math3132test2solutions_winter2015

DATE: March 12, 2015

EXAMINATION: Engineering Mathematical Analysis 3COURSE: MATH 3132

UNIVERSITY OF MANITOBATERM TEST 2

PAGE: 1 of 5TIME: 1 hour

EXAMINER: Harland

1. (a)[12] Determine series solutions to the differential equation

x2y′′ + xy′ + (x2 − 1)y = 0.

(The final series answer should be in the form of exponentials and factorials.)

Solution:

y =∞∑n=0

anxn ⇒ y =

∞∑n=1

nanxn−1 ⇒ y =

∞∑n=2

n(n− 1)anxn−2.

Thus

0 = x2y′′ + xy′ + (x2 − 1)y

= x2∞∑n=2

n(n− 1)anxn−2 + x

∞∑n=1

nanxn−1 + x2

∞∑n=0

anxn −

∞∑n=0

anxn

=∞∑n=2

n(n− 1)anxn +

∞∑n=1

nanxn +

∞∑n=0

anxn+2 −

∞∑n=0

anxn

=∞∑n=0

n(n− 1)anxn +

∞∑n=0

nanxn +

∞∑n=2

an−2xn −

∞∑n=0

anxn

=∞∑n=0

(n(n− 1)an + nan − an)xn +∞∑n=2

an−2xn

= −a0 + 0a1x+∞∑n=2

((n2 − 1)an + an−2

)xn

Hence a0 = 0, a1 is arbitrary and for n ≥ 2

an = − an−2(n− 1)(n+ 1)

Now a0 = 0 ⇒ a2 = 0 ⇒ ... and in general a2n = 0. Further a1 isarbitrary and thus

a3 = − a12 · 4

a5 = − a34 · 6

=a1

2 · 42 · 6

a7 = − a56 · 8

= − a12 · 42 · 62 · 8

and in general

a2n+1 = (−1)na1

2 · 42 · (2n)2 · (2n+ 2)= (−1)n

a1

22n(1 · 22 · n2 · (n+ 1)

) = (−1)na1

22nn!(n+ 1)!

for n ≥ 1.

Thus

y = a1x+ a1

∞∑n=1

(−1)nx2n+1

22nn!(n+ 1)!= a1

∞∑n=0

(−1)nx2n+1

22nn!(n+ 1)!





EXAMINER: Harland

(b)[4] Determine the radius of convergence of the solution found in (a).

Solution:

Let X = x2 thus the sum becomes

a1x

∞∑n=0

(−1)nXn

22n(n!)2

RX = limn→∞

∣∣∣∣ cncn+1

∣∣∣∣= lim

n→∞

∣∣∣∣∣(−1)n

22n(n!)2

(−1)n+1

22n+2((n+1)!)2

∣∣∣∣∣= lim

n→∞

122n(n!)2

122n+2((n+1)!)2

= limn→∞

22n+2((n+ 1)!)2

22n(n!)2

= limn→∞

4(n+ 1)2

=∞

Thus |X| <∞⇒ |x2| <∞⇒ |x| <∞⇒ Rx =∞.

(c)[3] Is this the general solution to the differential equation? If so, explain how youknow. If not, explain how we can tell from the original differential equationthat we may not get the general solution.

Solution: No this is not the general solution as a second order differ-ential equation should have 2 arbitrary constants and we only have one.This happened since x = 0 is a singular point and thus we have noguarantee that the Taylor series centered at x = 0 will yield a generalsolution.





EXAMINER: Harland

2. For the differential equation

(x2 + 4)y′′ + xy′ + (x2 − 1)y = 0

(a)[3] Determine all singular points to the differential equation.

Solution:

The equation can be rearranged to yields

y′′ +x

x2 + 4y′ +

x2 − 1

x2 + 4y = 0

which has singular points at x = ±2i. since that makes the rationalfunction infinite.

(b)[2] Determine the radius of convergence of the series solution centered at x = 0.(Hint: Don’t solve for the series solution.)

Solution:

The radius is the distance to the nearest singular point which would bethe distance from x = 0 to x = 2i in the complex plane. Thus R = 2.

(c)[2] Determine the radius of convergence of the series solution centered at x = 1.(Hint: Don’t solve for the series solution.)

Solution:

The radius is the distance to the nearest singular point which wouldbe the distance from x = 1 to x = 2i in the complex plane. ThusR =

√12 + 22 =

√5.





EXAMINER: Harland

3.[14] Calculate the Fourier series of

f(x) =

{2 + x −2 ≤ x < 02− x 0 ≤ x < 2

, f(x+ 4) = f(x).

(your final answer should not contain cos(nπ) nor sin(nπ) but rather should haveexponentials.)

Solution:

L = 2 and note that f is even. The fourier series is

a02

+∞∑n=1

an cos

(nπx

2

)+ bn sin

(nπx

2

)

Using that f(x) is even we get

a0 =1

2

ˆ 2

−2f(x) dx

=

ˆ 2

0

f(x) dx

=

ˆ 2

0

(2− x) dx

= 2x− x2

2

∣∣∣∣20

= 2

Using that f(x) cos(nπx/2) is even we get

an =1

2

ˆ 2

−2f(x) cos

(nπx

2

)dx

=

ˆ 2

0

f(x) cos

(nπx

2

)dx

=

ˆ 2

0

(2− x) cos

(nπx

2

)dx

Let u = 2 − x, dv = cos

(nπx2

)dx ⇒ du = −dx, v =

2

nπsin

(nπx2

)we get

that

an =

ˆ 2

0

(2− x) cos

(nπx

2

)dx

=2(2− x)

nπsin

(nπx

2

)∣∣∣∣20

+2

nπ

ˆ 2

0

sin

(nπx

2

)= 0− 4

nπsin 0− 4

n2π2cos

(nπx

2

)∣∣∣∣20

= − 4

n2π2cos(nπ) +

4

n2π2cos(0)

=4

n2π2

(1− (−1)n

)





EXAMINER: Harland

Using that f(x) sin(nπx/2) is odd we get

an =1

2

ˆ 2

−2f(x) sin

(nπx

2

)dx = 0

and thus the fourier series is

1 +∞∑n=1

(1− (−1)n

) 4

n2π2cos

(nπx

2

)or

1 +∞∑

n=1, n odd

8

n2π2cos

(nπx

2

)or

1 +∞∑m=0

8

(2m+ 1)2π2cos

((2m+ 1)πx

2

)

math3132test2solutions_winter2015

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