12th Commerce Maths 1 Chapter 8 Exercise 8.2 Answers Maharashtra Board

Differential Equation and Applications Class 12 Commerce Maths 1 Chapter 8 Exercise 8.2 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.

Std 12 Maths 1 Exercise 8.2 Solutions Commerce Maths

Question 1.
Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) y = Ae 3x + Be -3x
Solution:
y = Ae 3x + Be -3x ……(1)
Differentiating twice w.r.t. x, we get
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1i
This is the required D.E.

(ii) y = \(c_{2}+\frac{c_{1}}{x}\)
Solution:
y = \(c_{2}+\frac{c_{1}}{x}\)
∴ xy = c 2 x + c 1
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1ii

Maharashtra-Board-Solutions

(iii) y = (c 1 + c 2 x) e x
Solution:
y = (c 1 + c 2 x) e x
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1iii
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1iii.1
This is the required D.E.

(iv) y = c 1 e 3x + c 2 e 2x
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1iv
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1iv.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1iv.2
This is the required D.E.

(v) y 2 = (x + c) 3
Solution:
y 2 = (x + c) 3
Differentiating w.r.t. x, we get
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q1v
This is the required D.E.

Maharashtra-Board-Solutions

Question 2.
Find the differential equation by eliminating arbitrary constant from the relation x 2 + y 2 = 2ax.
Solution:
x 2 + y 2 = 2ax
Differentiating both sides w.r.t. x, we get
2x + 2y\(\frac{d y}{d x}\) = 2a
Substituting value of 2a in equation (1), we get
x 2 + y 2 = [2x + 2y \(\frac{d y}{d x}\)]x = 2x 2 + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) = y 2 – x 2 is the required D.E.

Question 3.
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution:
bx + ay = ab
∴ ay = -bx + ab
∴ y = \(-\frac{b}{a} x+b\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{b}{a} \times 1+0=-\frac{b}{a}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 0 is the required D.E.

Question 4.
Find the differential equation whose general solution is x 3 + y 3 = 35ax.
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.2-Q4

Maharashtra-Board-Solutions

Question 5.
Form the differential equation from the relation x 2 + 4y 2 = 4b 2 .
Sol ution:
x 2 + 4y 2 = 4b 2
Differentiating w.r.t. x, we get
2x + 4(2y\(\frac{d y}{d x}\)) = 0
i.e. x + 4y\(\frac{d y}{d x}\) = 0 is the required D.E.