Math 528
Homework 1 Solutions
Assignment:
Section 1.1: #1-7 odd, 19
Section 1.3: #12, 15
Section 1.4: #1-13 odd
Section 1.5: #1-13 odd, 30
Section 1.7: #1-5
Section 1.1
1.
dy
= −2 sin(2πx)
Z dx Z
dy = −2 sin(2πx)dx
y=
1
cos(2πx) + c
π
3.
Z
Z
dy
= dx
y
ln(y) = x + c̃
y = cex .
5. By integration by parts
Z
Z
−x
−x
y = 4e cos x dx = 4e sin x + 4e−x sin x dx
Z
−x
−x
= 4e sin x − 4e cos x − 4e−x cos x dx.
Thus,
Z
y=
7.
4e−x cos x dx = 2e−x sin x − 2e−x cos x + c.
Z
y=
cosh 5.13x dx =
1
sinh 5.13x + c.
5.13
19.
y 00 (t) = g
y 0 (0) = 0
y(0) = 0
y 00 (t) = g ⇒ y 0 (t) = gt + c1 .y 0 (0) = 0 ⇒ c1 = 0 ⇒ y 0 (t) = gt.
1
1
y 0 (t) = gt ⇒ y(t) = gt2 + c2 .y(0) = 0 ⇒ c2 = 0 ⇒ y(t) = gt2 .
2
2
Section 1.3
12. Solve the IVP:
Z
dy
=
1 + 4y 2
Z
dx
1
tan−1 (2y) = x + c̃
2
tan−1 (2y) = 2x + c
1
y = tan(2x + c)
2
y(1) = 0 ⇒
1
tan(2 + c) = 0 ⇒ 2 + c = 0 ⇒ c = −2.
2
1
y = tan(2x − 2).
2
Check:
1
tan(2x + c)
2
y 0 = sec2 (2x + c)
y=
p
1 + 4y 2
2y
2x + c
1
y 0 = sec2 (2x + c) = 1 + 4y 2 .
15. Solve IVP:
Z
Z
y dy = −4
x dx
1 2
y = −2x2 + c̃
2
√
y = ± −4x2 + c
√
y(2) = 3 ⇒ ± c − 16 = 3 ⇒ c = 25.
√
y = ± −4x2 + 25.
Check:
√
y = ± −4x2 + c
−4x
−4x
y0 = √
=
.
y
± −4x2 + c
Section 1.4
1. M = 2xy, N = x2 , and My = Nx = 2x, so the differential equation is exact.
Z
Z
u = M dx + k(y) = 2xy dx + k(y) = x2 y + k(y).
dk
∂u
= x2 +
= N = x2 .
∂y
dy
Hence, dk/dy = 0 so k = c̃. Thus,
u(x, y) = x2 y = c.
3. M = sin x cos y, N = cos x sin y, and My = Nx = cos x cos y, so the differential
equation is exact.
Z
Z
u = M dx + k(y) = sin x cos y dx + k(y) = − cos x cos y + k(y).
∂u
dk
= cos x sin y +
= N = cos x sin y.
∂y
dy
Hence, dk/dy = 0 so k = c̃. Thus,
u(x, y) = − cos x cos y = c.
5. P = x2 + y 2 , Q = −2xy, Py = 2y, and Qx = −2y so the differential equation
is not exact. Letting
∂Q
2
1 ∂P
−
=− ,
R=
Q ∂y
∂x
x
then the integrating factor is
Z
F (x) = exp
−
2
1
dx = 2 .
x
x
Multiplying the original differential equation by F we get the exact equation
2y
y2
1 + 2 dx − dy = 0.
x
x
Z y2
y2
u=
1 + 2 dx + k(y) = x −
+ k(y).
x
x
∂u
2y dk
2y
=− +
=− .
∂y
x
dy
x
Hence, dk/dy = 0 so k = c̃. Thus,
u(x, y) = x −
y2
= c.
x
7. Using methods similar to 5 we find this differential equation is not exact and
the integrating factor is
2
F (x) = ex .
Multiplying the original differential equation by F we get the exact equation
2
2
2xex tan y dx + ex sec2 y dy = 0.
Solving this using methods similar to the previous problems we find
2
u(x, y) = ex tan y = c.
9. Using methods similar to previous problems we can see this differential equation
is exact and
Z
u = 2e2x cos y dx + k(y) = e2x cos y + k(y).
∂u
dk
= −e2x sin y +
= −e2x sin y.
∂y
dx
Hence, dk/dy = 0 so k = c̃. Thus,
u(x, y) = e2x cos y = c.
y(0) = 0 ⇒ c = 1 ⇒ e2x cos y = 1.
11. Letting P = 2 cosh x cos y and P = − sinh x sin y we can see the differential
equation is not exact. We can find the integrating factor by first finding
∂Q
1 ∂P
−
R=
Q ∂y
∂x
1
=−
(−2 cosh x sin y + cosh x sin y)
sinh x sin y
cosh x
=
.
sinh x
Then the integrating factor is
Z
cosh x
dx = exp (ln (sinh x)) = sinh x.
F (x) = exp
sinh x
Multiplying the original differential equation by F we get the exact equation
2 sinh x cosh x cos y dx − sinh2 x sin y dy = 0.
Solving this using similar methods as previous problems we find
u(x, y) = sinh2 x cos y = c.
13. Multiplying the original differential equation by F we get the exact equation
ex dx + (−1 + ey )dy = 0.
Solving this using similar methods as previous problems we get
u(x, y) = ex + ey − y = c.
Section 1.5
1.
e− ln x = eln x
−1
= x−1 =
−1 )
e− ln(sec x) = eln((sec x)
=
1
.
x
1
= cos x.
sec x
3. p = −1, r = 5.2 and
Z
h=
Z
p dx =
−dx = −x.
From this we see
eh = e−x , e−h = ex , eh r = 5.2e−x ,
and the general solution of our equation is
Z
x
y(x) = e
5.2e−x dx + ce−x = −5.2 + cex .
5. p = k, r = e−kx so using methods similar to 3 we find
Z
−kx
y=e
dx + ce−kx = xe−kx + ce−kx .
7. This equation can be rewritten as
2
y 0 − y = x2 e x .
x
Using methods similar to 3 we find
Z
2
y=x
ex dx + cx2 = x2 ex + cx2 .
9. Using methods similar to 3 we find
Z
cos x
y=e
dx + cecos x = xecos x + cecos x .
y(0) = −2.5 ⇒ c = −2.5/e.
11. Using methods similar to 3 we find
Z
cos x
y = sin x −2 2 dx + c sin x = 2 + c sin x.
sin x
13. This equation can be separated
Z
Z
dy
= 6 tanh 1.5x dx
y − 2.5
ln(y − 2.5) = 4 ln(cosh 1.5x) + c
y = c cosh4 1.5x + 2.5.
Section 1.7
1. In this problem y 0 = f (x, y) = r(x) − p(x)y. This is continuous for all x in the
interval |x − x0 | ≤ a and bounded in any rectangle
R : |x − x0 | < a, |y − y0 | < b,
for any b. Now fy = −p(x) which is continuous and bounded on for all (x, y) in
the rectangle R. Thus, our theorems are satisfied. We did not need our theorems
in this case as we can solve a linear ODE explicitly (Section 1.5).
2. The given IVP does not have a solution since trying to solve it by separating we
get
y = c(x − 2)
so y(2) = 0 for all c. This does not contradict the theorems as f (x, y) =
not continuous on any rectangle in which x = 2.
y
x−2
is
3. If the theorems are satisfied on a strip then the solution will exist when |x−x0 | < a
as b/K will eventually be greater than a since we can satisfy the theorems for any
b and K is fixed.
4. Unless k = 0 we get no solutions by the same reasoning as 2. If k = 0 then we
get a set of nonunique solutions
y = c(x − 2).