12th Commerce Maths 1 Chapter 8 Exercise 8.4 Answers Maharashtra Board

Differential Equation and Applications Class 12 Commerce Maths 1 Chapter 8 Exercise 8.4 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.4 Questions and Answers.

Std 12 Maths 1 Exercise 8.4 Solutions Commerce Maths

Solve the following differential equations:

Question 1.
x dx + 2y dy = 0
Solution:
x dx + 2y dy = 0
Integrating, we get
∫x dx + 2 ∫y dy = c 1
∴ \(\frac{x^{2}}{2}+2\left(\frac{y^{2}}{2}\right)=c_{1}\)
∴ x 2 + 2y 2 = c, where c = 2c 1
This is the general solution.

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Question 2.
y 2 dx + (xy + x 2 ) dy = 0
Solution:
y 2 dx + (xy + x 2 ) dy = 0
∴ (xy + x 2 ) dy = -y 2 dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q2
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q2.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q2.2
This is the general solution.

Question 3.
x 2 y dx – (x 3 + y 3 ) dy = 0
Solution:
x 2 y dx – (x 3 + y 3 ) dy = 0
∴ (x 3 + y 3 ) dy = x 2 y dx
∴ \(\frac{d y}{d x}=\frac{x^{2} y}{x^{3}+y^{3}}\) ……(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q3
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q3.1
This is the general solution.

Maharashtra-Board-Solutions

Question 4.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q4
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q4.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q4.2
This is the general solution.

Question 5.
(x 2 – y 2 ) dx + 2xy dy = 0
Solution:
(x 2 – y 2 ) dx + 2xy dy = 0
∴ 2xy dy = -(x 2 – y 2 ) dx = (y 2 – x 2 ) dx
∴ \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) ………(1)
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q5
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q5.1

Question 6.
xy\(\frac{d y}{d x}\) = x 2 + 2y 2
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q6
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q6.1

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Question 7.
x 2 \(\frac{d y}{d x}\) = x 2 + xy – y 2
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q7
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-8-Differential-Equation-and-Applications-Ex-8.4-Q7.1