Question

primitive of xdx +ydy =0

Answer 1

x dx = -y dy

(1/2)(x^2) + C = (-1/2)(y^2) + C
x^2 + C = -y^2
C - x^2= y^2

y = sqrt (C - x^2)

Hope that helps

Answer 2

The primitive is $y =\sqrt{C - x^2}$.

Step-by-step explanation:

We have,

$xdx +ydy =0$

∴ $xdx +ydy =0$

⇒ $xdx =-ydy$     .....(1)

Integrating both sides in (1), we get

$\int {x} \, dx  =-\int {y} \, dy$

⇒ $\dfrac{x^{1+1} }{1+1} =-\dfrac{y^{1+1} }{1+1}+C$

[∵ $\int {x^n} \, dx =\frac{x^{n+1} }{n+1} +C$]

Where, C is called integration constant

⇒ $\dfrac{x^{2} }{2} =-\dfrac{y^{2} }{2}+C$

⇒ $\dfrac{y^{2} }{2} =C-\dfrac{x^{2} }{2}$

⇒$y =\sqrt{C - x^2}$

Hence, the primitive is $y =\sqrt{C - x^2}$.