Math 302 - Differential Equations (Metcalfe)Summer 2002June 17, 2002

Exam 2 - Practice Exercises

1. Find a solution to the initial value problem

y′′ − 2xy′ + 8y = 0; y(0) = 3, y′(0) = 0

The only technique that we have that applies in this situation is power series. Sincex = 0 is an ordinary point, we may center our series there. Let

y =∞∑

n=0

anxn

Thus,

y′ =∞∑

n=1

nanxn−1

y′′ =∞∑

n=2

n(n− 1)anxn−2

Plugging in yields:

∞∑n=2

n(n− 1)anxn−2 +∞∑

n=1

(−2nan)xn +∞∑

n=0

8anxn = 0

(2a2 + 8a0) +∞∑

k=1

[(k + 2)(k + 1)ak+2 − 2kak + 8ak]xk = 0

Thus, our recurrence relations are:

a2 = −4a0

andak+2 =

2k − 8(k + 2)(k + 1)

ak

Using the initial conditions, we see that a0 = 3 and a1 = 0. Then, a2 = −12, a4 =−412 a2 = 4, a6 = 0, a8 = 0, ... and a3 = 0, a5 = 0, ... Thus, the solution is:

y(x) = 3− 12x2 + 4x4

2. Find a general solution to(x2 − 1)y′′ + 4xy′ + 2y = 0

and give a lower bound on its radius of convergence.

1

Since x = 0 is an ordinary point, we may center our series there. Let

y =∞∑

n=0

anxn

Thus,

y′ =∞∑

n=1

nanxn−1

y′′ =∞∑

n=2

n(n− 1)anxn−2

Plugging in yields:

∞∑n=2

n(n− 1)anxn +∞∑

n=2

[−n(n− 1)an]xn−2 +∞∑

n=1

4nanxn +∞∑

n=0

2anxn = 0

(−2a2 +2a0)+(−6a3 +6a1)x+∞∑

n=2

[n(n−1)an− (n+2)(n+1)an+2 +4nan +2an]xn = 0

Thus,a2 = a0

a3 = a1

an+2 =n(n− 1) + 4n + 2

(n + 2)(n + 1)an = an

So, the general solution is:

y(x) = a0(1 + x2 + x4 + x6 + ...) + a1(x + x3 + x5 + x7 + ...) = a0

∞∑n=0

x2n + a1

∞∑n=0

x2n+1

3. Find the general solution to:xy′′ + y′ = 0

for x 6= 0.

This is an Euler equation (even though it doesn’t currently look like it). Multiplythrough by x to get:

x2y′′ + xy′ = 0

Here, the indicial equation is given by:

r(r − 1) + αr = r(r − 1) + r = r2

Since this has a repeated root of 0, the general solution is given by:

y(x) = c1 + c2 ln |x|

2

4. Find the general solution to:x2y′′ − 7xy′ + 41y = 0

for x > 0.

This is an Euler equation. The indicial equation is given by:

r(r − 1) + αr + β = r(r − 1)− 7r + 41 = r2 − 8r + 41

The roots of this equation are

8±√

64− 4 · 412

= 4± 5i

So, the general solution is given by:

y(x) = c1x4 cos(5 lnx) + c2x

4 sin(5 lnx)

5. Find the general solution to:x2y′′ + 5xy′ + 3y = 0

for x > 0.

This is an Euler equation whose indicial equation is:

r(r − 1) + 5r + 3 = r2 + 4r + 3 = (r + 3)(r + 1)

The roots of this equation are −3 and −1. Thus, the general solution is given by:

y = c1x−3 + c2x

−1

6. Solve the initial value problem:

y(4) − y = 0; y(0) = 1, y′(0) = 0, y′′(0) = −1, y′′′(0) = 0

Apply the Laplace transform:

L{y(4)} − L{y} = L{0} = 0

(s4 − 1)L{y} − s3 + s = 0

L{y} =s3 − s

s4 − 1=

s

s2 + 1Thus,

y = cos t

7. Prove the commutative property for convolutions:

f ∗ g = g ∗ f

3

f ∗ g =∫ t

0f(t− τ)g(τ) dτ

Now, let’s do the change of variable T = t− τ and −dT = dτ . So, this becomes:

−∫ 0

tf(T )g(t− T ) dT =

∫ t

0g(t− T )f(T ) dT = g ∗ f

8. Solve the initial value problem:

2y′′′ + 3y′′ − 3y′ − 2y = e−t; y(0) = 0, y′(0) = 0, y′′(0) = 1

Apply the Laplace transform:

2L{y′′′}+ 3L{y′′} − 3L{y′} − 2L{y} = L{e−t}

(2s3 + 3s2 − 3s− 2)L{y} − 1 =1

s + 1

L{y} =s + 2

(s + 1)(s− 1)(s + 2)(2s + 1)=

1(s + 1)(s− 1)(2s + 1)

=12

1s + 1

+16

1s− 1

−23

1s + 1/2

Thus,

y =12e−t +

16et − 2

3e−t/2

9. Solve the initial value problem:

y′ − 5y = f(t) =

{t2 if 0 ≤ t < 1,

0 if t ≥ 1

with y(0) = 1.

Here

f(t) = t2−u1(t)t2 = t2−u1(t)(t−1)2−2u1(t)t+u1(t) = t2−u1(t)(t−1)2−2u1(t)(t−1)−u1(t)

So, applying the Laplace transform, we get:

L{y′} − 5L{y} =2s3

− e−s 2s3

− e−s 2s2

− e−s 1s

(s− 5)L{y} − 1 =2s3

− e−s

[2s3

+2s2

+1s

]L{y} =

s3 + 2s3(s− 5)

− e−s

[2 + 2s + s2

(s3)(s− 5)

]L{y} =

2125

1s

+225

1s2

− 15

2s3

+127125

1s− 5

− e−s

[− 37

1251s− 12

251s2

− 15

2s3

+37125

1s− 5

]Thus,

y =2

125+

225

t− 15t2 +

127125

e5t −[− 37

125− 12

25(t− 1)− 1

5(t− 1)2 +

37125

e5(t−1)

]u1(t)

4

10. Solve the initial value problem:

y′′ + 4y′ + 13y = δ(t− π) + δ(t− 3π); y(0) = 1; y′(0) = 0

Let’s start by applying the Laplace transform:

L{y′′}+ 4L{y′}+ 13L{y} = L{δ(t− π)}+ L{δ(t− 3π)}

(s2 + 4s + 13)L{y} − s− 4 = e−πs + e−3πs

L{y} =1

s2 + 4s + 13[s + 4 + e−πs + e−3πs] =

1(s + 2)2 + 9

[(s + 2) + 2 + e−πs + e−3πs]

Now, applying the inverse transform:

y = e−2t cos 3t+23e−2t sin 3t+

13uπ(t)e−2(t−π) sin 3(t−π)+

13u3π(t)e−2(t−3π) sin 3(t−3π)

11. A force of 2 lb stretches a spring 1 ft. With one end held fixed, an 8-lb weight is attached tothe other end and the system lies on a table that imparts a frictional force numerically equal to 3/2times the instantaneous velocity. Initially the weight is displaced 4 in. above the equilibrium pointand is released from rest. Find the equation of motion if the motion takes place along a horizontalstraight line that is taken as the x-axis.

Going through the derivation from section 3.8 again, we can see that the equationfor the motion becomes:

mu′′ + γu′ + k(L + u) = 0

where L is the length that the spring is stretched at equilbrium position. However, sincethere is no horizontal force at equilibrium, L = 0. Thus, we get the same equation:

mu′′ + γu′ + ku = 0

Here, using the first piece of info, we see that k = 2 since mg− kL = 0. We, also, knowthat γ = 3/2 from the statement of the problem. Thus, the problem that we are solvingis:

14u′′ +

32u′ + 2u = 0; u(0) = −4; u′(0) = 0

Multiplying through by 4 we get:

u′′ + 6u′ + 8u = 0

whose characteristic equation is:

x2 + 6x + 8 = (x + 4)(x + 2) = 0

Thus,u(t) = c1e

−4t + c2e−2t

Using the intial data, we see that:

u(t) =13e−4y − 2

3e−2t

5

12. Find a general solution of2x2y′′ − x(x− 1)y′ − y = 0

on x > 0 about the regular singular point x0 = 0.

Rewriting the equation a bit gives

x2y′′ − 12(x− 1)(xy′)− 1

2y = 0

The associated Euler equation we get by ignoring all be the constant terms of xp(x)and x2q(x) is

x2y′′ +12xy′ − 1

2y = 0

The indicial equation for this is given by

r(r − 1) +12r − 1

2= 0

or2r2 − r − 1 = (2r + 1)(r − 1) = 0

Thus, we have r1 = −1/2 and r2 = 1 (which don’t differ by an integer).We now begin by assuming that we have a solution of the form

y = xr∞∑

n=0

anxn =∞∑

n=0

anxn+r

Thus,

y′ =∞∑

n=0

an(n + r)xn+r−1

y′′ =∞∑

n=0

an(n + r)(n + r − 1)xn+r−2

Plugging into the original equation then yields:

∞∑n=0

2an(n+r)(n+r−1)xn+r−∞∑

n=0

an(n+r)xn+r+1+∞∑

n=0

an(n+r)xn+r−∑n=0

anxn+r = 0

or after shifting indices

∞∑k=0

2ak(k+r)(k+r−1)xk+r−∞∑

k=1

ak−1(k+r−1)xk+r+∞∑

k=0

ak(k+r)xk+r−∞∑

k=0

akxk+r = 0

Combining these into one sum yields

a0xr[2r(r−1)+r−1]+

∞∑k=1

[2ak(k+r)(k+r−1)−ak−1(k+r−1)+ak(k+r)−ak]xk+r = 0

From the second part, we get the recurrence relation

ak =ak−1

2k + 2r + 1

6

For r1 = −1/2, that isak =

ak−1

2k

So, we have a1 = a02 , a2 = a1

4 = a08 , a3 = a2

6 = a048 , ...

For r2 = 1, the recurrence relation becomes

ak =ak−1

2k + 3

Thus, a1 = a05 , a2 = a1

7 = a035 , a3 = a2

9 = a0315 , ...

Combining these, we have the general solution

y = c1x−1/2

(1 +

x

2+

x2

8+

x3

48+ ...

)+ c2x

(1 +

x

5+

x2

35+

x3

315+ ...

)

13. Find a general solution of

x2y′′ + xy′ + (x2 − 4/9)y = 0

on x > 0 about the regular singular point x0 = 0. (Find the first 3 nonzero terms in eachindependent piece.)

The associated Euler equation we get by ignoring all be the constant terms of xp(x)and x2q(x) is

x2y′′ + xy′ − 49y = 0

The indicial equation for this is given by

r(r − 1) + r − 49

= 0

orr2 − 4

9= 0

Thus, we have r1 = 2/3 and r2 = −2/3 (which don’t differ by an integer).We now begin by assuming that we have a solution of the form

y = xr∞∑

n=0

anxn =∞∑

n=0

anxn+r

Thus,

y′ =∞∑

n=0

an(n + r)xn+r−1

y′′ =∞∑

n=0

an(n + r)(n + r − 1)xn+r−2

Plugging into the original equation then yields:

∞∑n=0

(n + r)(n + r − 1)anxn+r +∞∑

n=0

(n + r)anxn+r −∞∑

n=0

49anxn+r +

∞∑n=0

anxn+r+2

7

So, after shifting indices, we have

∞∑k=0

(k + r)(k + r − 1)akxk+r +

∞∑k=0

(k + r)akxk+r − 4

9

∞∑k=0

akxk+r +

∞∑k=2

ak−2xk+r = 0

or

[r(r − 1) + r − 4/9]a0xr + [(r + 1)r + (r + 1)− 4/9]a1x

r+1

+∞∑

k=2

[(k + r)(k + r − 1)ak + (k + r)ak − (4/9)ak + ak−2]xk+r = 0

So, we see that a1 = 0 andak = − ak−2

(k + r)2 − (4/9)

For r = 2/3, this is

ak = − ak−2

k(k + 4/3)= − 3ak−2

k(3k + 4)

So, a1 = 0, a2 = −3a020 , a3 = −3a1

39 = 0, a4 = −3a264 = 9a0

1380 , ...For r2 = −2/3, we similarly get

ak = − ak−2

k(k − 4/3)= − 3ak−2

k(3k − 4)

So, a1 = 0, a2 = −3a04 , a3 = 0, a4 = −3a2

32 = 9a0128 , ...

Thus, our general solution is:

y = c1x2/3

(1− 3

20x2 +

91380

x4 + ...

)+ c2x

−2/3

(1− 3

4x2 +

9128

x4 + ...

)

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