What is the solution of (y(dy/dx) + 2x)^2 = (y^2 + 2x^2)[1 + (dy/dx)^2]?

Category: QuestionsWhat is the solution of (y(dy/dx) + 2x)^2 = (y^2 + 2x^2)[1 + (dy/dx)^2]?
Editor">Editor Staff asked 11 months ago

What is the solution of (y(dy/dx) + 2x)^2 = (y^2 + 2x^2)[1 + (dy/dx)^2]?
 
(a) cx^±1/√2 = y/x + √(y^2 – 2x^2)/x^2
 
(b) cx^±√2 = y/x + √(y^2 + 2x^2)/x^2
 
(c) cx^±1/2√2 = y/x + √(y^2 – 2x^2)/x^2
 
(d) cx^±1/√2 = y/x + √(y^2 + 2x^2)/x^2
 
The question was asked during an internship interview.
 
My enquiry is from Linear First Order Differential Equations in portion Differential Equations of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Right answer is (d) cx^±1/√2 = y/x + √(y^2 + 2x^2)/x^2
 
Explanation: Here, y^2(dy/dx)^2 + 4x^2 + 4xy(dy/dx) = (y^2 + 2x^2)[1 + (dy/dx)^2]
 
=>dy/dx = y/x ± √(1/2(y/x)^2) + 1
 
Let, y = vx
 
=> v + x dv/dx = v ± √(1/2(v)^2) + 1
 
Integrating both sides,
 
±√dv/(√(1/2(v)^2) + 1) = ∫dx/x
 
cx^±1/√2 = y/x + √(y^2 + 2x^2)/x^2