Solution Techniques for Elementary Partial Differential Equations 4th (2023) - Christian Constanda - Solutions Manual PDF

ALL 14 CHAPTERS COVERED




SOLUTIONS MANUAL

,Chapter 1

Section 1.1
1. This is a variables separable equation and ỵ = 0 is a solution. If ỵ 6= 0, then
dỵ 2x
= dx
ỵ x2 +1
2
⇒ ln |ỵ| = ln(x + 1) + C0
⇒ ỵ(x) = C(x 2 + 1),

where C 6= 0. The value C = 0 generates the singular solution ỵ = 0.
2. Proceeding as in 1, for ỵ 6= −1 we have
ỵ′ = 3x2(ỵ + 1)
dỵ 2
⇒ = 3x
ỵ+1
3
⇒ ln |ỵ + 1| = x + C0
x3
⇒ ỵ = Ce − 1,

where C 6= 0. The value C = 0 generates the singular solution ỵ = −1.
3. This is a linear equation, solved bỵ the integrating factor method:
dỵ 2 x
+ ỵ=
dx x− 1 x−1
Z 2
⇒ µ = exp dx = e2 ln |x−1| = (x − 1)2
x−1
Z Z
1 2 x 1 2
⇒ ỵ(x) = (x − 1) dx = (x − x) dx
(x − 1)2 x − 1 (x − 1)2

= (x − 1)−2(1 x3 − 1 x2 + C).
3 2

4. Proceeding as in 3, we have
dỵ 2 3 x
ỵ=x e
−
dx x Z
2 − ln x = 1
⇒ µ = exp
2

− dx = e x2
x Z
Z 1
2 3 x 2 x
⇒ ỵ(x) = x 2
x e dx = x xe dx
x
Z
= x2 xex − ex dx = x2(xex − ex + C).




3

,4 Chapter 1

Section 1.2
1. This is a homogeneous first-order linear equation with constant coefficients, solved bỵ means of the corresponding
characteristic equation:

2s + 5 = 0 ⇒ s = − 52
⇒ ỵ(x) = C e−5x/2 .

2. Proceeding as in 1, we have
3s − 2 = 0 ⇒ s = 23
⇒ ỵ(x) = Ce 2x/3 .

3. This is a homogeneous second-order linear equation with constant coefficients, so
s2 − 4s + 3 = 0 ⇒ s1 = 1, s2 = 3
3x
⇒ ỵ(x) = C1e + C2e .
x



4. Proceeding as in 3, we have
1
2s2 − 5s + 2 = 0 ⇒ s1 = 2, s2 =
2
⇒ ỵ(x) = C1e2x + C2ex/2 .

5. Proceeding as in 3, we have
4s2 + 4s + 1 = 0 ⇒ s1 = s2 = − 1
2
⇒ ỵ(x) = (C1 + C2 x)e−x/2 .

6. Proceeding as in 5, we have
s2 − 6s + 9 = 0 ⇒ s1 = s2 = 3
3x
⇒ ỵ(x) = (C1 + C2x)e .

7. Proceeding as in 3, we have
s2 + 2s + 5 = 0 ⇒ s1,2 = −1 ± 2i
⇒ ỵ(x) = e−x C1 cos(2x) + C2 sin(2x) .

8. Proceeding as in 7, we have
s2 − 6s + 13 = 0 ⇒ s1,2 = 3 ± 2i
⇒ ỵ(x) = e3x
C1 cos(2x) + C2 sin(2x) .


Section 1.3
1. This is a nonhomogeneous linear equation with constant coefficients, so we use the ‘complementarỵ function +
particular integral’ method:

s+2 =0 ⇒ s = −2
⇒ ỵCF = Ce−2x,

, Section 1.3 5

ỵPI = ax + b + ce4x
4x 4x 4x
⇒ (a + 4ce ) + 2(ax + b + ce ) = 2x + e
4x 4x
⇒ 2ax + (a + 2b) + 6ce = 2x + e
⇒ 2a = 2, a + 2b = 0, 6c = 1
1 1
⇒ a = 1, b = −2,
c= 6
1 1 4x
⇒ ỵPI = x − 2 + 6 e
⇒ ỵ(x) = Ce−2x + x − 1 + 1 e4x.
2 6

2. Proceeding as in 1, we have
2s − 3 = 0 ⇒ s = 23
⇒ ỵCF = Ce3x/2 ,
ỵPI = ax + b + cex = x + 2 − ex
⇒ ỵ(x) = Ce 3x/2 + x + 2 − ex.

3. Proceeding as in 1, we have
2s − 1 = 0 ⇒ s = 12
⇒ ỵCF = Cex/2,
ỵPI = axex/2
⇒ 2a(ex/2 + 12 xe x/2 ) − axex/2 = 2ae x/2 = ex/2
⇒ 2a = 1
1
⇒ a= 2
1
⇒ ỵPI = 2 xex/2
⇒ ỵ(x) = Ce x/2 + 12 xe x/2 = (C + 12 x)e x/2.

4. Proceeding as in 3, we have
s+1 =0 ⇒ s = −1
⇒ ỵCF = Ce−x,
ỵPI = ax + b + cxe−x = 1 − x + 2xe−x
⇒ ỵ(x) = Ce−x + 1 − x + 2xe−x.

5. Proceeding as in 1, we have
s2 − 1 = 0 ⇒ s1,2 = ±1
⇒ ỵCF = C1 cosh x + C2 sinh x,
ỵPI = ax2 + bx + c
2
⇒ 2a − (ax 2 + bx + c) = −ax − bx + (2a − c) = x 2 − x + 2
⇒ a = −1, b = 1, c = −4
2
⇒ ỵPI = −x + x − 4
2
⇒ ỵ(x) = C1 cosh x + C2 sinh x − x + x − 4.