Expert Answer to "2xydx+(x2+2y)dy=0,x2y+y2=1 The dx and dy in this notation..."

Rui Baldwin

Answered question

2021-01-28

2 x y d x + (x^{2} + 2 y) d y = 0, x^{2} y + y^{2} = 1

The dx and dy in this notation throws me off! Do I solve thegiven function for x and y and also their derivatives. Thenplug into the DE.

Macsen Nixon

Skilled2021-01-29Added 117 answers

2 x y d x + (x^{2} + 2 y) d y = 0

we can rewrite this equation as

2 x y + (x^{2} + 2 y) (\frac{d y}{d x}) = 0

now the given equation is

x^{2} y + y^{2} = 1

differentiating with respect to x implicilty

2 x y + x^{2} (\frac{d y}{d x}) + 2 y (\frac{d y}{d x}) = 0

2 x y + (x^{2} + 2 y) (\frac{d y}{d x}) = 0

which is given differential equation so the equation

x^{2} y + y^{2} = 1

is the solution of the diff. equation

2 x y d x + (x^{2} + 2 y) d y = 0

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