What is the solution of the equation(y + x(√xy)(x + y))dx – (y + y(√xy)(x + y))dy = 0?

Category: QuestionsWhat is the solution of the equation(y + x(√xy)(x + y))dx – (y + y(√xy)(x + y))dy = 0?
Editor">Editor Staff asked 11 months ago

What is the solution of the equation(y + x(√xy)(x + y))dx – (y + y(√xy)(x + y))dy = 0?
 
(a) x^2 + y^2 = 2tan^-1(√(y/x)) + c
 
(b) x^2 + y^2 = 4tan^-1(√(y/x)) + c
 
(c) x^2 + y^2 = tan^-1(√(y/x)) + c
 
(d) x^2 + y^2 = 2tan^-1(√(x/y)) + c
 
The question was asked in my homework.
 
My question 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

The correct answer is (b) x^2 + y^2 = 4tan^-1(√(y/x)) + c
 
Best explanation: The given equation can be written as on simplifying,
 
xdx + ydy + (ydx – xdy)/(√xy)(x + y) = 0
 
the above equation can be written as,
 
 or, 1/2d(x^2 + y^2) = (xdy – ydx)/(x^2 (√y/x) (1 + y/x))
 
     = 2/(1 + y/x)* d(√y/x)
 
Thus, x^2 + y^2 = 4tan^-1(√(y/x)) + c