Download Suggested practice problems from recent sections: • 10.4 and more Lecture notes Differential Equations in PDF only on Docsity!
Suggested practice problems from recent sections:
- 10.4: 3, 4, 7, 8, 13, 14, 19, 20
- 10.5: 17, 18, 27, 28, 35, 36
- 10.6: 1, 2, 3, 4, 40, 41, 42, 43
a
Approximate
1 e
xdx as a Riemann sum with 3 equal intervals, choosing the
left endpoint of each rectangle to be its height. Solution: 2(e + e^3 + e^5 )
b
Give the general solution of the differential equation
dy dt
ey^ − 1 ey^
t^2.
Solution: If y = 0, the right side is 0, so y = 0 is a solution. If y 6 = 0, we have ey ey^ − 1
dy = t^2 dt.
Integrating both sides (and using u = ey^ to integrate the left), we have
ln |ey^ − 1 | =
t^3 3
+ C.
Therefore |ey^ − 1 | = e t 33 +C
and so ey^ − 1 = e
t 33 +C or ey^ = −e
t 33 +C .
Solving for y, we get
y = ln
e
t 33 +C
or y = ln
−e
t 33 +C
or y = 0.
c
Recall that arcsin x =
∫ (^) x 0 √^1 1 −t^2 dt. Show that when^ y >^ 0,
arcsin
y^2
≤ y
y^2
Solution: arcsin
1 − (^) y^12 =
q 1 − (^) y^12 0 √^1 1 −t^2 dt. Since^ √^1 1 −t^2 is increasing, the
area under the curve √ 11 −t 2 as t goes from 0 to
1 − (^) y^12 is contained inside the
rectangle with corners (0, 0) and (
1 − (^) y^12 , y). This box has area y
1 − (^) y^12.
d
You know that 2 ≤ f (x) ≤ 3 for all x. Is it possible that
2 f^ (x)dx^ = 4? Solution: No.
0 f^ (x)dx^ contains the rectangle with corners (2,^ 0) and (5,^ 2), which has area 6, so 6 ≤
0 f^ (x)dx.
e
What is
− 1
cos x x dx? Solution: 0, since the bounds are equal.
f
Find a value a > 0 such that
∫ (^) a 1
sin(x−2) (x−2)^2 dx^ = 0. Solution: We can’t find the indefinite integral, so we must use geometry. This function is anti-symmetric around 2: sin(2(2−−c−c−2)2) = −^ sinc 2 c= − sin(2+(2+c−c−2)2) 2. Therefore we need 1 and a to be symmetric around 2, so a = 3.
g
Define F (x) =
∫ (^) x 0
sin t t dt. What is^
d dx F^ (ln^ x)? Solution: By FTC, F ′(x) = sinx^ x, so by the chain rule, (^) dxd F (ln x) = sin lnx ln x^ x.
h
Water is flowing into a container at a rate of W (t)gal/sec (where t is the time). Express the amount of water that enters the container between t = 0 and t = 4. Solution:
0 W^ (t)dt.
i
What is the partial fraction decomposition of
1 (x^2 + 4)^3 (x^2 + 1)^2 (x − 1)^3 (x + 2)
Solution: Ax + B x^2 + 4
Cx + D (x^2 + 4)^2
Ex + F (x^2 + 4)^3
Gx + H x^2 + 1
Ix + J (x^2 + 1)^2
K
x − 1
L
(x − 1)^2
M
(x − 1)^3
N
x + 2
j
Find and solve the partial fraction decomposition for
1 (x^2 + 1)(x^2 − 1)
∫ (^) x √ 1 −x 2 dx
arcsin x dx
x^4 − 1 dx
4 x^2 +8x+29 dx
1
ln x x dx
1 e
xdx
Solution:
- u = ln x^3 , dv = x^2 dx, du = 3x^2 /x^3 = 3/x, v = x^3 /3,
x^2 ln x^3 dx = (x^3 /3) ln x^3 −
x^2 dx = (x^3 /3) ln x^3 − x^3 / 3
- u = 1 − x^2 , du = − 2 x,
∫ (^) x √ 1 −x^2 dx^ =^ −^
1 2
u−^1 /^2 du = −
u = −
1 − x^2
- u = arcsin x, dv = dx, du = √ 11 −x 2 , v = x,
arcsin x dx = x arcsin x − ∫ (^) x √ 1 −x^2 dx^ =^ x^ arcsin^ x^ +^
1 − x^2
x^4 − 1 dx^ =^
− (^) 2(x (^21) +1) − (^) 4(x^1 +1) + (^) 4(x^1 −1) dx = − 12 arctan x − 14 ln(x + 1) + 1 4 ln(x^ −^ 1)
- Since (2x + 2)^2 = 4x^2 + 8x + 4,
4 x^2 +8x+29 dx^ =^
(2x+2)^2 +25 dx^ = 1 25
(2x+2)^2 25 +^
dx = 251
( 2 x 25 +2 )^2 +1 dx.^ Substituting^ u^ =^
2 x+ 5 , du^ =^
2 5 dx, we get
4 x^2 +8x+29 dx^ =^
1 10
u^2 +1 du^ =^
1 10 arctan^ u^ =^
1 10 arctan^
2 x+
1
ln x x dx^ = lima→∞
∫ (^) a 1
ln x x dx.^ u^ = ln^ x, du^ =^ dx/x, so
lim a→∞
∫ (^) a
1
ln x x
dx = lim a→∞
∫ (^) ln a
0
udu
= lim a→∞ u^2 / 2
∣ln^ a 0 = lim a→∞ (ln^2 a/ 2 − 0)
Since lima→∞ ln^2 a/2 = ∞, this integral does not exist.
1 e
xdx = − ∫^1 −∞ e
xdx = lima→∞ − ∫^1 −a e
xdx = lima→∞ − ex|^1 −a = lima→∞ e−a^ − e = −e.
l
Describe the domain, range, and level curves of ln(x^2 + y^2 − 1). Solution: ln u is undefined for u ≤ 0, so this is only defined when x^2 +y^2 − 1 > 0, which is when x^2 + y^2 > 1. The level curves are solutions c = ln(x^2 + y^2 − 1), so x^2 + y^2 = ec^ + 1, which are circles of radius
ec^ + 1.
m
Find the following partial derivatives:
- (^) ∂x∂ (x^3 + xy + ln x)
- (^) ∂y∂ exe xy
2 ∂x∂y e
xexy
- (^) ∂y∂ ln xy
- ∂
3 ∂y∂x∂y e
x^2 y^2
- (^) ∂z∂ ln(xy + xz + yz)
Solution:
- 3x^2 + y + 1/x
- x^2 exy^ exe
xy
- 2xexy^ exe
xy
xy
xy (exy^ + xyexy^ )
- 1/y
- (^) ∂y∂ ex (^2) y 2 = 2x^2 yex (^2) y 2 , ∂
2 ∂x∂y e
x^2 y^2 = 4xyex^2 y^2 + 4x (^3) y (^3) ex^2 y^2 , ∂^3 ∂y∂x∂y e
x^2 y^2 =
4 xex
(^2) y 2
(^2) y 2
(^2) y 2
(^2) y 2
- (^) xy+xxz+y+yz
n
Indicate whether the following statements are (A)lways True, (S)ometimes True, or (N)ever True.
- A function that is continuous at (x, y) is also differentiable at (x, y)
- If f is differentiable at (x, y) then the partial derivative ∂f∂x is exists at (x, y)
- If f is differentiable and ∇f 6 = 0, ∇f is the direction in which f decreases most rapidly
- If ∂f∂x and ∂f∂y both exist at (x, y) then ∇f (x, y) is defined
- If f, fx, fy , fxy , fyx are both defined and continuous at (x, y) then the mixed partials are equal at (x, y)
Solutions:
At the origin, almost all these terms disappear, and we have
D(0, 0) = −(e^0 − 2 e^0 )^2 < 0
so the origin is a saddle point. On the hyperbola, we can simplify a bit. We have
D(x, y) = x^2 y^2 e^2 xy^ − 8 x^2 y^2 e^3 xy^ + 16x^2 y^2 e^4 xy^ − (exy^ − 2 e^2 xy^ + xyexy^ − 4 xye^2 xy^ )^2.
In particular, we never use x or y alone, just xy or x^2 y^2 = (xy)^2. Note that e−^ ln 2^ = eln 2 − 1 = 1/2 and that e^2 xy^ = (exy^ )^2 = 1/4, e^3 xy^ = 1/8 and e^4 xy^ = 1 /16. So on the hyperbola we have
D(x, y) = (ln^2 2)/ 4 −(ln^2 2)+(ln^2 2)−(1/ 2 − 1 / 2 −ln 2/2+ln 2)^2 = (ln^2 2)/4+2−(ln 2/2)^2 = 0
so the points on the hyperbola cannot be classified.
q
Find the candidates for where exy^ achieves its minimum on the circle x^2 +y^2 = 1. Solution: [yexy^ , xexy^ ]′^ = c[2x, 2 y]′, which gives the equations yexy^ = 2cx and xexy^ = 2cy. Multiplying by y and x respectively, y^2 exy^ = 2cxy = x^2 exy^. Since exy^ is never 0, x^2 = y^2 , and therefore the candidates are (± 1 /
r
Find the candidates for where exy^ achieves its minimum on the hyperbola x = 1 /y. Solution: We solve the second equation to give xy = 1. But exy^ is constantly equal to e when xy = 1, so all points are maxima and minima! (If we used Lagrange multipliers, we’d find that [yexy^ , xexy^ ]′^ = c[y, x]′, so yexy^ = cy, xexy^ = cx, and therefore exy^ = c and xy = 1. Then, wtih c = e, this is true everywhere xy = 1.)
s
Find the candidates for where x^2 + y^2 achieves its minimum on the hyperbola x = 1/y. Solution: We solve the second equation to give xy = 1. [2x, 2 y] = c[y, x] so 2 x = cy, 2 y = cx, and so 2x^2 = cxy = 2y^2 , and therefore x^2 = y^2. Since also x = 1/y, 1/y^2 = y^2 , so 1 = y^4 , so y = ±1. Since xy = 1, the candidates are (1, 1) and (− 1 , −1).
t
- Find and classify as stable or unstable the equilibria of dy dt
= (y − 3)(ey^ − e).
- y 0 is a solution with y 0 (0) = 0. What is limt→∞ y 0?
- y 1 is a solution with y 1 (0) = 1. What is limt→∞ y 1?
- y 2 is a solution with y 2 (0) = 2. What is limt→∞ y 2?
- y 3 is a solution with y 3 (0) = 3. What is limt→∞ y 3?
- y 4 is a solution with y 4 (0) = 4. What is limt→∞ y 4?
Solution:
- g(y) = (y − 3)(ey^ − e). g(y) = 0 means y = 3 or ey^ = e, which means y = 1. g′(y) = (y − 3)ey^ + ey^ − e. g′(3) = e^3 − e > 0, so 3 is an unstable equilibrium. g′(1) = − 3 e < 0, so 1 is a stable equilibrium.
- When y 0 (0) = 0, y′^ = g(0) > 0, so y 0 is increasing towards the equilibrium, so limt→∞ y 0 = 1.
- When y 1 (0) = 1, y′^ = g(1) = 0, so y 1 is constantly equal to 1, so limt→∞ y 1 = 1.
- When y 2 (0) = 2, y′^ = g(2) < 0, so y 2 is decreasing towards the equilib- rium, so limt→∞ y 2 = 1.
- When y 3 (0) = 3, y′^ = g(3) = 0, so y 3 is constantly equal to 3, so limt→∞ y 3 = 3.
- When y 4 (0) = 4, y′^ = g(4) > 0, so y 4 is increasing away from the equilib- rium, so limt→∞ y 0 = ∞.
u
Give an example of an autonomous differential equation which has x^3 as a solution. Solution: When y = x^3 , y′^ = 3x^2. We must express y′^ as a function of y, which is easily done by setting y′^ = 3y^2 /^3.
