Math246 Exercises

3/16/16, 10:23 PMMath246 Exercises

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Higher-Order Linear Ordinary Differential Equations Homogeneous Equations: General Methods and Theory

Find the general solution to the following differential equations.

Exercise 1Consider the second-order differential equation . The four functions , , , and areall solutions to it. Is that weird? Why (or why not)?

SolutionIt is not weird, because we can express two of these functions in terms of the other two, for instance using theidentities

So there are only two linearly independent functions in play here.

Exercise 2Consider the fourth-order differential equation . Check that the functions and

both are solutions to it. There should be two other fundamentally different solutions; can you findthem by inspection?

Solution

The following table lists derivatives of sine and cosine: Indeed the fourth

derivatives match the functions themselves, so they are solutions. One further solution comes from (if it is itsown derivative then it is its own fourth derivative). The last solution is .

Another possibility you may have guessed after doing the previous problem is and which are alsofundamentally different from and , and are independent solutions.

= xx′′ cosh(u) sinh(u) eu e−u

cosh(u) = and sinh(u) = .+eu e−u

2−eu e−u

2

= yy (4) (t) = cos(t)y1(t) = sin(t)y2

n

01234

cos(t)dn

dtn

cos(t)− sin(t)− cos(t)

sin(t)cos(t)

sin(t)dn

dtn

sin(t)cos(t)

− sin(t)− cos(t)

sin(t)et

e−t

sinh(t) cosh(t)sin(t) cos(t)



Exercise 3Check that and are solutions to . Then find and so that

is a solution to the differential equation satisfying the initial conditions , .

SolutionThe check is routine. To find the constants, note that and , so the desiredconstants come from the system of equations

whose solution is , . (So .)

Exercise 4Check that and are solutions to . Then find and so that

is a solution to the differential equation satisfying the initial conditions , .

SolutionWe’ll check , and the other one will be similar. The product rule tells us

and

Therefore

so it checks out.

To find the constants, note that since . Immediately we get . Next

sin(t) cos(t)

e3z e4z − 7 + 12w = 0w′′ w′ c1 c2W(z) = +c1e3z c2e4z W(0) = 2

(0) = 0W ′

W(0) = +c1 c2 (0) = 3 + 4W ′ c1 c2

+c1 c2

3 + 4c1 c2

= 2= 0,

= 8c1 = −6c2 W(z) = 8 − 6e3z e4z

sin(x)ex cos(x)ex − 2 + 2y = 0y ′′ y ′ c1 c2Y (x) = sin(x) + cos(x)c1ex c2ex

Y (0) = 3 (0) = −2Y ′

f(x) = sin(x)ex

(x) = sin(x) + cos(x)f ′ ex ex

(x) = ( sin(x) + cos(x)) + ( cos(x) − sin(x)) = 2 cos(x).f ′′ ex ex ex ex ex

− 2 + 2f = 2 cos(x) − 2( sin(x) + cos(x)) + 2 sin(x) = 0,f ′′ f ′ ex ex ex ex

Y (0) = c2 sin(0) = 0 = 3c2

(x) = sin(x) + cos(x) + cos(x) − sin(x),Y ′ c1ex c1ex c2ex c1ex

(0) = +′1 2 (0) = −2′



and plugging in destroys the terms with sine, leaving . Since and ,we have .

The solution is .

Exercise 5a. Check that and are solutions to the equation

b. Find and so that is a solution to the differential equation satisfyingthe initial conditions , .

c. On what interval(s) is the function you found in part (b) a solution to the equation?

Solutiona. All the derivatives of are equal to , so when we plug it in we get

The polynomial is slightly more interesting. Its derivative is and its second derivative is, so putting it into the equation gives

so everything checks out.

b. Putting into and leads to the system of equations

Subtracting second equation from the first brings us to , and then putting that into the firstequation and rearranging gives us the unflattering other constant . (So the solution is

.)

c. In normal form the equation we’re working with is

whose coefficients have a discontinuity at . So the solution is valid on the half-line .

x = 0 (0) = +Y ′ c1 c2 (0) = −2Y ′ = 3c2= −5c1

Y (x) = −5 sin(x) + 3 cos(x)ex ex

ex + 2x + 2x2

x − (x + 2) + 2y = 0.y ′′ y ′

c1 c2 Y (x) = + ( + 2x + 2)c1ex c2 x2

Y (1) = e (1) = e + 2Y ′

Y

ex ex

x − (x + 2) + 2 = 0.ex ex ex

+ 2x + 2x2 2x + 22

x(2) − (x + 2)(2x + 2) + 2( + 2x + 2) = 2x − (2 + 6x + 4) + (2 + 4x + 4) = 0,x2 x2 x2

x = 1 Y Y ′

e + 5c1 c2

e + 4c1 c2

= e

= e + 2.

= −2c2= + 1c1

10e

( + 1) − 2( + 2x + 2)10e ex x2

− + y = 0,y ′′ x + 2x

y ′ 2x

x = 0 (0, ∞)



Exercise 6Check that , , and are solutions to the third-order differential equation

Then find constants , , and so that are solutions to the equation on satisfying , , and .

SolutionThe check is straightforward, though the numbers get large. For , only the last two terms matter, and we get

. For we get . For , we get .

The initial conditions lead to the following system of equations:

Solving the system can be done in a variety of ways; one option is to subtract the third equation from the secondto get , then plug that into the first equation to get , then multiply it by three andthe third equation by two, and then subtract those. That should give , or . From there and follow painlessly. The equation desired is

Exercise 7One fundamental set of solutions for the differential equation is . Find a solution

which satisfies the general initial conditions , .

SolutionThe system of equations that and satisfy is

w w3 w5

− 6 + 15w − 15z = 0 .w3z′′′ w2z′′ z′

c1 c2 c3 Z(w) = w + +c1 c2w3 c3w5

(0, ∞) Z(2) = 0 (2) = 0Z ′ (2) = 1Z ′′

w15w − 15w = 0 w3 (6 − 6 + 15 − 15) = 0w3 w5

(60 − 6 ⋅ 20 + 15 ⋅ 5 − 15) = 0w5

2 + 8 + 32c1 c2 c3

+ 12 + 80c1 c2 c3

12 + 160c2 c3

= 0= 0= 1.

= 80 − 1c1 c3 8 + 192 = 2c2 c3256 = 4c3 =c3

164 =c1

14

= −c218

− + .w

4w3

8w5

64

− 2 − 8y = 0y ′′ y ′ { , }e−2w e4w

Y = +c1e−2w c2e4w Y (0) = y0 (0) =Y ′ y1

c1 c2

+c1 c2

−2 + 4c1 c2

= ,y0

= .y1



By solving the first equation for we find . By plugging this result into the second equation we find . Therefore we obtain


that satisfies the general initial conditions , .

SolutionBecause and the system of equations that and satisfy is

Therefore and .


that satisfies the general initial conditions , .

SolutionThe system of equations that and satisfy is

From this we see that and so . (Note that the interval of definition of the solution is because the normal form of the differential equation has coefficients that are undefined at .)

Exercise 10

c1 = −c1 y0 c2= +c2

13 y0

16 y1

c1

c2

= − ,23

y016

y1

= + .13

y016

y1

+ 9y = 0y ′′ {cos(3t), sin(3t)}Y = cos(3t) + sin(3t)c1 c2 Y ( ) =π

2 y0 ( ) =Y ′ π2 y1

cos(3 ) = 0π2 sin(3 ) = −1π

2c1 c2

y0

y1

= − ,c2

= 3 .c1

=c113 y1 = −c2 y0

+ 4u = 0u2x′′ x′ {1, }u−3

X(u) = +c1 c2u−3 X(1) = x0 (1) =X ′ x1

c1 c2

x0

x1

= + ,c1 c2

= −3 .c2

= −c213 x1 = +c1 x0

13 x1

(0, ∞) u = 0

− = 0′′ 1 ′ {1, }2



a. One fundamental set of solutions for the differential equation is . Find a solution that satisfies the general initial conditions , .

b. What is the interval of definition for your solution in (a)?

c. What happens if you try to solve for the initial conditions , ?

Solutiona. The system of equations we have to solve is

Adding these equations gives , and of course .

b. Because the coefficient of the differential equation is undefined at and continuous elsewhere,while the initial point is , the interval of definition of the solution is .

c. The differential equation is undefined at so that is not a valid initial point. Therefore there is nosuch solution. Had you not seen this fact and tried to solve for and the system of equations thatresults is

The second equation shows that you have made a mistake.

Exercise 11Compute the Wronskian of the functions and . Is this defined for all points ?

Solution

Yes, the Wronskian of and is defined for all real .

Exercise 12Compute the Wronskian of the functions , , and . Notice thisalso explains what was going on in Exercise #1.

− = 0y ′′ 1x y ′ {1, }x2

Y (x) = +c1 c2x2 Y (−2) = y0 (−2) =Y ′ y1

y(0) = 1 (0) = 2y ′

y0

y1

= + 4 ,c1 c2

= −4 .c2

= +c1 y0 y1 = −c214 y1

−1/x x = 0x = −2 (−∞, 0)

x = 0 x = 0c1 c2

12= ,c1

= 0 .

(z) = zW1 (z) = cos(z)W2 z

W [ , ](z) = det ( ) = −z sin(z) − cos(z).W1 W2z

1cos(z)

− sin(z)

W1 W2 z

(u) = cosh(u)X1 (u) = sinh(u)X2 (u) =X3 eu



Solution

since this matrix has two copies of the same row.

If you didn’t notice this, we can get to the long way too. We’ll need the following: , , and . If you expand the determinant out by using the first row, you

get

The identities above tell us that and . Using that in thefirst two terms of the sum and using in the third one, we simplify all the way down to

and indeed , so everything cancels.

Exercise 13(Continuation of Exercise #3) Show that is a fundamental set of solutions for the second-orderdifferential equation .

SolutionCompute the Wronskian:

Since this is not identically zero, they form a fundamental set.

Exercise 14(Continuation of Exercise #4) Show that is a fundamental set of solutions for the second-order differential equation .

W [ , , ](u) = det = 0,X1 X2 X3

⎛⎝⎜

cosh(u)sinh(u)cosh(u)

sinh(u)cosh(u)sinh(u)

eu

eu

eu

⎞⎠⎟

0 cosh(u) = +eu e−u

2sinh(u) = −eu e−u

2 (u) − (u) = 1cosh2 sinh2

cosh(u) ( cosh(u) − sinh(u)) − sinh(u) ( sinh(u) − cosh(u)) + ( (u) − (u)) .eu eu eu eu eu sinh2 cosh2

cosh(u) − sinh(u) = e−u sinh(u) − cosh(u) = −e−u

(u) − cosh(u = −1sinh2 )2

cosh(u) + sinh(u) − ,eu

cosh(u) + sinh(u) = eu

{ , }e3t e4t

− 7 + 12y = 0y ′′ y ′

W [ , ](t) = det [ ] = 4 − 3 = .e3t e4te3t

3e3t

e4t

4e4te7t e7t e7t

{ sin(x), cos(x)}ex ex

− 2 + 2y = 0y ′′ y ′



SolutionCompute the Wronskian:

Since this is not identically zero, they form a fundamental set.

Exercise 15(Continuation of Exercise #6) Show that is a fundamental set of solutions for the third-orderdifferential equation

SolutionAnother Wronskian calculation:

To calculate this determinant, we’ll expand down the first column to exploit the zero. Doing so gives us

so these functions do form a fundamental set.

Exercise 16Show that the functions , , and are linearly independent on .

SolutionApproach one: suppose that for all , and plug in three values for . I’ll choose ,

, and , but of course any numbers will work. This gives the system

W [ sin(x), cos(x)](x)ex ex

= det [ ]sin(x)ex

sin(x) + cos(x)ex ex

cos(x)ex

cos(x) − sin(x)ex ex

= x (sin(x) cos(x) − x) − (sin(x) cos(x) + (x)) = − .e2 sin2 e2x cos2 e2x

{x, , }x3 x5

− 6 + 15x − 15y = 0.x3y ′′′ x2y ′′ y ′

W [x, , ](x) = det .x3 x5⎡⎣⎢

x

10

x3

3x2

6x

x5

5x4

20x3

⎤⎦⎥

x (60 − 30 ) − 1 (20 − 6 ) + 0 = 30 − 14 = 16 ≠ 0,x5 x5 x6 x6 x6 x6 x6

et e2t e3t (−∞, ∞)

+ + = 0c1et c2e2t c3e3t t t t = 0t = ln(2) t = ln(3)

0



Now the first equation says , so the second equation says , so the third equation says . Then , so , so they’re all zero. This proves linear independence.

Approach two: Compute the Wronskian:

Since this is not identically zero, they are linearly independent.

Exercise 17Show that , , and are linearly dependent over .

SolutionThe linear dependence of is seen from the fact that , whichshows that

The Wronskian of is zero, but you cannot conclude that these functions are linearlydependent from this fact alone.

Exercise 18(Continuation of Exercise #3) Find a natural fundamental set for the differential equation associated with the initial time .

SolutionEarlier we saw that a fundamental set of solutions for this equation is . If we seek a solution

that satisfies the general initial conditions , then we obtain thesystem of equations

000

= + +c1 c2 c3

= 2 + 4 + 8c1 c2 c3

= 3 + 9 + 27 .c1 c2 c3

= − −c1 c2 c3 = −3c2 c36 = 0c3 = 0c2 = 0c1

W [ , , ](t)et e2t e3t = det⎡⎣⎢

et

et

et

e2t

2e2t

4e2t

e3t

3e3t

9e9t

⎤⎦⎥

= (18 − 12 ) − (9 − 3 ) + (4 − 2 ) = 2 .et e5t e5t e2t e4t e4t e3t e3t e3t e6t

log(z) log(5z) 1 (0, ∞)

{log(z), log(5z), 1} log(5z) = log(5) + log(z)

log(z) − log(5z) + ln(5) ⋅ 1 = 0 .

{log(z), log(5z), 1}

− 7 + 12y = 0y ′′ y ′

0

{ , }e3t e4t

Y (t) = +c1e3t c2e4t Y (0) = y0 (0) =Y ′ y1



These have the solution

The solution is thereby

From this we can read off that the natural fundamental set of solutions for is and .

Notice that , , , and .

Exercise 19(Continuation of Exercise #9) Find a natural fundamental set for the differential equation associated with the initial time .

SolutionWe’ll use the generic initial conditions from a previous problem to get the desired functions. In said Problem 9 wefound that the general initial conditions , are satisfied by the constants and . Tallying up the contributions to and , we obtain the functions and

.

Exercise 20(Continuation of Exercise #8) Find a natural fundamental set for the differential equation associatedwith the initial time .

SolutionIn Problem 8 we found that the general initial conditions , are satisfied by the constants

and . These contributions of and imply that the functions we’re looking for are and .

+c1 c2

3 + 4c1 c2

= ,y0

= .y1

c1

c2

= 4 − ,y0 y1

= −3 + .y0 y1

Y (t) = (4 − ) + (−3 + ) = (4 − 3 ) + ( − ) .y0 y1 e3t y0 y1 e4t e3t e4t y0 e4t e3t y1

t = 0 (t) = 4 − 3N0 e3t e4t

(t) = −N1 e4t e3t

(0) = 1N0 (0) = 12 − 12 = 0N ′0 (0) = 0N1 (0) = −3 + 4 = 1N ′

1

+ 4w = 0w2z′′ z′

1

Z(1) = z0 (1) =Z ′ z1 = +c1 z0w

3 z1

= −c2w

3 z1 z0 z1 (w) = 1N0

(w) = −N1w

3w

3 w−3

+ 9y = 0y ′′π

2

Y ( ) =π

2 y0 ( ) =Y ′ π

2 y1

=c1y1

3 = −c2 y0 y0 y1

(t) = − sin(3t)N0 (t) = cos(3t)N113



Exercise 21Check that is a solution to , and then use reduction of order to find a fundamental set.

SolutionThe check is straightforward, so let’s dive in with the reduction of order. Let the other solution be for an unknown function . Now we differentiate:

and

Adding everything up, we get

Now is never zero, so we can divide it away, leaving us with . If we set , then the differentialequation is , which has the solution (any constant will do). Integrating both sides tells us that (up to a constant), so the other solution is , a fundamental set of solutions is ,and the general solution is .

If you love constants, you might have said is the solution to . If you did, that’s fine; you’ll get , and that will give you the complete solution straightaway.

1. Check the Wronskian if you don’t believe me.↩

Exercise 22It’s clear that solves the differential equation . Find another solution by reduction of order.

SolutionLet the other solution be . Plugging into the equation gives . Make up anew variable , and put the equation in normal form to arrive at . Now the first-order recipeleads us to the solution , and from there we get . (So a general solution is .)

1. The integrating factor is , which then says , so , and .↩

ez − 2 + w = 0w′′ w′

w(z) = u(z)ez

u

(z) = u(z) + (z)w′ ez ezu′

(z) = u(z) + 2 (z) + (z).w′′ ez ezu′ ezu′′

0 = − 2 + w = u(z) + 2 (z) + (z) − 2 u(z) − 2 (z) + u(z) = (z).w′′ w′ ez ezu′ ezu′′ ez ezu′ ez ezu′′

ez (z) = 0u′′ k = u′

= 0k′ k = 1 u = zw(z) = u(z) = zez ez { , z }ez ez 1

W(z) = + zc1ez c2 ez

k = c1 = 0k′

u(z) = z +c1 c2 W(z)

(t) = 1Y1 t + = 0y ′′ y ′

y(t) = (t)u(t) = u(t)Y1 y t + = 0u′′ u′

w = u′ + w = 0w′ 1t

=w′ 1t

1 u = log(t) + log(t)c1 c2

µ(t) = t [tw] = 0ddt

tw = c w = c 1t

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Exercise 23If is a positive integer, then is also a solution to . Complete a fundamental set for thisequation. [Hint. Your answer will have an in it somewhere.]

SolutionLet the other solution be . Plugging into the equation gives . Makeup a new variable , and put the equation in normal form to arrive at . Now the first-orderrecipe tells us to use the integrating factor , which gives the equation , or .Therefore . A candidate for the other slot in a fundamental set is .

Exercise 24Check that is a solution to . Then use reduction of order tocomplete a fundamental set for it.

SolutionYou should calculate that and . Thingscheck out when those are plugged in.

To reduce order, let be the second solution, with a to-be-determined function. We have and , and using the derivatives from the previous paragraph

we get the following:

Plugging this into the initial equation will wipe out the first term in each line when taken together. When the dustsettles, we wind up with

which is a first-order equation in . Let’s use a new letter for that, say . We can divide by because it’snever zero, bringing us to

or in normal form

n (t) = 1Y1 − = 0y ′′ nx y ′

n

y(x) = (x)u(x) = u(x)Y1 y − = 0u′′ nx u′

w = u′ − w = 0w′ nx

µ(x) = x−n [w ] = 0ddt

x−n w = cxn

u = cn+1 xn+1 xn+1

(u) = cos(u)Z1 e2u z − 4Dz + 5z = 0D2

(u) = 2 cos(u) − sin(u)Z1 e2u e2u (u) = 3 cos(u) − 4 sin(u)Z1 e2u e2u

z(u) = (u)q(u)Z1 q(u) = q +z Z1 Z1 q (u) = q + 2 +z Z1 Z1 q Z1 q

z(u)

(u)z

(u)z

= cos(u)q(u)e2u

= (2 cos(u) − sin(u))q(u) + cos(u) (u)e2u e2u e2u q

= (3 cos(u) − 4 sin(u))q(u) + (4 cos(u) − 2 sin(u)) (u) + cos(u) (u) .e2u e2u e2u e2u q e2u q

0 = cos(u) (u) − 2 sin(u) (u) ,e2u q e2u q

q w = q e2u

cos(u) (u) − 2 sin(u)w(u) = 0 ,w

(u) − 2 tan(u)w(u) = 0 .



The proper integrating factor is , because (perform asubstitution to get this). This simplifies to . Upon multiplying through by the integrating factor, theequation becomes

so . Now we integrate both sides of this to get ; an antiderivative of is given by . So up to choice of constants, is a fine example, and so

is the other element of the fundamental set.

Exercise 25Check that is a solution to . Then use reduction of order to find asecond linearly independent solution.

SolutionThe check isn’t hard—it boils down to the fact that the coefficient on is and the coefficient on is .Let’s worry more about the reduction of order.

We suppose the second solution is with a function to be discovered. Then we have and , so when we plug into the equation we get

As usual the terms with drop out. If we expand what remains we get

a first-order differential equation in . Set , so that this equation is (in normal form)

Now to get our integrating factor we need to be able to integrate this rational function of . Thankfully we weregiven a factorization of the denominator—it’s . The resulting partial fractions decompositionlooks like this:

(I’ll spare you the details.) When we antidifferentiate, we get

(u) − 2 tan(u)w(u) = 0 .w

q(u) = exp(2 ln(cos(u))) ∫ tan(u) du = − ln(cos(u))q(u) = (u)cos2

D[w(u) (u)] = 0 ,cos2

w(u) = c (u)sec2 q(u) (u)sec2

tan(u) q(u) = tan(u)z(u) = (u)q(u) = cos(u) tan(u) = sin(u)Z1 e2u e2u

(w) = wZ1 (w − 1)(w − 2) − w + z = 0z z

z +1 wz −1

z(w) = wu(w) u(w) = w (w) + u(w)z u (w) = w (w) + 2 (w)z u u

(w − 1)(w − 2)(w + 2 ) − w(w + u) + wu = 0 .u u u

u(w)

( − 3 + 2w) + ( − 6w + 4) = 0 ,w3 w2 u w2 u

u′ k(w) = (w)u

+ k = 0 .k− 6w + 4w2

− 3 + 2ww3 w2

ww(w − 1)(w − 2)

= + + .− 6w + 4w2

− 3 + 2ww3 w2

2w

1w − 1

−2w − 2

∫ dw = 2 log(w) + log(w − 1) − 2 log(w − 2) .− 6w + 42



Therefore, our integrating factor is

Multiplying our differential equation by and using the first-order recipe, we get , so for some constant . Then we integrate both sides to arrive at , which means our unknown

function is given by the integral

The constant is just going to get passed around, and because any solution linearly independent from willsatisfy the problem, let’s just ignore it for now. It’s partial fractions time again! This time the decomposition is

(There could have been a term in the expansion, but it wound up having a coefficient .) So, finally, we havedetermined

so the other solution we were looking for is

If you are like the person writing this solution, you will have to check this to believe that it works, but it does.

1. Try to look enthused.↩

Exercise 26As suggested in the text, give a proof of Abel’s theorem for the third-order case. That is to say, suppose ,

, and are three solutions to the differential equation

then show that their Wronskian satisfies the first-order differential equation

∫ dw = 2 log(w) + log(w − 1) − 2 log(w − 2) .− 6w + 4w2

− 3 + 2ww3 w2

µ(w) = exp (2 log(w) + log(w − 1) − 2 log(w − 2)) = exp (log ( )) = .(w − 1)w2

(w − 2)2

(w − 1)w2

(w − 2)2

µ(w) [ku] = 0Ddw

k(w) = c

u(w)c u(w)

u(w) = c ∫ dw .(w − 2)2

(w − 1)w2

c w1

= − .(w − 2)2

(w − 1)w2

1w − 1

4w2

1w 0

u(w) = log(w − 1) + ,4w

z(w) = wu(w) = w log(w − 1) + 4 .

(t)Y1(t)Y2 (t)Y3

(t) + (t) (t) + (t) (t) + (t)y(t) = 0,y ′′′ a1 y ′′ a2 y ′ a3

W(t) = W [ , , ](t)Y1 Y2 Y3

+ (t)W = 0.′1





SolutionWe start by writing down what the Wronskian is:

which for simplicity I’ll expand at this juncture. I’ll also try to improve legibility by putting the subscripts inincreasing order in every term:

Now hold your breath and differentiate this. Each term will spawn three terms when we do this, and I’ll write eachon its own line.

Twelve terms match up and cancel, leaving only six to contend with, which I’ll rearrange a little bit:

The rearrangement was so that I could group terms as follows:

which is starting to look like the original line for at the top of the solution. Notice that each term that survivedhas a third derivative in it somewhere. We know from the differential equation that, for , , and ,

Plug this fact in for all three of the third derivatives. We arrive at

+ (t)W = 0.W ′ a1

W(t) = ( − ) − ( − ) + ( − ) ,Y1 Y ′2 Y ′′

3 Y ′′2 Y ′

3 Y2 Y ′1 Y ′′

3 Y ′′1 Y ′

3 Y3 Y ′1 Y ′′

2 Y ′′1 Y ′

2

W(t) = − − + + − .Y1Y ′2 Y ′′

3 Y1Y ′′2 Y ′

3 Y ′1 Y2Y ′′

3 Y ′′1 Y2Y ′

3 Y ′1 Y ′′

2 Y3 Y ′′1 Y ′

2 Y3

(t)W ′ = ( + + )Y ′1 Y ′

2 Y ′′3 Y1Y ′′

2 Y ′′3 Y1Y ′

2 Y ′′′3

− ( + + )Y ′1 Y ′′

2 Y ′3 Y1Y ′′′

2 Y ′3 Y1Y ′′

2 Y ′′3

− ( + + )Y ′′1 Y2Y ′′

3 Y ′1 Y ′

2 Y ′′3 Y ′

1 Y2Y ′′′3

+ ( + + )Y ′′′1 Y2Y ′

3 Y ′′1 Y ′

2 Y ′3 Y ′′

1 Y2Y ′′3

+ ( + + )Y ′′1 Y ′′

2 Y3 Y ′1 Y ′′′

2 Y3 Y ′1 Y ′′

2 Y ′3

− ( + + ) .Y ′′′1 Y ′

2 Y3 Y ′′1 Y ′′

2 Y3 Y ′′1 Y ′

2 Y ′3

(t) = − − + + − .W ′ Y ′′′1 Y2Y ′

3 Y ′′′1 Y ′

2 Y3 Y1Y ′′′2 Y ′

3 Y ′1 Y ′′′

2 Y3 Y1Y ′2 Y ′′′

3 Y ′1 Y2Y ′′′

3

(t) = ( − ) − ( − ) + ( − ),W ′ Y ′′′1 Y2Y ′

3 Y ′2 Y3 Y ′′′

2 Y ′1 Y3 Y1Y ′

3 Y ′′′3 Y1Y ′

2 Y ′1 Y2

Wi = 1 2 3

(t) = − (t) (t) − (t) (t) − (t) (t).Y ′′′i a1 Y ′′

i a2 Y ′i a3 Yi

(t)W ′ = (− − − )( − )a1Y ′′1 a2Y ′

1 a3Y1 Y2Y ′3 Y ′

2 Y3

−(− − − )( − )a1Y ′′2 a2Y ′

2 a3Y2 Y ′1 Y3 Y1Y ′

3

+(− − − )( − ).a1Y ′′3 a2Y ′

3 a3Y3 Y1Y ′2 Y ′

1 Y2



Now expand all of this. All the terms with an or an pair up and cancel, while all of the terms with an haveno mates. Therefore we can pull out from the entire expression. What remains is

and the expression in the big parentheses is up to rearrangement exactly the that we started with. So we’veshown that , or , as Abel’s theorem predicted.

Exercise 27Suppose we have a second-order homogeneous differential equation and we happen to know one of the solutions.Then the method of reduction of order will always give us a first-order differential equation whose solution is alinearly independent solution to the equation. In the problems above, the first-order differential equation is solvable,but this doesn’t happen in general—often we wind up with an integral that we can’t solve. This is not to say all islost; having an integral (or even just a differential equation in the first place) opens up the possibility of determiningvalues of the function by use of numerical methods, after all.

Consider the differential equation

One of the homogeneous solutions is . Check this. Then use the method of reduction of order tocome up with an expression for another solution to the homogeneous equation, linearly independent from . [Youshould guess from the paragraph preceeding this that you probably will get stuck at an integral.]

SolutionWe start with the check. We have , , and, as we hoped,

Now suppose the other solution to the equation is , where is to bedetermined. We calculate:

We therefore have

a2 a3 a1(t)a1

(t) = − (t)( − − + + − ),W ′ a1 Y ′′1 Y2Y ′

3 Y ′′1 Y ′

2 Y3 Y1Y ′′2 Y ′

3 Y ′1 Y ′′

2 Y ′3 Y1Y ′

2 Y ′′3 Y ′

1 Y2Y ′′3

W= − WW ′ a1 + W = 0W ′ a1

− 2(2 + 1)y = 0.y ′′ x2

(x) =Y1 ex2

Y1

(x) =Y1 ex2(x) = 2xY ′

1 ex2

(x) = 2 + 4 = (4 + 2) = 2(2 + 1) .Y ′′1 e

x2x2e

x2x2 e

x2x2 e

x2

(x) = u(x) (x) = u(x)Y2 Y1 ex2u(x)

(x)Y ′2

(x)Y ′′2

= (x) + 2xu(x)u′ ex2

ex2

= (4 + 2)u(x) + 4x (x) + (x) .x2 ex2

u′ ex2

u′′ ex2

0 = (x) − 2(2 + 1) = (x) + 4x (x) .Y ′′2 x2 Y2 u′′ e

x2u′ e

x2

+ 4x w = 02 ′ 2 2



Writing , we see that satisfies the first-order equation . The justclutters things up; as it is never zero, there’s no harm in dividing it away. The differential equation we need to solveis , which doesn’t look too scary. The coefficient on is which has as anantiderivative , so an appropriate integrating factor is . We then get

and a member of the solution family is . This function doesn’t have a nice antiderivative, so the bestwe can say for our friend is that

and is our other solution.

Exercise 28

Show that is a fundamental set of solutions for the second-order differential equation

. Make sure to check to see if the Wronskian of and is defined everywhere. (Note:you will need first, to verify that each of the two functions satisfies the differential equation and

second, to show that they form a fundamental set of solutions.)

Solution

We first check that and satisfy the differential equation (this involves just taking the first andsecond-order derivatives and checking that they satisfy the above differential equation).

Second, to verify whether and form a fundamental set of solutions, let us compute the Wronskian of thetwo:

which is never zero for any nonzero x. Therefore, and do indeed form a fundamental set of solutions.

Exercise 29Without solving, determine the Wronskian of two solutions evaluated at for the following differentialequation:

w(x) = (x)u′ w + 4x w = 0ex2

w′ ex2

ex2

+ 4xw = 0w′ w a(x) = 4xA(x) = 2x2 µ(x) = e2x2

[w ] = 0,d

d xe2x2

w(x) = e−2x2

u

u(x) = ∫ dx,e−2x2

(x) = ∫ dxY2 ex2e−2x2

{ , }x−1 x32

2 + x − 3y = 0x2y ′′ y ′ x1 x2

{ , }x−1 x32

=Y1 x−1 =Y2 x32

Y1 Y2

W [ , ](x) = det = 2 = ,Y1 Y2

⎛⎝⎜

x−1

−x−2

x32

32

x12

⎞⎠⎟ x

−12

2x)(√

(x)Y1 (x)Y2

t = 1

− 2 − y = 0 , y(1) = 5 , = 10.5 2 7



What is the interval of definition of this particular solution to the initial-value problem?

SolutionWe first need to put the second-order, homogeneous linear equation into normal form. The normal form of theequation looks like:

Now, using Abel’s Theorem, we obtain that the Wronskian is:

Then .

Also note that the only point where a solution to would attain a singularity is at . Theinterval of definition of this particular solution to the initial-value problem is , because the initial time point

falls into that time interval.

Exercise 30Without solving, determine the Wronskian of two solutions evaluated at for the following differentialequation:

Is the Wronskian defined for all ?

SolutionWe first need to put the second-order, homogeneous linear equation into normal form. The normal form of theequation looks like:

Now, using Abel’s Theorem, we obtain that the Wronskian is:

− 2 − y = 0 , y(1) = 5 , = 10.t5 y t2 y t7 y

− 2 − y = 0.y t−3 y t2

W(t) = c =e∫ dt

2t−3

W(t) = c .et−2

W(1) = c ⋅ e

− 2 − y = 0y t−3 y t2 t = 0(0, +∞)

t = 1

u = 4

2 + u − 3y = 0.u2y ′′ y ′

u

+ − = 0.y ′′ 12u

32u2

W(u) = c =e∫ − dt

12u

W(u) = c = c = c .− log(u)log

⎛⎝ ⎞⎠



The Wronskian is defined for all . Also,

Exercise 31The following problem is an application of the Method of Linear Superposition, . Assume that

and are both solutions of the equation , for a certain polynomial and a certainfunction .

(a) Write down a nonzero solution of the equation .

(b) Write down a solution of such that .

We haven’t discussed how to solve nonhomogeneous second-order linear differential equations yet,but we don’t need to know that yet!

Solution(a) By linearity, . So a possible solution tothe homogeneous equation could be .

(b) All we need to do is to ensure that a solution to the differential equation satisfies . For example, wecan ensure this by considering , or using linear suposition, , where isany real parameter.

Exercise 32Prove the following statement: ``If and are linearly independent solutions of a second-order linearhomogeneous differential equation on an interval , then and cannot have a maximum at the same location in

."

( This is a more theoretical argument than what you’ve seen before. Think of an argument by contradiction.Your proof should include complete sentences.)

SolutionWe will approach this problem through an argument via contradiction.

Suppose not, i.e. suppose that both and attain their maximum at the same point (call it in ), and letus study what happens in this case. The Wronskian evaluated at then becomes

W(u) = c = c = c .e− log(u)12 e

log⎛⎝ 1

u)(√⎞⎠ 1

u)(√

u > 0 W(4) = 2c.

Theorem 2.1cos(x) x p(D)w = q(x) p(x)

q(x)

p(D)w = 0

w(x) p(D)w = q(x) w(0) = 2

Remark

p(D) = (cos(x) − x) = p(D) cos(x) − p(D)x = q(x) − q(x) = 0w(x) = cos(x) − x

w(0) = 2w(x) = 2 cos(x) w(x) = 2 cos(x) − ax a

f(t) g(t)I f g

I

Hint :

f(t) g(t) t∗ It = t∗

W( ) = det ( ) = det ( ) = f( ) ⋅ 0 − 0 ⋅ g( ) = 0,∗ ∗ ∗



since maximum for both and implies that it is also a critical point, i.e.

Thus, we have obtained that and since is in , guarantees that everywhere in . However, this is a contradiction with the fact that the Wronskian of linearly independentsolutions of a second-order linear homogeneous differential equation cannot equal zero, see . Thisimplies that our initial assumption, that and can indeed have a maximum at the same location inside , is false.

Exercise 33Verify that and are solutions to the differential equation for .Then show that is not in general a solution of this equation. Can you explain why this result doesn’tcontradict the Method of Linear Superposition in ?

SolutionPlugging in the derivatives of and into the differential equation should yield that both aresolutions to the differential equation.

Now, the derivatives of are and . Plugging this into thedifferential equation yields

For the right hand side to equal , we need to set either or to . Thus, the differential equation is notsatisfied in general for any linear combination of the solutions and .

This does not contradict the method of linear superposition, because the differential equation for is not a linear homogeneous differrential equation, which is a requirement for to hold!

Exercise 34

If the functions and are linearly independent solutions of , determine what thenecessary and sufficient conditions are such that the functions and alsoform a linearly independent set of solutions.

W( ) = det ( ) = det ( ) = f( ) ⋅ 0 − 0 ⋅ g( ) = 0,t∗ f( )t∗

( )f ′ t∗g( )t∗

( )g ′ t∗f( )t∗

0g( )t∗

0t∗ t∗

t∗ f g ( ) = ( ) = 0.f ′ t∗ g ′ t∗

W( ) = 0t∗ t∗ I Theorem 2.3 W(t) = 0I 2

Theorem 2.6f g I

(x) = 1y1 (x) =y2 x12 y + ( = 0y ′′ y ′ )2 x > 0

+c1 c2x12

Theorem 2.1

(x) = 1y1 (x) =y2 x12

x) = +y( c1 c2x12 =y ′ 1

2 c2x− 12 = −y ′′ 1

4 c2x− 32

y + ( = − .y ′′ y ′ )2 14

c1c2x− 32

0 c1 c2 0y1 y2

y + ( = 0y ′′ y ′ )2

x > 0 Theorem 2.1

More exploration of the Wronskian :)

w1 w2 + p(u) + q(u)w = 0w′′ w′

= α + βw3 w1 w2 = γ + ϵw4 w1 w2



SolutionWe know that and form a linearly set of solutions of the second-order homogeneous differential equation inthe problem. Therefore by , the Wronkian of the two functions is nonzero, i.e.

for all t in the interval where the solution to the differential equation is defined.

Let us now study what the Wronskian of and looks like.

The second term is nonzero, so in order to ensure that , we need to impose the followingcondition, that is nonzero.

Then, by , we obtain that if and only if is nonzero, can and defined as in theproblem form a linearly indepedent set of solutions.

Exercise 35

(a) Consider the differential equation . Show that one of the solutions of the equation is .

(b)Use Abel’s Formula to show that the Wronskian of any two solutions of the given equaton is

where is a constant.

(c) Consider from part (a) and use the result in (b) to obtain a differential equation satisfied by thesecond solution . Then solve this equation to show that

Solution(a) The characteristic equation is . The characteristic root is with multiplicity . Thus, onepossible solution to the differential equation would look like (you can also plug in this solution toverify that it indeed satisfies the differential equation).

(b) Abel’s Formula states that the Wronskian satisfies the differential equation , where Solving this differential equation, we obtain that .

w1 w2Theorem 2.6

W( , )(t) = − ≠ 0,w1 w2 w1w′2 w′

1w2

w3 w4

W( , )(t) = − = (α + β )(γ + ϵ ) − (α + β )(γ + ϵ )w3 w4 w3w′4 w′

3w4 w1 w2 w′1 w′

2 w′1 w′

2 w1 w2

= (αϵ − γβ) + (βγ − αϵ) = (αϵ − βγ)( − ) = (αϵ − βγ)W( , )(t).w1w′2 w2w′

1 w1w′2 w2w′

1 w1 w2

W( , )(t) ≠ 0w3 w4αϵ − βγ

Theorem 2.6 αϵ − βγ w3 w4

A clever use of Abe s Theorem : l′

+ 2a + z = 0z′′ z′ a2

e−au

W(u) = (u) (u) − (u) (u) = c ,z1 z′2 z′

1 z2 e−2au

c

(u) =z1 e−au

(u)z2 (u) = u .z2 e−au

+ 2ar + = 0r2 a2 −a 2(u) =z1 c1e−au

+ 2aW = 0W ′

W [ , ](u) = c.z1 z2 W(u) = c = ce− ∫ 2adu e−2au

(u) =1−au W(u) = (u) (u) − (u) (u) = c ,1 2

−2au (u)2



(c) Plugging in into we obtain that mustsatisfy . Using the method of integrating factors from , we obtainthat , for any constant . Setting , we obtain that a possible solution to thedifferential equation is . This solution is precisely the solution that would yield from our usual wayof solving (a). This was just a detour!

When we learn about the methods of Key Identity Evaluations and Undetermined Coefficients, wewill be able to solve through methods other than just using integrating factors!

Exercise 36Use the method of order reduction to find a second solution of the differential equation

.

SolutionFirst of all, we should verify that indeed satisfies the differential-equation. This should be a quickcalculation.

Second, using the suggestion in the problem to utilize the method of order reduction, let us now suppose there isanother solution to the differential equation of the form . Substituting this into the differentialequation, we obtain that

Setting , the equation becomes . Upon simplification, we obtain that . Thisis a separable equation which we can solve to obtain that . Alternatively, we could also use the methodof integrating factors to get to the same result.

Thus, meaning that by integrating we obtain that . We are not done yet - thesecond solution to the differential equation in the problem is thus , by taking and

in the equation above. You can also plug in into to verify that isindeed another solution of the differential equation.

(u) =z1 e−au W(u) = (u) (u) − (u) (u) = c ,z1 z′2 z′

1 z2 e−2au (u)z2+ a = cz′

2 z2 e−au Chapter 1 Section 8(u) = u + cuz2 e−au e−au c c = 0

(u) = uz2 e−au

Remark : + a = cz′

2 z2 e−au

+ 2z − 2w = 0 , z > 0 , (z) = zz2w w w1

(z) = zw1

(z) = zv(z)w2

( z + 2 ) + 2z( z + z) − 2zv = + 4 .z2 v v v z3 v z2 z

= uv + 4 u = 0z3u z2 z + 4u = 0uu = cz−4

= u = cv z−4 v = − + kc3 z−3

(z) = zv(z) =w2 z−2 c = −3k = 0 (z) =w2 z−2 + 2z − 2w = 0z2w w w2

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