Solution:
Given that, $\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{x}=\text{C}$
$\Rightarrow\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{C}-\text{x}$
$\Rightarrow\text{ydy}=(\text{C}-\text{x})\text{dx}$
On integrating both sides, we get
$\int\text{ydy}=\int(\text{C}-\text{x})\text{dx}$
$\Rightarrow\frac{\text{y}^2}{2}=\text{Cx}-\frac{\text{x}^2}{2}+\text{k}$
$\Rightarrow\frac{\text{x}^2}{2}+\frac{\text{y}^2}{2}=\text{Cx}+\text{k}$
$\Rightarrow\frac{\text{x}^2}{2}+\frac{\text{y}^2}{2}-\text{Cx}=\text{k}$
which represent family of circles.
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$STATEMENT-1$: $y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$ and
$STATEMENT-2$ : $\mathrm{y}(\mathrm{x})$ is given by $\frac{1}{\mathrm{y}}=\frac{2 \sqrt{3}}{\mathrm{x}}-\sqrt{1-\frac{1}{\mathrm{x}^2}}$
Statement$-1$ If $f R \rightarrow R$ and $c \in R$ is such that $f$ is increasing in $(c - \delta , c)$ and $f$ is decreasing in $(c, c + \delta )$ then $f$ has a local maximum at $c$. Where $\delta$ is a sufficiently small positive quantity.
Statement $-2$ Let $f (a, b) \rightarrow \,R, c \in (a, b)$. Then $f$ can not have both a local maximum and a point of inflection at $x = c.$
Statement $-3 $ The function $f (x) = x^2 | x |$ is twice differentiable at $x = 0.$
Statement $-4$ Let $f [c - 1, c + 1] \rightarrow [a, b]$ be bijective map such that $f$ is differentiable at $c$ then $f^{-1}$ is also differentiable at $f (c)$.