Balbharati Maharashtra State BoardStd 12 Commerce Statistics Part 1 Digest PdfChapter 8 Differential Equation and Applications Miscellaneous Exercise 8 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8
(I) Choose the correct option from the given alternatives:
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(a) 3, 1
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(c) 3, 3
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(b)
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(a)
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(d) y – x = c
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(c) x3 + y3 = c
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(d) y = eax
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(b) 6 hours
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(c) ex
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(a) ye-x = x + c
(II) Fill in the blanks:
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order
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degree
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Answer: particular
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positive
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e-x
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(III) State whether each of the following is True or False:
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True
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True
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True
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False
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False
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True
(IV) Solve the following:
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The given differential equation is ∴ This D.E. has highest order derivative with power 3 ∴ order = 3 and degree = 3
(ii)
Solution:
The given differential equation is
This D.E. has highest order derivative with power 2.
∴ order = 1, degree = 2.
Solution & Step-by-Step Answer:
y = log x + c Differentiating both sides w.r.t. x, we get ∴ x = 1 Differentiating again w.r.t. x, we get ∴ This shows that y = log x + c is a solution of the D.E.
Solution & Step-by-Step Answer:
= 1 + x + y + xy ∴ = (1 + x) + y(1 + x) = (1 + x)(1 + y) ∴ dy = (1 + x) dx Integrating, we get ∫ dy = ∫(1 + x) dx ∴ log|1 + y| = x + + c This is the general solution.
(ii)
Solution:
∴ from (1), the general solution is
y = x log x – x + c, i.e. y = x(log x – 1) + c.

(iii) dr = ar dθ – θ dr
Solution:
dr = ar dθ – θ dr
∴ dr + θ dr = ar dθ
∴ (1 + θ) dr = ar dθ
∴
On integrating, we get
∴ log |r| = a log |1 + θ| + c
This is the general solution.
(iv) Find the differential equation of the family of curves y = ex(ax + bx2), where a and b are arbitrary constants.
Solution:
y = ex(ax + bx2)
ax + bx2= ye-x…….(1)
Differentiating (1) w.r.t. x twice and writing as y1and as y2, we get
This is the required differential equation.


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y dx – x dy = -log x dx ∴ y dx – x dy + log x dx = 0 ∴ x dy = (y + log x) dx ∴ ∴ …….(1) This is the linear differential equation of the form This is the general solution.


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……..(1) This is a linear differential equation of the form This is the general solution.

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Let P be the population at time t years. Then the rate of growth of the population is which is proportional to P. ∴ ∝ P ∴ = kP, where k is a constant ∴ = k dt On integrating, we get ∴ log P = kt + c The population doubled in 25 years and present population is 1,00,000. ∴ initial population was 50,000 i.e. when t = 0, P = 50000 ∴ log 50000 = k × 0 + c ∴ c = log 50000 ∴ log P = kt + log 50000 When t = 25, P = 100000 ∴ log 100000 = k × 25 + log 50000 ∴ 25k = log 100000 – log 50000 = log() ∴ k = log 2 ∴ log P = log 2 + log 50000 If P = 400000, then log 400000 = log 2 + log 50000 ∴ log 400000 – log 50000 = log 2 ∴ log() = ∴ log 8 = ∴ 8 = ∴ = (2)3 ∴ = 3 ∴ t = 75 ∴ the population will be 400000 in (75 – 25) = 50 years.
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Let V be the value of the machine after x years. Then rate of change of the value is which is 2200(x – 10) ∴ = 2200(x – 10) ∴ dV = 2200(x – 10) dx On integrating, we get ∫dV = 2200∫(x – 10) dx ∴ V = 2200[ – 10x] + c Initially, i.e. at x = 0, V = 120000 ∴ 120000 = 2200 × 0 + c = c ∴ c = 120000 ∴ V = 2200[ – 10x] + 120000 …….(1) This gives value of the machine in terms of initial value and age x. We have to find V when x = 10. When x = 10, from (1) V = 2200[ – 100] + 120000 = 2200 [-50] + 120000 = -110000 + 120000 = 10000 Hence, the value of the machine after 10 years will be ₹ 10000.
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(x + 2y3) = y ∴ x + 2y3 = y This is the general solution.

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y dx – x dy + log x dx = 0 ∴ (y + log x) dx = x dy This is the general solution.


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= log x dx ∴ dy = log x dx On integrating, we get ∫dy = ∫log x. 1 dx ∴ y = (log x) ∫1 dx – ∴ y = (log x). x – ∴ y = x log x – ∫1 dx ∴ y = x log x – x + c This is the general solution.
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