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This set of practice questions focuses on first-order differential equations, covering various types including separable equations, homogeneous equations, and linear equations. Students will practice identifying the type of differential equation, solving using appropriate methods, and verifying solutions. These questions aim to reinforce understanding of key concepts in differential equations and their applications.
Solution:
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)
Hence, the radius after time t =
Solution:
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
Solution:
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
Solution:
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|
Solution:
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 = e0.6
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 = 16.66Loge|2|
t = 11.55 years
Solution:
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 = Loge|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| = e2.19722
|P/P0| = 9
P = 9P0
At 'T' time, a number of bacteria become 10 times.
At t = T, P = 10P0
Log|10P0/P0| = Loge|3|/5T
Log|10| = T(Loge|3|/5)
Solution:
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
Solution:
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
Solution:
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|
e0.08 = |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%
Solution:
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
Solution:
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|2| = kT
T = (1/k)Log|2|
Related Articles:
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.6
Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.7
Exercise 22.11 | Set 1 in RD Sharma's Class 12 Differential Equations chapter focuses on solving first-order differential equations, particularly exact differential equations and those that can be made exact using integrating factors. This exercise introduces students to the concept of exactness in differential equations, methods to verify exactness, and techniques to solve exact equations. It also covers the process of finding integrating factors to transform non-exact equations into exact ones. The problems in this set range from straightforward exact equations to more complex scenarios requiring integrating factors, providing a comprehensive practice in handling various forms of first-order differential equations.
1. Verify that the following differential equation is exact and solve it:
(2x + y) dx + (x - 3y) dy = 0
2. Find the integrating factor and solve:
(x² + xy) dx + (x² - y²) dy = 0
3. Solve the exact differential equation:
(y² cos x + 2xy) dx + (x² + 2y sin x) dy = 0
4. Show that the following equation is not exact. Find an integrating factor and solve:
(3x² + 2xy) dx + (x² + y²) dy = 0
5. Verify the exactness and solve:
(e^y sin x + y) dx + (e^y cos x + x) dy = 0
6. Find the general solution of:
(2xy + y²) dx + (x² + 2xy) dy = 0
7. Solve the differential equation:
(x² + y²) dx - xy dy = 0
8. Find the integrating factor and solve:
y dx - x dy = x√(x² + y²) dy
9. Verify that the following is an exact differential equation and solve it:
(3x² + 4xy) dx + (2x² + 3y²) dy = 0
10. Solve the equation after finding a suitable integrating factor:
(2x - y) dx + (2y - x) dy = 0
