Chapter 8 Differential Equation and Applications Ex 8.3 Solutions

Balbharati Maharashtra State BoardStd 12 Commerce Statistics Part 1 Digest PdfChapter 8 Differential Equation and Applications Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.3

(ii)
Solution:


This is the general solution.

(iii) (x²– yx²) dy + (y²+ xy²) dx = 0
Solution:
(x²– yx²) dy + (y²+ xy²) dx = 0
∴ x²(1 – y) dy + y²(1 + x) dx = 0
∴
Integrating, we get


This is the general solution.

(iv)
Solution:


∴ 2y²log |x + 1| = 2cy²– 1 is the required solution.

(ii) (x + 1) -1 = 2e^-y, when y = 0, x = 1.
Solution:

∴ log |2 + e^y| = log |c(x + 1)|
∴ 2 + e^y= c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰= c(1 + 1)
∴ 3 = 2c
∴ c =
∴ the particular solution is
2 + e^y= (x + 1)
∴ 4 + 2e^y= 3x + 3
∴ 3x – 2e^y– 1 = 0

(iii) y(1 + log x) – x log x = 0, when x = e, y = e².
Solution:

∴ from (1), the general solution is
log |x log x| – log |y| = log c, where c₁= log c
∴ log || = log c
∴ = c
∴ x log x = cy
This is the general solution.
Now, y = e², when x = e
e log e = ce²
1 = ce ……[∵ log e = 1]
c =
∴ the particular solution is x log x = () y
∴ y = e^xlog x

(iv) = 4x + y + 1, when y = 1, x = 0.
Solution:
= 4x + y + 1
Put 4x + y + 1 = v

∴ log |v + 4| = x + c
∴ log |4x + y + 1 + 4| = x + c
i.e. log |4x + y + 5| = x + c
This is the general solution.
Now, y = 1 when x = 0
∴ log|0 + 1 + 5| = 0 + c,
i.e. c = log 6
∴ the particular solution is
log |4x + y + 5| = x + log 6
∴ = x