CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The solution of differential equation (x2+y2)dx2xydy=0 is
  1. |x2y2|=|x|
  2. |x2y2|=k|x|, where k is a positive constant.
  3. |x2y2|=k, where k is a positive constant.
  4. |x2y2|=k|xy|, k is a positive constant.


Solution

The correct option is B |x2y2|=k|x|, where k is a positive constant.
(x2+y2)dx2xydy=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+v22vxdvdx=1+v22vv=1v22v2v1v2dv=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|(1v2)|log|x|=log|c||x(1v2)|=|c|=k (k > 0)
x(x2y2)x2=k.|x2y2|=k|x|. which is the required solution.
 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image