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.

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



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
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)


This is the general solution.

Question 4.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:



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)

Question 6.
xy\(\frac{d y}{d x}\) = x ² + 2y ²
Solution:

Question 7.
x ² \(\frac{d y}{d x}\) = x ² + xy – y ²
Solution: