1 , The Particular Solution Of The D.E. (1 + Y^2)dx + (1 + X^2)dy = 0 Is

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SolveGuidesJoin / LoginUse appLogin0You visited us 0 times! Enjoying our articles? Unlock Full Access!Standard XIIMathematicsQuestionWhen x=0,y=−1, the particular solution of the D.E. (1+y2)dx+(1+x2)dy=0 is:
  1. x+y+xy=1
  2. x+y+1=xy
  3. x+y+xy+1=0
  4. x−y+xy=1
Ax+y+1=xyBx+y+xy+1=0Cx−y+xy=1Dx+y+xy=1Open in AppSolutionVerified by Toppr

(1+y2)dx+(1+x2)dy=0⇒−dx1+x2=dy1+y2⇒−tan−1x=tan−1y−cc=tan−1x+tan−1y⇒c=tan−1(x+y1−xy)⇒tanc=(x+y1−xy)But given particular solution (x,y)=(0,1)⇒tanc=−1⇒−1=x+y1−xyxy−1=x+y⇒x+y+1=xy

Was this answer helpful?18Similar QuestionsQ1Find the Particular solution of the differential equationsx2dy+(xy+y2)dx=0;y=1 when ,x=1View SolutionQ2When x=0,y=−1, the particular solution of the D.E. (1+y2)dx+(1+x2)dy=0 is:View SolutionQ3Find a particular solution of each of the following differential equations: (i) 1+x2dydx+2xy=11+x2; y=0, when x=1 (ii) (x + y) dy + (x − y) dx = 0; y = 1 when x = 1 (iii) x2 dy + (xy + y2) dx = 0; y = 1 when x = 1View SolutionQ4Solve the following differential equation:x2dy+(xy+y2)dx=0, when x=1 and y=1View SolutionQ5Solution of y2dx=(xy−x2)dy, given that y=1 when x=1, is:View Solution

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