Maharashtra Board Text books

Chapter 6 Differential Equations Miscellaneous Exercise 6

(I) Choose the correct option from the given alternatives:

Question 1.
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respectively……..
(a) 2, 1
(b) 1, 2
(c) 3, 2
(d) 2, 3
Answer:
(d) 2, 3

Question 2.

Question 3.

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Question 4.
The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is
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Question 5.

(a) circles
(b) parabolas
(c) ellipses
(d) hyperbolas
Answer:
(a) circles

Hint:
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Question 6.
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Question 7.
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Question 8.
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Question 9.
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Question 10.
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Question 11.
The solution of the differential equation dy/dx= sec x – y tan x is…….

(a) y sec x + tan x = c
(b) y sec x = tan x + c
(c) sec x + y tan x = c
(d) sec x = y tan x + c
Answer:
(b) y sec x = tan x + c

Hint:
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Question 13.
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Question 14.
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……

Question 15.
If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..

(a) 51.7°C
(b) 54.7°C
(c) 52.7°C
(d) 50.7°C
Answer:
(b) 54.7°C

(II) Solve the following:

Question 1.

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Solution:
The given D.E. is


Differentiating both sides w.r.t. x, we get


Solution:

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Solution:

y = 3 cos(log x) + 4 sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get
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Solution:
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Differentiating both sides w.r.t. y, we get
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Question 3.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:

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Differentiating again w.r.t. x, we get
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This is the required D.E.

(ii) y = a sin(x + b)
Solution:

y = a sin(x + b)
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This is the required D.E.

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Differentiating both sides w.r.t. x, we get
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Differentiating twice w.r.t. x, we get
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This is the required D.E.

Question 4.
Form the differential equation of:
(i) all circles which pass through the origin and whose centres lie on X-axis.
Solution:

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Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.
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This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.
Solution:

Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.
Then equation of the parabola is
(x – h = 4b(y – k) ……. (1)
where h and k are arbitrary constants.
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Differentiating both sides of (1) w.r.t. x, we get
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This is the required D.E.

(iii) an ellipse whose major axis is twice its minor axis.
Solution:

Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
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This is the required D.E.

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∴ a = 4√k, b = 6√k
∴ l(transverse axis) = 2a = 8√k
and l(conjugate axis) = 2b = 12√k
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = 4√k and 2B = b = 6√k
∴ A = 2√k and B = 3√k
∴ equation of the required hyperbola is
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This is the required D.E.

Question 5.
Solve the following differential equations:

Solution:

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (i)


Solution:


Solution:
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(iv) x dy = (x + y + 1) dx
Solution:
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This is the general solution.


Solution:
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Solution:
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Question 6.
Find the particular solution of the following differential equations:

Solution:

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (i)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (i).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (i).2


Solution:
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This is the general solution.
When x = 2, y = 1, we have
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This is the linear differential equation of the form
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(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y
Solution:
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Solution:
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Question 7.

Solution:

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Question 8.
The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Solution:

Let P(x, y) be a point on the curve y = f(x).
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∴ equation of the normal is
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This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
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Question 9.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Solution:
Let r be the radius and V be the volume of the spherical balloon at any time t.
Then the rate of change in volume of the spherical balloon is  which is a constant.
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Hence, the radius of the spherical balloon after t seconds is (63t+27 units.

Question 10.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in  years from the start.
Solution:

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Integrating both sides, we get
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