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If x3y2 is an integrating factor of (6y2 + axy)dx + (6xy + bx2)dy =0, a, b ∈ ℝ then
- a)3a - 5b = 0
- b)2a - b = 0
- c)3a + 5b = 0
- d)2a + b = 0
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If x3y2is an integrating factor of (6y2+ axy)dx + (6xy + bx2)dy =0, a,...
Given Information:
- Integrating factor: x^3y^2
- Differential equation: (6y^2 + axy)dx + (6xy + bx^2)dy = 0
Explanation:
Step 1: Finding the Integrating Factor
- To find the integrating factor, we use the formula: μ = e^(∫M_y - N_x dx)
- Given equation: (6y^2 + axy)dx + (6xy + bx^2)dy = 0
- Now, M_y = 12y + ax and N_x = 6x + bx
- Integrating factor μ = x^3y^2
Step 2: Applying the Integrating Factor
- Multiply the given differential equation by the integrating factor: x^3y^2[(6y^2 + axy)dx + (6xy + bx^2)dy] = 0
- Simplify the equation: 6x^3y^4 + ax^4y^3 + 6x^3y^3 + bx^5y^2 = 0
Step 3: Comparing Coefficients
- Comparing coefficients of x^4y^3 and x^5y^2:
- From x^4y^3: a = 3
- From x^5y^2: b = -5
Conclusion:
- Therefore, the relationship between a and b is 3a - 5b = 0
- The correct option is: 3a - 5b = 0, which is option 'A'
- Integrating factor: x^3y^2
- Differential equation: (6y^2 + axy)dx + (6xy + bx^2)dy = 0
Explanation:
Step 1: Finding the Integrating Factor
- To find the integrating factor, we use the formula: μ = e^(∫M_y - N_x dx)
- Given equation: (6y^2 + axy)dx + (6xy + bx^2)dy = 0
- Now, M_y = 12y + ax and N_x = 6x + bx
- Integrating factor μ = x^3y^2
Step 2: Applying the Integrating Factor
- Multiply the given differential equation by the integrating factor: x^3y^2[(6y^2 + axy)dx + (6xy + bx^2)dy] = 0
- Simplify the equation: 6x^3y^4 + ax^4y^3 + 6x^3y^3 + bx^5y^2 = 0
Step 3: Comparing Coefficients
- Comparing coefficients of x^4y^3 and x^5y^2:
- From x^4y^3: a = 3
- From x^5y^2: b = -5
Conclusion:
- Therefore, the relationship between a and b is 3a - 5b = 0
- The correct option is: 3a - 5b = 0, which is option 'A'
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Community Answer
If x3y2is an integrating factor of (6y2+ axy)dx + (6xy + bx2)dy =0, a,...
Given (6y2 + axy)dx + (6xy + bx2)dy = 0
I. F. = x3y2
multiplying by I.F. in (1). We get
(6y4x3 + ax4y3)dx + (6x4y3 + bx5y2)dy = 0 be exact. Then 
⇒ 24x3y3 + 3ax4y2 = 24a3y3 + 5bx4y2
⇒ 3a = 5b
⇒ 3a - 5b = 0
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If x3y2is an integrating factor of (6y2+ axy)dx + (6xy + bx2)dy =0, a, b ∈ thena)3a - 5b = 0b)2a - b = 0c)3a + 5b = 0d)2a + b = 0Correct answer is option 'A'. Can you explain this answer?
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