 |
VOOZH | about |
|
VOOZH | about |
Exercise 22.11 Set 3 in RD Sharma's Class 12 Mathematics textbook continues to explore the method of variation of parameters for solving non-homogeneous second-order linear differential equations. This set likely presents more challenging problems, building upon the skills developed in Sets 1 and 2. Students will encounter a diverse range of right-hand side functions, including combinations of trigonometric, exponential, and polynomial terms.
These problems are designed to enhance students' proficiency in applying the variation of parameters technique and to deepen their understanding of its versatility in solving complex differential equations. As students progress through this set, they'll gain valuable experience in handling increasingly sophisticated equations, preparing them for advanced applications in physics, engineering, and other scientific fields.
Solution:
Slope is given by, (dy/dx)
We have,
(dy/dx) = x + xy
(dy/dx) = x(y + 1)
dy/(y + 1) = xdx
On integrating both sides, we get
∫dy/(y + 1) = ∫xdx
Log|y + 1| = (x2/2) + c ...(i)
Since the curve is passing through (0, 1)
Log|2| = c
Log|y + 1| = (x2/2) + Log|2|
Log|(y + 1)/2| = x2/2
Solution:
Tangent of the curve is given by,
Y - y = (dy/dx)(X - x)
If P be perpendicular from the origin, then
x2(dy/dx)2 - 2xy(dy/dx) + y2 = x2 + x2(dy/dx)2
y2 - 2xy(dy/dx) - x2 = 0
Hence Proved.
Now we find the curve
So, we have y2 - 2xy(dy/dx) - x2 = 0
(dy/dx) = (y2 - x2)/2xy
Above equation is homogeneous equation.
Let, y = vx
On differentiating both sides we have,
(dy/dx) = v + x(dv/dx)
v + x(dv/dx) = (v2x2 - x2)/2xvx
v + x(dv/dx) = (v2 - 1)/2v
x(dv/dx) = [(v2 - 1)/2v] - v
x(dv/dx) = (v2 - 1 - 2v2)/2v
x(dv/dx) = -(1 + v2)/2v
2vdv/(1 + v2) = -(dx/x)
On integrating both sides
∫2vdv/(1 + v2) = -∫(dx/x)
log|1 + v2| = -log|x| + log|c|
log|x(1 + v2)| = log|c|
x(1 + y2/x2) = c
(x2 + y2) = cx
Solution:
Let us considered
the point of contact of tangent = P(x, y)
and the curve is y = f(x).
The equation of tangent of the curve is given by,
Y - y = (dy/dx)(X - x)
Where (X, Y) is arbitrary point on the tangent.
Putting Y = 0,
0 - y = (dy/dx)(X - x)
(X - x) = -y(dx/dy)
X = x - y(dx/dy)
Coordinates at contact of x-axis = [x - y(dx/dy), 0]
The distance between the foot of the ordinate of the point of
contact and the point of intersection of the tangent with x-axis is equal to 2x.
y(dx/dy) = 2x
(dx/x) = 2(dy/y)
On integrating both sides
∫(dx/x) = 2∫(dy/y)
log|x| = 2log|y| + log|c| ...(i)
Curve is passing through (1, 2)
log|1| = 2log|2| + log|c|
log|c| = -2log|2|
On putting the value of log|c| in equation (i)
log|x| = 2log|y| - 2log|2|
log|x| = log|y2/4|
x = (y2/4)
y2 = 4x
Solution:
Let us considered the point on the curve = P(x, y).
The equation of tangent of the curve is given by,
Y - y = -(dx/dy)(X - x) ...(i)
It passes through (3, 0) So,
0 - y = -(dx/dy)(3 - x)
ydy = 3dx - xdx
On integrating both sides
∫ydy = 3∫dx - ∫xdx
(y2/2) = 3x - (x2/2) + c
It passes through (3, 4)
(16/2) = 9 - (9/2) + c ...(ii)
c = (16/2) - (9/2)
c = (7/2)
(y2/2) = 3x - (x2/2) + (7/2)
or
y2 = 6x - x2 + 7
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
∫(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
Count of bacteria becomes doubled in 6 hours.
At t = 6, P = 2P0
Log|2P0/P0| = 6g
g = Log|2|/6
Log|P/P0| = [Log|2|/6] × t
After t = 18 hours count of bacteria is equal to
Log|P/P0| = [Log|2|/6] × 18
Log|P/P0| = 3Log|2|
Log|P/P0| = Log|2|3
(P/P0) = 8
P = 8P0
Hence proved
Solution:
Let us considered
the original amount of radium = P0
and the amount of radium 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
∫(dP/P) = -∫kdt
Log|P| = -kt + c ...(i)
At t = 0, P = P0
Log|P0| = 0 + c
c = log|P0|
Log|P| = -kt + Log|P0|
Log|P/P0| = -kt ...(ii)
According to the question,
In 25 years bacteria decomposes 1.1%.
So, P = (100 - 1.1)%P0
P = 0.989P0
Log|P/P0| = -kt
Log|0.989| = -25k
k = -(1/25)Log|0.989|
On putting the value of k in equation (ii)
Log|P/P0| = (1/25)Log|0.989| × t
Time 'T' for one-half of the original amount of radium(i.e., P0 = P/2)
Log|P0/(P0/2)| = (1/25)Log|0.989| × t
Log|2| = (1/25)Log|0.989| × t
t = (25Log|2|/Log|0.989|)
t = (25×0.69311)/(0.01106)
t = 1566.70
t = 1567 years
Solution:
We have,
(dy/dx) = (x2 + y2)/2xy
The given equation is a homogeneous equation,
So, let us considered, y = vx
On differentiating both sides we have,
(dy/dx) = v + x(dv/dx)
(dy/dx) = (x2 + v2x2)/2xvx
v + x(dv/dx) = (x2 + v2x2)/2xvx
v + x(dv/dx) = (1 + v2)/2v
x(dv/dx) = [(1 + v2)/2v] - v
x(dv/dx) = (1 + v2 - 2v2)/2v
x(dv/dx) = (1 - v2)/2v
2vdv/(1 - v2) = (dx/x)
On integrating both sides
∫2vdv/(1 - v2) = ∫(dx/x)
-log|1 - v2| = log|x| - log|c|
-log|(1 - v2)| = -[log|c| - log|x|]
-log|(1 - v2)| = -log|c/x|
(1 - v2) = c/x
(1 - y2/x2) = c/x
(x2 - y2)/x2 = c/x
(x2 - y2) = cx
This is the required equation of a rectangular hyperbola.
Solution:
The equation of tangent of the curve is given by,
(dy/dx) = x + y
(dy/dx) - y = x
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = -1, Q = -x
So, I.F = e∫Pdx
= e∫-dx
= e-x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx+c
y(e-x) = ∫(x).(e-x)dy+c
y(e-x) = x∫e-xdx-∫{(dx/dx)∫e-xdx}dx+c
y(e-x) = -xe-xx + ∫e-x + c
y(e-x) = -xe-x - e-x + c
y(e-x) = -e-x(x + 1) + c (i)
Since the curve is passing through origin. So,
0 = -e-0(1) + c
c = 1
On putting the value of c in equation (i)
y(e-x) = -e-x(x + 1) + c
(x + y + 1) = c.ex
Put c = 1
x + y + 1 = ex
Solution:
The equation of tangent of the curve is given by
(dy/dx) = x + xy
(dy/dx) - xy = x
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = -x, Q = x
So, I.F = e∫Pdx
= e-∫xdx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
Let,
Let,x2/2 = z
On differentiating both sides,
xdx = dz
I = ∫e-zdz
I = -e-z
So,
Curve is passing through point (0, 1).
1e0 = -e0 + c
c = 2
Solution:
According to the question,
(dy/dx) = x2
dy = x2dx
On integrating both sides
∫dy = ∫x2dx
y = (x3/3) + c ...(i)
Curve is passing through the point (-1, 1)
1 = -(1/3) + c
c = (4/3)
On putting the value of c in equation (i)
y = (x3/3) + (4/3)
3y = x3 + 4
Solution:
According to the question,
y(dy/dx) = x
dy = x2dx
On integrating both sides
∫ydy = ∫xdx
(y2/2) = (x2/2) + c ...(i)
Curve is passing through the point (0, a)
(a2/2) = c
c = a2/2
On putting the value of c in equation (i)
(y2/2) = (x2/2) + (a2/2)
x2 - y2 = -a2
Solution:
Slope at any point is given by P = (dy/dx)
According to the question,
Slope at any point is equal to ordinate
We have,
(dy/dx) = y
dy/y = dx
On integrating both sides
∫(dy/y) = ∫(dx)
log|y| = x + log|c|
log|y| = log|ex| + log|c|
y = c.ex ...(i)
Curve is passing through the point (1, 1)
1 = c.e
c = e-1
On putting the value of c in equation (i)
y = ex.e-1
y = e(x-1)
Exercise 22.11 | Set 3 of RD Sharma's Class 12 Differential Equations chapter delves into advanced techniques for solving various types of first-order differential equations. This set focuses on applying methods such as integrating factors, exact equations, and Bernoulli's equation to solve complex problems. Students learn to identify the type of differential equation and choose the appropriate solution method. The exercise emphasizes the importance of understanding the relationship between the equation's form and its solution technique. It also introduces practical applications, including orthogonal trajectories, which have significant relevance in fields like physics and engineering. By mastering these concepts, students develop a strong foundation for tackling more complex differential equations in higher mathematics and real-world problem-solving scenarios.
