Differential Equations Question 205
Question: The differential equation representing the family of curves $ y^{2}=2c(x+\sqrt{c}), $ where c is a positive parameter, is of
[IIT 1999; AIEEE 2005; MP PET 2002]
Options:
A) Order 1
B) Order 2
C) Degree 3
D) Degree 4
Show Answer
Answer:
Correct Answer: A
Solution:
Given curve is $ y^{2}=2c(x+\sqrt{c}). $
Differentiate w.r.t. x, $ 2y\frac{dy}{dx}=2c $
Therefore $ c=y\frac{dy}{dx} $
Hence differential equation is
$ y^{2}=2y\frac{dy}{dx}( x+\sqrt{y\frac{dy}{dx}} ) $
Therefore $ \frac{y}{2dy/dx}-x=\sqrt{y\frac{dy}{dx}} $
Squaring and multiplying by $ {{( \frac{dy}{dx} )}^{2}} $
$ y{{( \frac{dy}{dx} )}^{3}}-x^{2}{{( \frac{dy}{dx} )}^{2}}+xy( \frac{dy}{dx} )-\frac{y^{2}}{4}=0 $
Hence order is 1 and degree is 3.
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