Differential Equations
The differential equation of the family of parabolas y2=4a(x+a) is:
Select the correct option:
Solution
y(y′)2+2xy′−y=0
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Start with the equation: y2=4ax+4a2.
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Differentiate with respect to x: 2yy′=4a⟹a=2yy′.
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Eliminate parameter 'a' by substituting back: y2=4(2yy′)x+4(2yy′)2 y2=2xyy′+y2(y′)2.
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Simplify: Since parabolas are non-degenerate, y is not identically 0. Divide by y: y=2xy′+y(y′)2 y(y′)2+2xy′−y=0.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- differential equations
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
y(y′)2+2xy′−y=0
-
Start with the equation: y2=4ax+4a2.
-
Differentiate with respect to x: 2yy′=4a⟹a=2yy′.
-
Eliminate parameter 'a' by substituting back: y2=4(2yy′)x+4(2yy′)2 y2=2xyy′+y2(y′)2.
-
Simplify: Since parabolas are non-degenerate, y is not identically 0. Divide by y: y=2xy′+y(y′)2 y(y′)2+2xy′−y=0.
This medium difficulty mathematics question is from the chapter differential equations, covering the topic of differential equations. It appeared in the 2025 exam.
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