Question

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

Answer 1

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