# Explicit solution of -2x^2y+y^2=1; 2xydx+(x^2-y)dy=0?

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How would I go about solving this??? I did find the second problem is an implicit solution of the firsto ne

**Update :**I'm finding explicit solution, not implicit. Implicit solution is ...show more

**Update 2:**again please help me out for the explicit solution...not the implicit...i ...show more

### Other Answers (2)

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Given -2x^2y+y^2=1

Differentiating Both Sides

-2(2xy + x^2dy/dx) + 2ydy/dx = 0

-4xy - 2x^2dy/dx + 2ydy/dx = 0

- 2x^2dy/dx + 2ydy/dx = 4xy

dy/dx = 4xy/(2y - 2x^2) = 2xy/(y-x^2)

Equation 2) 2xydx+(x^2-y)dy=0

Checking For dy/dx

2xydx = -(x^2-y)dy

dy/dx = 2xy/(y-x^2) = Previously Obtained dy/dx

Hence, Second Equation Is The Implicit Solution Of The First -
2xydx+x^2dy-ydy=0

on integration

2y.x^2/2 + x^2y - y^2/2 = 0

=> 2 x^y - y^2/2 = 0

=> 4x^2y - y^2 = 0

=> 4 x^2 - y = 0

=> x^2 = y /4

putting this value in 1st giving equetion

-2. y/4.y + y^2 = 1

=> y^2/2 = 1

=> y = 1/ sqrt 2

so x^2 = 1/ 4 sqrt 2

explicit solution of -2x^2y+y^2=1; 2xydx+(x^2-y)dy=0?

How would I go about solving this??? I did find the second problem is an implicit solution of the firsto ne

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