The solution of the differential equation $x d x + y d y = 0$ is $f (x, y) = c$ passes through (1, 1) then value of c is :

-2

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-1

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Solution

The correct option is D 1
$x d x + y d y = 0$
$\Rightarrow \frac{x^{2}}{2} + \frac{y^{2}}{2} = c$ $\Rightarrow f (x, y) = \frac{x^{2} + y^{2}}{2}$
Given f(x,y) passes through(1,-1)
$\Rightarrow \frac{1}{2} + \frac{1}{2} = c$
$\Rightarrow c = 1$

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