Checkpoint
5 5
y = 2 x 2 + 3 x + 2 y = 2 x 2 + 3 x + 2
y = 1 3 x 3 − 2 x 2 + 3 x − 6 e x + 14 y = 1 3 x 3 − 2 x 2 + 3 x − 6 e x + 14
v ( t ) = −9.8 t v ( t ) = −9.8 t
The equilibrium solutions are y=−2y=−2 and y=2.y=2. For this equation, y=−2y=−2 is an unstable equilibrium solution, and y=2y=2 is a semi-stable equilibrium solution.
row: nn | xnxn | yn=yn−1+hf(xn−1,yn−1)yn=yn−1+hf(xn−1,yn−1)
row: 00 | 11 | −2−2
row: 11 | 1.11.1 | y1=y0+hf(x0,y0)=−1.5y1=y0+hf(x0,y0)=−1.5
row: 22 | 1.21.2 | y2=y1+hf(x1,y1)=−1.1419y2=y1+hf(x1,y1)=−1.1419
row: 33 | 1.31.3 | y3=y2+hf(x2,y2)=−0.8387y3=y2+hf(x2,y2)=−0.8387
row: 44 | 1.41.4 | y4=y3+hf(x3,y3)=−0.5487y4=y3+hf(x3,y3)=−0.5487
row: 55 | 1.51.5 | y5=y4+hf(x4,y4)=−0.2442y5=y4+hf(x4,y4)=−0.2442
row: 66 | 1.61.6 | y6=y5+hf(x5,y5)=0.0993y6=y5+hf(x5,y5)=0.0993
row: 77 | 1.71.7 | y7=y6+hf(x6,y6)=0.5099y7=y6+hf(x6,y6)=0.5099
row: 88 | 1.81.8 | y8=y7+hf(x7,y7)=1.0272y8=y7+hf(x7,y7)=1.0272
row: 99 | 1.91.9 | y9=y8+hf(x8,y8)=1.7159y9=y8+hf(x8,y8)=1.7159
row: 1010 | 22 | y10=y9+hf(x9,y9)=2.6962y10=y9+hf(x9,y9)=2.6962
y = 2 + C e x 2 + 3 x y = 2 + C e x 2 + 3 x
y = 4 + 14 e x 2 + x 1 − 7 e x 2 + x y = 4 + 14 e x 2 + x 1 − 7 e x 2 + x
Initial value problem:
d u d t = 2.4 − 2 u 25 , u ( 0 ) = 3 d u d t = 2.4 − 2 u 25 , u ( 0 ) = 3
Solution:u(t)=30−27e−2t/25Solution:u(t)=30−27e−2t/25 Concentration: 30-27e-2t25Concentration: 30-27e-2t25
Initial value problem dTdt=k(T−70),T(0)=450dTdt=k(T−70),T(0)=450
T(t)=70+380ektT(t)=70+380ekt
Approximately 114114 minutes.
dPdt=0.04(1−P750),P(0)=200dPdt=0.04(1−P750),P(0)=200
P(t)=3000e.04t11+4e.04tP(t)=3000e.04t11+4e.04t
After 1212 months, the population will be P(12)≈278P(12)≈278 rabbits.
y′+15x+3y=10x−20x+3;p(x)=15x+3y′+15x+3y=10x−20x+3;p(x)=15x+3 and q(x)=10x−20x+3q(x)=10x−20x+3
y = x 3 + x 2 + C x − 2 y = x 3 + x 2 + C x − 2
y = - 2 x - 5 2 + 1 2 e 2 x y = - 2 x - 5 2 + 1 2 e 2 x
dvdt=−v−9.8v(0)=0dvdt=−v−9.8v(0)=0
v(t)=9.8(e−t−1)v(t)=9.8(e−t−1)
limt→∞v(t)=limt→∞(9.8(e−t−1))=−9.8m/s≈−21.922mphlimt→∞v(t)=limt→∞(9.8(e−t−1))=−9.8m/s≈−21.922mph
Initial-value problem:
8 q ′ + 1 0.02 q = 20 sin 5 t , q ( 0 ) = 4 8 q ′ + 1 0.02 q = 20 sin 5 t , q ( 0 ) = 4
q ( t ) = 10 sin 5 t − 8 cos 5 t + 172 e −6.25 t 41 q ( t ) = 10 sin 5 t − 8 cos 5 t + 172 e −6.25 t 41
Section 4.1 Exercises
1 1
3 3
1 1
1 1
y = 4 + 3 x 4 4 y = 4 + 3 x 4 4
y = 1 2 e x 2 y = 1 2 e x 2
y = 2 e − 1 / x y = 2 e − 1 / x
u = sin −1 ( e −1 + t ) u = sin −1 ( e −1 + t )
y = − x + 1 1 − x − 1 y = − x + 1 1 − x − 1
y = C − x + x ln x − ln ( cos x ) y = C − x + x ln x − ln ( cos x )
y = C + 4 x ln ( 4 ) y = C + 4 x ln ( 4 )
y = 2 3 t 2 + 16 ( t 2 + 16 ) + C y = 2 3 t 2 + 16 ( t 2 + 16 ) + C
x = 2 15 4 + t ( 3 t 2 + 4 t − 32 ) + C x = 2 15 4 + t ( 3 t 2 + 4 t − 32 ) + C
y = C x y = C x
y = 1 − t 2 2 , y = − t 2 2 − 1 y = 1 − t 2 2 , y = − t 2 2 − 1
y = e − t , y = − e − t y = e − t , y = − e − t
y = 2 ( t 2 + 5 ) , t = 3 5 y = 2 ( t 2 + 5 ) , t = 3 5
y = 10 e −2 t , t = − 1 2 ln ( 1 10 ) y = 10 e −2 t , t = − 1 2 ln ( 1 10 )
y=14(41−e−4t),y=14(41−e−4t), never
Solution changes from increasing to decreasing at y(0)=0y(0)=0
Solution changes from increasing to decreasing at y(0)=0y(0)=0
v ( t ) = −32 t + a v ( t ) = −32 t + a
00 ft/s
52.35452.354 meters
x=50t−15π2cos(πt)+3π2,2x=50t−15π2cos(πt)+3π2,2 hours 11 minute
y = 4 e 3 t y = 4 e 3 t
y = 3 − 2 t + t 2 y = 3 − 2 t + t 2
y=1k(ekt−1)y=1k(ekt−1) and y=xy=x
Section 4.2 Exercises
y=0y=0 is a stable equilibrium
y=0y=0 is a stable equilibrium and y=2y=2 is unstable
General solution is y=et+Cy=et+C.
General solution is y=et(t-1)+Cy=et(t-1)+C.
ਈ
ਏ
ਬੀ
ਏ
ਸੀ
੨੨
੫(e−1)੫(e−1)
੦੦
੧ln(੨)੧ln(੨)
−੧/੨−੧/੨
y ′ = 2 e t 2 / 2
੨
੩.੨੭੫੬
੨e
ਕਤਾਰ: ਕਦਮ ਦਾ ਆਕਾਰ | ਸੰਬੰਧਿਤ ਤਰੁੱਟੀ
ਕਤਾਰ: h=੦.੧ | ੦.੩੯੩੫
ਕਤਾਰ: h=੦.੦੧ | ੦.੦੬੧੬੩
ਕਤਾਰ: h=੦.੦੦੧ | ੦.੦੦੬੬੧੨
ਕਤਾਰ: h=੦.੦੦੦੧ | ੦.੦੦੦੬੬੬੧
੪.੦੭੪੧ e −੧੦
ਧਾਰਾ ੪.੩ ਅਭਿਆਸ
y = e t − ੧
y = ੧ + C e − t
y = C x e −੧ / x
y = 1 C − x 2 y = 1 C − x 2
y = − 2 C + ln x y = − 2 C + ln x
y = C e x ( x + 1 ) + 1 y = C e x ( x + 1 ) + 1
y = sin ( ln t + C ) y = sin ( ln t + C )
y = − ln ( e − x ) y = − ln ( e − x )
y = 1 2 − e x 2 y = 1 2 − e x 2
y = tanh −1 ( x 2 2 ) y = tanh −1 ( x 2 2 )
x = sin ( 1 - t + t ln t ) x = sin ( 1 - t + t ln t )
y = ln ( ln ( 5 ) ) − ln ( 2 − 5 x ) y = ln ( ln ( 5 ) ) − ln ( 2 − 5 x )
y=Ce−2x+12y=Ce−2x+12
y=12C−exy=12C−ex
y=Ce−xxxy=Ce−xxx
y = r d ( 1 − e − d t ) y = r d ( 1 − e − d t )
y ( t ) = 10 − 9 e − x / 50 y ( t ) = 10 − 9 e − x / 50
134.3134.3 kilograms
720720 seconds
2424 hours 5757 minutes
T ( t ) = 20 + 50 e −0.125 t T ( t ) = 20 + 50 e −0.125 t
T ( t ) = 20 + 38.5 e −0.125 t T ( t ) = 20 + 38.5 e −0.125 t
y = ( c + b a ) e a x − b a y = ( c + b a ) e a x − b a
y ( t ) = c L + ( I − c L ) e − r t / L y ( t ) = c L + ( I − c L ) e − r t / L
y=40(1−e−0.1t),40y=40(1−e−0.1t),40 g/cm2
Section 4.4 Exercises
P=0P=0 semi-stable
P = 10 e 10 x e 10 x + 4 P = 10 e 10 x e 10 x + 4
P ( t ) = 10000 e 0.02 t 150 + 50 e 0.02 t P ( t ) = 10000 e 0.02 t 150 + 50 e 0.02 t
6969 hours 55 minutes
88 years 1111 months
P1P1 semi-stable
P2>0P2>0 stable
P1=0P1=0 is semi-stable
P t = 3500 4 + 3 e - 035 t P t = 3500 4 + 3 e - 035 t
P ( t ) = 850 + 500 e 0.009 t 85 + 5 e 0.009 t P ( t ) = 850 + 500 e 0.009 t 85 + 5 e 0.009 t
1313 years months
31.46531.465 days
September 20082008
K + T 2 K + T 2
r = 0.0405 r = 0.0405
α = 0.0081 α = 0.0081
Logistic: 361,361, Threshold: 436,436, Gompertz: 309.309.
Section 4.5 Exercises
Yes
Yes
y ′ − x 3 y = sin x y ′ − x 3 y = sin x
y ′ + ( 3 x + 2 ) x y = − e x y ′ + ( 3 x + 2 ) x y = − e x
d y d t − y x ( x + 1 ) = 0 d y d t − y x ( x + 1 ) = 0
e e x e e x
− ln ( cosh x ) − ln ( cosh x )
y = C e 3 x − 2 3 y = C e 3 x − 2 3
y = C x 3 + 6 x 2 y = C x 3 + 6 x 2
y = C e x 2 / 2 − 3 y = C e x 2 / 2 − 3
y = C tan ( x 2 ) − 2 x + 4 tan ( x 2 ) ln ( sin ( x 2 ) ) y = C tan ( x 2 ) − 2 x + 4 tan ( x 2 ) ln ( sin ( x 2 ) )
y = C x 3 − x 2 y = C x 3 − x 2
y = C ( x + 2 ) 2 + 1 2 y = C ( x + 2 ) 2 + 1 2
y = C x + 2 sin ( 3 t ) y = C x + 2 sin ( 3 t )
y = C ( x + 1 ) 3 − x 2 − 2 x − 1 y = C ( x + 1 ) 3 − x 2 − 2 x − 1
y = C e sinh −1 x − 2 y = C e sinh −1 x − 2
y = x + 4 e –x − 1 y = x + 4 e –x − 1
y = − 3 x 2 ( x 2 − 1 ) y = − 3 x 2 ( x 2 − 1 )
y = 1 − e tan −1 x y = 1 − e tan −1 x
y = ( x + 2 ) ln ( x + 2 2 ) y = ( x + 2 ) ln ( x + 2 2 )
y = 2 e 2 x − 2 x − 2 x − 1 y = 2 e 2 x − 2 x − 2 x − 1
v ( t ) = g m k ( 1 − e − k t / m ) v ( t ) = g m k ( 1 − e − k t / m )
40.45140.451 seconds
g m k g m k
y = C e x − a ( x + 1 ) y = C e x − a ( x + 1 )
y = C e x 2 / 2 − a y = C e x 2 / 2 − a
y = e k t − e t k − 1 y = e k t − e t k − 1
Review Exercises
F
T
y ( x ) = 2 x ln ( 2 ) + x cos −1 x − 1 − x 2 + C y ( x ) = 2 x ln ( 2 ) + x cos −1 x − 1 − x 2 + C
y ( x ) = ln ( C − cos x ) y ( x ) = ln ( C − cos x )
y ( x ) = e e C + x y ( x ) = e e C + x
y ( x ) = 4 + 3 2 x 2 + 2 x − sin x y ( x ) = 4 + 3 2 x 2 + 2 x − sin x
y ( x ) = − 2 1 + 3 ( x 2 + 2 sin x ) y ( x ) = − 2 1 + 3 ( x 2 + 2 sin x )
y ( x ) = −2 x 2 − 2 x − 1 3 − 2 3 e 3 x y ( x ) = −2 x 2 − 2 x − 1 3 − 2 3 e 3 x
y(x)=Ce−x+lnxy(x)=Ce−x+lnx
Euler: 0.6939,0.6939, exact solution: y(x)=3x−e−2x2+ln(3)y(x)=3x−e−2x2+ln(3)
40494049 second
x(t)=5000+2459−493t−2459e−5/3t,t=307.8x(t)=5000+2459−493t−2459e−5/3t,t=307.8 seconds
T ( t ) = 200 ( 1 − e − t / 1000 ) T ( t ) = 200 ( 1 − e − t / 1000 )
P ( t ) = 1600000 e 0.02 t 9840 + 160 e 0.02 t P ( t ) = 1600000 e 0.02 t 9840 + 160 e 0.02 t