第 12 类 RD Sharma 解 – 第 22 章微分方程 – 练习 22.11 |设置 1

问题 1. 被充气的气球的表面积以与时间 t 成正比的速率变化。如果最初它的半径是 1 个单位，3 秒后是 2 个单位，求时间 t 后的半径。

解决方案：

Let us considered radius = r

and the surface area of the balloon at a particular time ‘t’ = S

Surface area is given by,

S = 4πr²   …..(1)

We have,

(dS/dt)∝ t

(dS/dt) = kt (where k is proportional constant)

On differentiating eq(1), we get

d/dt(4πr²) = kt

8πr(dr/dt) = kt

ktdt = 8πrdr

On integrating both sides, we get

∫ktdt = ∫8πrdr

kt²/2 = 8π(r²/2) + c    …..(2)

At t = 0, r = 1 unit and at t = 3sec, r = 2units

0 = 4π + c

c = -4π

And,

k(3)²/2 = 8π(2²/2) – 4π

(9/2)k = 12π

k = (8π/3)

On putting the values of k in equation (2)

(8π/3)(t²/2) = 8π(r²/2) – 4π

4t²/3 = 4r²-4

r² = 1 + (t²/3)

$r=\sqrt{1+\frac{t^2}{3}}$

Hence, the radius after time t = $\sqrt{1+\frac{t^2}{3}}$

问题 2. 人口以每年 5% 的速度增长。人口翻倍需要多长时间？

解决方案：

Let us considered the initial population = P₀ 

and the population at a particular time ‘t’ = P’

We have,

dP/dt = 5%P

dP/dt = 5P/100 

dP/P = 0.05dt

On integrating both sides, we get

∫(dP/P) = ∫0.05dt

Log|P| = 0.05t + c

At t = 0, P = P₀ 

log|P₀| = c

Log|P| = 0.05t + log|P₀|

Log|P/P₀| = 0.05t

Now we find the time population becomes double,

P = 2P₀ 

Log|2P₀/P₀| = 0.05t

Log|2| = 0.05t

t = 20Log|2| years

问题3：人口增长率与一个城市的现有人口数量成正比，在过去的25年里，现在的人口是10万，这个城市什么时候有50万的人口？

解决方案：

Let us considered

The initial population = P₀

the population at a particular time ‘t’ = P

and the growth of population = g’

We have,

dP/dt = gP

dP/P = gdt

On integrating both sides, we get

∫(dP/P) = g∫dt

Log|P| = gt + c

At t = 0, P = P₀

log|P₀| = c

Log|P| = gt + log|P₀|

Log|P/P₀| = gt

Population of city is doubled in 25 years.

At t = 25, P = 2P₀ 

Log|2P₀/P₀| = 25g

g = Log|2|/25

g = 0.0277

Log|P/P₀| = 0.0277t

For P = 500000 and P₀ = 100000

Log|500000/100000| = 0.0277t

t = Log|5|/0.0277

t = 58.08 year

t = 58 year

问题 4. 在一个培养物中，细菌数为 100000。数量在 2 小时内增加了 10%。如果细菌的生长速度与存在的数量成正比，那么计数将在多少小时内达到 200000？

解决方案：

Let us assume the number of bacteria count at a particular time ‘t’ = P

We have,

dP/dt ∝ P

(dP/dt) = kP (where k is proportional constant)

(dP/P) = kdt

On integrating both sides, we get

∫(dP/P) = ∫kdt

Log|P| = kt + c

At t = 0, P = 100000

Log|100000| = c

Log|P| = kt + Log|100000|

After t = 2 hours number is increases by 10%.

Therefore, P = 100000 + (100000)(5/100)

P = 110000

Log|110000| = 2k + Log|100000|

k = (1/2)Log|11/10|

Log|P| = (t/2)Log|11/10| + Log|100000|      …(i)

Putting the value of k in equation (i)

Let at t = T P = 200000

Log|200000| = (T/2)Log|11/10| + Log|100000|

Log|2| = (T/2)Log|11/10|

$T=2\frac{log2}{log\frac{11}{10}}$

问题 5. 如果利息以每年 6% 的速度连续复利，10 年后 RS 1000 的价值是多少？ RS翻倍需要多长时间。 1000？

解决方案：

Let us assume be the initial amount = P₀ 

and the amount at a particular time ‘t’ = P

dP/dt = 6%P

dP/dt = 6P/100

dP/P = 0.06dt

On integrating both sides, we get

∫(dP/P) = ∫0.06dt

Log|P| = 0.06t + c

At t = 0, P = P₀

log|P₀| = c

Log|P| = 0.06t + log|P₀|

Log|P/P₀| = 0.06t

At t = 10 years find the amount

Log|P/P₀| = 0.06 × 10

Log|P/P₀| = 0.6

P/P₀ = e^0.6

P = P0 × 1.8221

P = 1000 × 1.8221

P = 1822

At what time amount becomes double,

P = 2P₀

Log|2P₀/P₀| = 0.06t

Log|2| = 0.06t

t = 16.66Log_e|2|

t = 11.55 years

问题 6. 某种细菌培养物中细菌数量的增加率与存在的数量成正比，给定 5 小时内的数量增加三倍，求 10 小时后将存在多少细菌。另外，找出细菌数量达到初始存在数量的 10 倍所需的时间。

解决方案：

Let us considered

The initial count of bacteria = P₀

the count of bacteria at a particular time ‘t’ = P

and the growth of bacteria = g times.

We have,

dP/dt ∝ P

dP/dt = gP

dP/P = gdt

On integrating both sides, we get

∫(dP/P) = g∫dt

Log|P| = gt + c

At t = 0, P = P₀

log|P₀| = c

Log|P| = gt + log|P₀|

Log|P/P₀| = gt

At t = 5 hours, P = 3P₀

Log|3P₀/P₀| = 5g

g = Log_e|3|/5

g = 0.219722

Log|P/P₀| = 0.219722t

At t = 10 hours find the numbers of bacteria.

Log|P/P₀| = 0.219722 × 10

|P/P₀| = e^2.19722

|P/P₀| = 9

P = 9P₀

At ‘T’ time, a number of bacteria become 10 times.

At t = T, P = 10P₀

Log|10P₀/P₀| = Log_e|3|/5T

Log|10| = T(Log_e|3|/5)

$T=\frac{5Log10}{Log3}$

问题 7. 城市人口的增长速度与任何时间 t 的居民人数成正比。如果这个城市的人口在 1990 年是 20 万，到 2000 年是 25 万，那么 2010 年的人口是多少？

解决方案：

Let us considered

The initial population = P₀

the population at a particular time ‘t’ = P

and the growth of population = g times.

We have,

dP/dt ∝ P

dP/dt = gP

dP/P = gdt

On integrating both sides, we get

∫(dP/P) = g∫dt

Log|P| = gt + c                …(i)

At t = 1990, P = 200000 and at t = 2000, P = 250000

Log|200000| = 1990g + c          …(ii)

Log|250000| = 2000g + c          …(iii)

On subtracting eq (iii) from (ii)

10g = Log|250000/200000|

g = (1/10)Log|5/4|

On putting the value of ‘g’ in equation (i)

Log|200000| = 1990 × (1/10)Log|5/4| + c

c = Log|200000| – 199 × Log|5/4|

Population in 2010,

Log|P| = (1/10)Log|5/4| × 2010 + Log|200000| – 199 × Log|5/4|

Log|P| = 201Log|5/4| – 199Log|5/4| + Log|200000|

Log|P| = Log|5/4|²⁰¹ – Log|5/4|¹⁹⁹ + Log|200000|

Log|P| = Log|(5/4)²⁰¹(4/5)¹⁹⁹| + log|200000|

Log|P| = Log|5/4|² + log|200000|

Log|P| = Log|(25/16)200000|

Log|P| = Log|312500|

P = 312500

问题 8. 如果制造某个项目的边际成本由 C'(x) = (dC/dx) = 2 + 0.15x 给出。给定 C(0) = 100，求总成本函数C(x)

解决方案：

We have,

dC/dx = 2 + 0.15x

dC = (2 + 0.15x)dx

On integrating both sides, we get

∫dC = ∫(2 + 0.15x)dx

C(x) = 2x + (0.15/2)x² + c₁

At C(0) = 100, we have

100 = 2(0) + (0.15/2)(0)² + c₁

c₁ = 100

C(x) = 0.075x² + 2x + 100

问题九、银行以连续复利的方式支付利息，即将利率视为本金的瞬时变化率。假设一个账户的利息以每年 8% 的速度累积，并连续复利。计算一年内此类账户的增长百分比。

解决方案：

Let us considered

The initial population = P₀

and the population at a particular time ‘t’ = P

We have,

dP/dt = (8/100)P

dP/dt = (2/25)P

dP/P = (2/25)dt

On integrating both sides, we get

∫(dP/P) = (2/25)∫dt

Log|P| = (2/25)t + c            …(i)

At t = 0, P = P₀

Log|P₀| = 0 + c

c = Log|P₀|

On putting the value of c in equation (i)

Log|P| = (2/25)t + Log|P₀|

Log|P/P₀| = (2t/25)

Amount after 1 year,

Log|P/P₀| = (2/25)

e^(2/25) = |P/P₀|

e^0.08 = |P/P₀|

1.0833 = |P/P₀|

P = 1.0833P₀ 

Percentage increase = [(P – P₀)/P₀] × 100%

= [(1.0833P₀ – P₀)/P₀] × 100%

= 0.0833 × 100%

= 8.33%

问题 10. 在一个由电阻 R、自感 L 和电压 E 组成的简单电路中，任何时候的电流 i 由 L(di/dt) + Ri = E 给出。如果 E 是恒定的并且最初没有电流通过电路，证明 i = (E/R){1 – e ^-(R/L)t }

解决方案：

We have,

L(di/dt) + Ri = E

(di/dt) + (R/L)i = E/L

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (R/L), Q = E/L

So, I.F = e^∫Pdi

= e^∫(R/L)di

= e^(R/L)t

The solution of a differential equation is,

i(I.F) = ∫Q(I.F)dt + c

e^(R/L)t × i = (E/L)∫e^(R/L)tdt + c

e^(R/L)t × i = (E/L)(L/R)e^(R/L)t + c

e^(R/L)t × i = (E/R)e^(R/L)t + c      …(i)

At t = 0, i = 0

e⁰ × 0 = (E/R)e⁰ + c

c = -(E/R)

On putting the value of c in equation (i)

e^(R/L)t × i = (E/R)e^(R/L)t – (E/R)

i = (E/R) – (E/R)e^-(R/L)t

$i=\frac{E}{R}(1-e^{-\frac{R}{L}t})$

问题 11. 半径在任何时间 t 的衰减率与其当时的质量成正比。找出质量将减半的时间。

解决方案：

Let us considered 

initial radius = R₀

and radius at a particular time ‘t’ = R 

We have,

dR/dt ∝ R

dR/dt = -kR

dR/R = -kdt

On integrating both sides, we get

∫(dR/R) = -k∫dt

Log|R| = -kt + c         …(i)

At t = 0, R = R₀

Log|R₀| = 0 + c 

c = Log|R₀|

Log|R| = -kt + Log|R₀|

kt = Log|R₀/R|

At time ‘T’ mass becomes R₀/2

$Log|\frac{R0}{\frac{R0}{2}}|=kT$

Log|2| = kT

T = (1/k)Log|2|