

infinsumn=0

[2(n + r)(n + r minus 1) + 7(n + r) + 3]cn

(r)xn+r +infinsumn=0

2cn

(r)xn+r+1 = 0

infinsumn=0

[(2n + 2r + 3)(n + r + 1)]cn

(r)xn+r +infinsumn=1

2cnminus1(r)xn+r = 0

(2r + 3)(r + 1)]c0(r)xr +infinsumn=1

[(2n + 2r + 3)(n + r + 1)]cn

(r) + 2cnminus1(r)xn+r = 0

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(2r + 3)(r + 1)]c0(r) = 0 [(2n + 2r + 3)(n + r + 1)]cn

(r) + 2cnminus1 = 0 n ge 1

Since c0(r) = 1 = 0 the above first equation results in our indicial equation (2 r +3)(r +

1) = 0 The second equation gives

cn

(r) = minus2(2n + 2r + 3)(n + r + 1)

cnminus1(r) n ge 1 (3)

Step 4 Find y1 Take r = r1 = minus1 by (3)

cn(minus1) = minus2

n(2n + 1)cnminus1(minus1) n ge 1 rArr

cn

(1) = (minus2)n

n(2n + 1) n ge 1

Therefore

y1 = xminus1 +infinsumn=1

(minus2)n

n(2n + 1)xnminus1 = xr1

9830801 +

infinsumn=1

(minus2)n

n(2n + 1)xn

983081

Step 5 Find y2 Take r = r2 = minus15 by (3)

cn

(minus15) = minus2

n(2n minus 1)cnminus1(minus15) rArr

cn

(minus15) = (minus2)n

n(2n minus 1)

Therefore

y2 = xminus15 +infinsumn=1

(minus2)n

n(2n minus 1)xnminus15

Step 6 The general solution is y(x) = c1y1 + c2y2 where c1 and c2 are constants

4 Find two linearly independent solutions x2yprimeprime(x) + xyprime(x) + (3x2 minus 4)y(x) = 0 valid for

x gt 0

Solution Note that λ2 = 3 and ν 2 = 4 rArr λ =radic

3 and ν = 2 Hence

y1(x) = J 2(radic 3x) y2(x) = Y 2(radic 3x)

5 Find two linearly independent solutions x2yprimeprime(x) + xyprime(x) + (4x2 minus 3)y(x) = 0 valid for

x gt 0

Solution Note that λ2 = 4 and ν 2 = 3 rArr λ = 2 and ν =radic

3 Hence

y1(x) = J radic 3(2x) y2(x) = J minus

radic 3(2x)

3

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(2r + 3)(r + 1)]c0(r) = 0 [(2n + 2r + 3)(n + r + 1)]cn

(r) + 2cnminus1 = 0 n ge 1

Since c0(r) = 1 = 0 the above first equation results in our indicial equation (2 r +3)(r +

1) = 0 The second equation gives

cn

(r) = minus2(2n + 2r + 3)(n + r + 1)

cnminus1(r) n ge 1 (3)

Step 4 Find y1 Take r = r1 = minus1 by (3)

cn(minus1) = minus2

n(2n + 1)cnminus1(minus1) n ge 1 rArr

cn

(1) = (minus2)n

n(2n + 1) n ge 1

Therefore

y1 = xminus1 +infinsumn=1

(minus2)n

n(2n + 1)xnminus1 = xr1

9830801 +

infinsumn=1

(minus2)n

n(2n + 1)xn

983081

Step 5 Find y2 Take r = r2 = minus15 by (3)

cn

(minus15) = minus2

n(2n minus 1)cnminus1(minus15) rArr

cn

(minus15) = (minus2)n

n(2n minus 1)

Therefore

y2 = xminus15 +infinsumn=1

(minus2)n

n(2n minus 1)xnminus15

Step 6 The general solution is y(x) = c1y1 + c2y2 where c1 and c2 are constants

4 Find two linearly independent solutions x2yprimeprime(x) + xyprime(x) + (3x2 minus 4)y(x) = 0 valid for

x gt 0

Solution Note that λ2 = 3 and ν 2 = 4 rArr λ =radic

3 and ν = 2 Hence

y1(x) = J 2(radic 3x) y2(x) = Y 2(radic 3x)

5 Find two linearly independent solutions x2yprimeprime(x) + xyprime(x) + (4x2 minus 3)y(x) = 0 valid for

x gt 0

Solution Note that λ2 = 4 and ν 2 = 3 rArr λ = 2 and ν =radic

3 Hence

y1(x) = J radic 3(2x) y2(x) = J minus

radic 3(2x)

3