Solutions for Diff.Eq. Homework 3

Xiangjin Xu

1 Section 2.8

In Problem 3 and 5 let φ0(t) = 0 and use the method of successive approximations tosolve the given initial value problem.

(a). Determine φn(t) for arbitrary value of n.(c). Express limn→∞ φn(t) = φ(t) in terms of elementary functions, that is solve the

given initial value problem.

Problem 3. y′ = 2(y + 1), y(0) = 0Solution: (a). Since φ0(t) = 0, and

φn(t) =∫ t

02(φn−1(s1) + 1)ds1

= 2∫ t

0φn−1(s)ds+ 2t

= 22∫ t

0

∫ s1

0φn−2(s2)ds2ds1 +

(2t)2

2!+ 2t

=n∑

k=1

(2t)k

k!

that is

φn(t) =n∑

k=1

(2t)k

k!

(b). limn→∞

φn(t) = limn→∞

n∑k=1

(2t)k

k!

= exp(2t)− 2

that isφ(t) = exp(2t)− 2

. Q.E.D.

1

Problem 5. y′ = −y/2 + t, y(0) = 0Solution: (a). Since φ0(t) = 0, and

φn(t) =∫ t

0(−φn−1(s1)/2 + t)ds1

= −1

2

∫ t

0φn−1(s)ds+

t2

2

=1

22

∫ t

0

∫ s1

0φn−2(s2)ds2ds1 −

1

2

∫ t

0

s21

2ds1 +

t2

2!

=1

22

∫ t

0

∫ s1

0φn−2(s2)ds2ds1 + 4

(−t)3

233!+ 4

(−t)2

222!

= 4n∑

k=1

(−t)k

2kk!

that is

φn(t) = 4n∑

k=1

(−t)k

2kk!

(b). limn→∞

φn(t) = limn→∞

4n∑

k=1

(−t)k

2kk!

= 4 exp(−t/2) + 2t− 4

that isφ(t) = 4 exp(−t/2) + 2t− 4

. Q.E.D.

In Problem 7 let φ0(t) = 0 and use the method of successive approximations to solve thegiven initial value problem.

(a). Determine φn(t) for arbitrary value of n.(b). Observe whether the iterates appear to be converging.

Problem 7. y′ = ty + 1, y(0) = 0Solution: (a). Since φ0(t) = 0, and

φn(t) =∫ t

0(s1φn−1(s1) + 1)ds1

=∫ t

0s1φn−1(s)ds+ t

=∫ t

0s1

∫ s1

0s2φn−2(s2)ds2ds1 +

∫ t

0s21ds1 + t

=∫ t

0s1

∫ s1

0φn−2(s2)ds2ds1 +

t3

3+ t

=n∑

k=1

t2k−1

(2k − 1)!!

2

that is

φn(t) =n∑

k=1

t2k−1

(2k − 1)!!

(b). By radius test of convergence, we see φn(t) convergence everywhere.Q.E.D.

Problem 14. Consider that the sequence φn(x) = 2nxe−nx2, 0 ≤ x ≤ 1.

(a). Show that limn→∞ φn(x) = 0 for 0 ≤ x ≤ 1; hence∫ 1

0lim

n→∞φn(x)dx = 0

Proof: For x = 0, φn(x) = 0, and for fixed 0 < x ≤ 1, by L’Hopital’s rule, we have

limn→∞

φn(x) = limn→∞

2nx

enx2 = limn→∞

2x

x2enx2 = limn→∞

2

xenx2 = 0

Q.E.D.(b). Show that

∫ 10 2nxe−nx2

dx = 1− e−n, hence

limn→∞

∫ 1

0φn(x)dx = 1

Thus, in this example,

limn→∞

∫ b

aφn(x)dx 6=

∫ b

alim

n→∞φn(x)dx,

even though limn→∞ φn(x) exists and is continuous.Proof: By changing variable y = nx2, we have∫ 1

02nxe−nx2

dx =∫ n

0e−ydy = 1− e−n

Q.E.D.

Problem 15. If ∂f/∂y is continuous in the rectangle D, show that there is a positiveconstant K such that

|f(t, y1)− f(t, y2)| ≤ K|y1 − y2|,where (t, y1) and (t, y2) are any two points in D having the same t coordinate. Thisinequality is known as a Lipschitz condition.

Proof: By the mean value theorem, we have

f(t, y1)− f(t, y2) = fy(t, y)(y1 − y2)

3

where y is between y1 and y2. Since ∂f/∂y is continuous in the rectangle D, defineK = sup(t,y)∈D |fy(t, y)|, then we have

|f(t, y1)− f(t, y2)| = |fy(t, y)||y1 − y2| ≤ K|y1 − y2|

Q.E.D.Problem 19. In this problem we deal with the question of uniqueness of the solution

of the integral equation (3),

φ(t) =∫ t

0f [s, φ(s)]ds,

(a). Suppose that φ and psi are two solutions of Eq.(3). Show that for t ≥ 0,

φ(t)− ψ(t) =∫ t

0{f [s, φ(s)]− f [s, ψ(s)]}ds.

Proof: Since φ and psi are two solutions of Eq.(3), plug the integral equation in, weshow the identity.

Q.E.D.(b). Show that

|φ(t)− ψ(t)| ≤∫ t

0|f [s, φ(s)]− f [s, ψ(s)]|ds.

Proof: From (a), we have

|φ(t)− ψ(t)| = |∫ t

0{f [s, φ(s)]− f [s, ψ(s)]}ds| ≤

∫ t

0|f [s, φ(s)]− f [s, ψ(s)]|ds.

Q.E.D.(c). Use the result of Problem 15 to show that

|φ(t)− ψ(t)| ≤ K∫ t

0|φ(s)]− ψ(s)|ds.

where K is an upper bound for |∂f/∂y| in D.Proof: From (b) and Problem 15, we have

|φ(t)−ψ(t)| ≤∫ t

0|f [s, φ(s)]−f [s, ψ(s)]|ds ≤

∫ t

0K|φ(s)]−ψ(s)|ds ≤ K

∫ t

0|φ(s)]−ψ(s)|ds.

Q.E.D.

2 Section 3.1

In Problem 1, 3, 5, find the general solution of the given differential equation.Problem 1. y′′ + 2y′ − 3y = 0

4

Solution: The characteristic equation is r2 + 2r− 3 = 0, solve this, we get two rootsare 1 and −3. Hence the general solution is

y(t) = c1et + c2e

−3t.

Q.E.D.Problem 3. 6y′′ − y′ − y = 0Solution: The characteristic equation is 6r2 − r− 1 = 0, solve this, we get two roots

are 1/3 and −1/2. Hence the general solution is

y(t) = c1e−t/2 + c2e

t/3.

Q.E.D.Problem 5. y′′ + 5y′ = 0Solution: The characteristic equation is r2 + 5r = 0, solve this, we get two roots are

0 and −5. Hence the general solution is

y(t) = c1 + c2e−5t.

Q.E.D.Problem 10. Find the solution of the given initial value problem, and describe its

behavior as t increases.

y′′ + 4y′ + 3y = 0, y(0) = 2, y′(0) = −1

Solution: The characteristic equation is r2 + 4r + 3 = 0, solve this, we get two rootsare −1 and −3. Hence the general solution is

y(t) = c1e−t + c2e

−3t.

andy′(t) = −c1e−t − 3c2e

−3t

, then by the initial value we have

c1 + c2 = 2, −c1 − 3c2 = −1

solve these two linear equations we have c1 = 52

and c2 = −12. Hence the solution is

y(t) =5

2e−t − 1

2e−3t

y(t) tends to 0 as t increases.Q.E.D.

Problem 17. Find a differential equation whose general solution is y = c1e2t +c2e

−3t.Solution: From the general solution we can see the characteristic equation is 0 =

(r − 2)(r + 3) = r2 + r − 6, hence the differential equation is

y′′ + y′ − 6y = 0.

5

Q.E.D.Problem 19. Find the solution of the initial value problem

y′′ − y = 0, y(0) =5

4, y′(0) = −3

4,

determine its minimum value in 0 ≤ x ≤ 2.Solution: The characteristic equation is r2 − 1 = 0, solve this, we get two roots are

1 and −1. Hence the general solution is

y(t) = c1et + c2e

−t.

andy′(t) = c1e

t − c2e−t

, then by the initial value we have

c1 + c2 = 5/4, c1 − c2 = −3/4

solve these two linear equations we have c1 = 1/4 and c2 = 1. Hence the solution is

y(t) =1

4et + e−t

Since y′(t) = 14et − e−t = 0 imply that t = ln2, hence ymin = min{y(0), y(ln2), y(2)} = 3

4.

Q.E.D.Problem 22. Solve the initial value problem 4y′′ − y = 0, y(0) = 2, y′(0) = β. Then

find β so that the solution approaches zero as t→∞.Solution: The characteristic equation is 4r2 − 1 = 0, solve this, we get two roots are

1/2 and −1/2. Hence the general solution is

y(t) = c1et/2 + c2e

−t/2.

andy′(t) =

c12et/2 − c2

2e−t/2

, then by the initial value we have

c1 + c2 = 4, c1 − c2 = 2β

solve these two linear equations we have c1 = 2 + β and c2 = 2− β. Hence the solution is

y(t) = (2 + β)et/2 + (2− β)e−t/2

Since the solution approaches zero as t→∞, we have 2 + β = 0, that is, β = −2.Q.E.D.

6

3 Section 3.2

In the Problem 1, 4, 5, 6, find the Wronskian of the given pair of functions.Problem 1. e2t, e−3t/2

Solution:

W [f, g] = fg′ − f ′g = e2t(−3/2)e−3t/2 − 2e2te−3t/2 =1

2e2te−3t/2

Q.E.D.Problem 4. x, xex

Solution:W [f, g] = fg′ − f ′g = x(ex + xex)− xex = x2ex

Q.E.D.Problem 5. et sin t, et cos tSolution:

W [f, g] = fg′ − f ′g = et sin t(et cos t− et sin t)− (et sin t+ et cos t)et cos t = −e2t

Q.E.D.Problem 6. cos2 θ, , 1 + cos 2θSolution:

W [f, g] = fg′ − f ′g = cos2 θ(−2 sin 2θ)− (−2 sin θ cos θ)(1 + cos 2θ)

= −2 cos2 θ sin 2θ + 2 sin 2theta cos2 θ = 0

Q.E.D.Problem 7. Determine the longest interval in which the given initial value problem

is certain to have a unique twice differentiable solution.

ty′′ + 3y = t, y(1) = 1, y′(1) = 2

Solution: First change the form of equation to

y′′ +3

ty = 1

From Theorem 3.2.1, we know the intervals in which the given initial value problem iscertain to have a unique twice differentiable solution are the same as the intervals onwhich p,q and g are continuous. Since for this problem, the intervals are (−∞, 0) and(0,∞), since the initial point is for point t = 0, we know the longest interval is (0,∞).

Q.E.D.In Problem 21, 22, find the fundamental set of the solutions specified by Theorem

3.2.5 for the given differential equation and initial point.Problem 21. y′′ + y′ − 2y = 0, t0 = 0

7

Solution: The characteristic equation is r2 + r − 2 = 0, solve this, we get two rootsare 1 and −2. Hence the general solution is

y(t) = c1et + c2e

−2t.

andy′(t) = c1e

t − 2c2e−2t

First for initial value y(0) = 1, y′(0) = 0, we have

c1 + c2 = 1, c1 − 2c2 = 0

solve above two linear equation, we have c1 = 2/3, c2 = 1/3, hence

y1(t) =2

3et +

2

3e−2t

Second for initial value y(0) = 0, y′(0) = 1, we have

c1 + c2 = 0, c1 − 2c2 = 1

solve above two linear equation, we have c1 = 1/3, c2 = −1/3, hence

y2(t) =1

3et − 1

3e−2t

Hence the fundamental set of solutions is (y1, y2) as above.Q.E.D.

Problem 22. y′′ + 4y′ + 3y = 0, t0 = 1Solution: The characteristic equation is r2 + 4r + 3 = 0, solve this, we get two roots

are −1 and −3. Hence the general solution is

y(t) = c1e−t + c2e

−3t.

andy′(t) = −c1e−t − 3c2e

−3t

First for initial value y(1) = 1, y′(1) = 0, we have

c1e−1 + c2e

−3 = 1, −c1e−1 − 3c2e−3 = 0

solve above two linear equation, we have c1 = 3e2, c2 = −e3

2, hence

y1(t) =3e

2e−t +

−e3

2e−3t

Second for initial value y(1) = 0, y′(1) = 1, we have

c1e−1 + c2e

−3 = 0, −c1e−1 − 3c2e−3 = 1

8

solve above two linear equation, we have c1 = e2, c2 = −e3

2, hence

y1(t) =e

2e−t +

−e3

2e−3t

Hence the fundamental set of solutions is (y1, y2) as above.Q.E.D.

Problem 24. Verify that the functions y1, y2 are solutions of the given differentialequation. Do they constitute a fundamental set of solutions?

y′′ − 2y′ + y = 0; y1(t) = et, y2(t) = tet

Solution: For y1(t) = et, we have

y′′ − 2y′ + y = et − 2et + et = 0

and for y2(t) = tet, we have

y′′ − 2y′ + y = (etet + tet)− 2(et + tet) + tet = 0

hence y1(t) and y2(t) are both the solution of the equation. Since y1(t) = et > 0 andy′1(t) = et > 0 are always true, there is no initial point t0 such that y1(t0) = 1 andy′1(t0) = 0. Hence y1(t) and y2(t) don’t constitute a fundamental set of solutions.

Q.E.D.

9