Question

General solution of (x2+y2)dx−2xydy=0 is:

• A. (A) y2+x2=cx2
• B. (B) x2−y2=cx
• C. (C) x2=cx(x2−y2)
• D. (D) None of these

Answer 1

Answer: (C)

(C) x2=cx(x2−y2)

Answer 2

Explanation:
Given,(x2+y2)dx=2xydydydx=(x2+y2)2xy is homogeneous.Put, y=vxdydx=v+xdvdx⇒v+xdvdx=x2+v2x22vx2=x2(1+v2)2vx2⇒xdvdx=1+v22v−v=1−v22v⇒2v1−v2dv=1xdxBy integrating both sides we get,∫2v1−v2dv=∫1xdx……(1)Now, we integrate ∫2v1−v2dvLet, 1−v2=aOn differentiating both sides w.r.t. v, we getddv(1−v2)=ddva−2v=dadvdv=−da2vNow,putting these values on equation (1), we get∫2va×−da2v=∫1xdx⇒∫−daa=∫1xdxWe know that,∫1xdx=log⁡xNow,−log⁡(1−v2)=log⁡x+log⁡c⇒log⁡[1(1−v2)]=log⁡x+log⁡c⇒[1(1−v2)]=cx⇒[1(1−y2x2)]=cx⇒[x2(x2−y2)]=cx⇒x2=cx(x2−y2) is the general solution of the differential equation.Hence, the correct option is (C).