Question

# The solution of differential equation (x2+y2)dx−2xydy=0 is|x2−y2|=|x||x2−y2|=k|x|, where k is a positive constant.|x2−y2|=k, where k is a positive constant.|x2−y2|=k|x−y|, k is a positive constant.

Solution

## The correct option is B |x2−y2|=k|x|, where k is a positive constant.(x2+y2)dx−2xydy=0 is the given equation. It can be re-written as dydx=x2+y22xy. Now x2+y22xy is a homogeneous function of degree zero. So it is a homogeneous differential equation. To solve, put y = vx. So dydx=v+xdvdx(Chain rule of differentiation) v+xdvdx=x2(1+v2)2x2v=1+v22v⇒xdvdx=1+v22v−v=1−v22v⇒2v1−v2dv=dxx             ⋯(1) Now we have two variables v and x. All the terms including v and x can be written separately. So we can use variable separable here onwards. Integrating (1) we get −log|(1−v2)|−log|x|=−log|c||x(1−v2)|=|c|=k (k > 0) ⇒∣∣∣x(x2−y2)x2∣∣∣=k.⇒|x2−y2|=k|x|. which is the required solution.

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