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junior college 2 | H2 Maths
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Differential equation involving x and y
Date Posted: 4 years ago
Views: 306
Alternative to substitution :
dy/dx + (8x+4y+1)/(4x+2y+1) = 0
dy/dx = -(8x+4y+1)/(4x+2y+1)
(4x + 2y + 1) dy/dx = -8x - 4y - 1
4x dy/dx + (2y+1) dy/dx = -8x - 4y - 1
Bring some of the right side terms to the left,
(4x dy/dx + 4y) + (2y+1) dy/dx = -8x - 1
Notice that the terms on the left side are directly integrable since they are already the result of applying a product rule or chain rule.
Integrate both sides with respect to x,
∫ ( (4x dy/dx + 4y) + (2y+1) dy/dx ) dx = ∫(-8x - 1)dx
4xy + y² + y = -4x² - x + c, c is a constant.
Sub x = 1, y = 1,
4(1)(1) + 1² + 1 = -4(1)² - 1 + c
c = 4 + 1 + 1 + 4 + 1 = 11
Therefore,
4xy + y² + y = -4x² - x + 11
4xy + y² + y + 4x² + x = 11
y(4x + 1) + y² + x(4x + 1) = 11
(4x + 1)(y + x) + y² = 11
dy/dx + (8x+4y+1)/(4x+2y+1) = 0
dy/dx = -(8x+4y+1)/(4x+2y+1)
(4x + 2y + 1) dy/dx = -8x - 4y - 1
4x dy/dx + (2y+1) dy/dx = -8x - 4y - 1
Bring some of the right side terms to the left,
(4x dy/dx + 4y) + (2y+1) dy/dx = -8x - 1
Notice that the terms on the left side are directly integrable since they are already the result of applying a product rule or chain rule.
Integrate both sides with respect to x,
∫ ( (4x dy/dx + 4y) + (2y+1) dy/dx ) dx = ∫(-8x - 1)dx
4xy + y² + y = -4x² - x + c, c is a constant.
Sub x = 1, y = 1,
4(1)(1) + 1² + 1 = -4(1)² - 1 + c
c = 4 + 1 + 1 + 4 + 1 = 11
Therefore,
4xy + y² + y = -4x² - x + 11
4xy + y² + y + 4x² + x = 11
y(4x + 1) + y² + x(4x + 1) = 11
(4x + 1)(y + x) + y² = 11
See 1 Answer
done
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clear
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This method is called substitution.
Date Posted:
4 years ago
Thank u so much sir...
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