Question

The solution of $\frac{x^{3}dx+y{x}^{2}dy}{\sqrt{x^2+y^2}}=ydx-xdy$, $y\left(1\right)=1$ is

• A. $\sqrt{x^2+y^2}=(\sqrt{2}-1)x$
• B. $\sqrt{x^2+y^2}+\frac{y}{x}=1+\sqrt{2}$
• C. $\sqrt{x^2+y^2}+\frac{y}{x^2}=\sqrt{2}-1$
• D. $(x^2+y^2{)}^2+x{y}^2=\sqrt{2}$

Answer 1

The correct option is B $\sqrt{x^2+y^2}+\frac{y}{x}=1+\sqrt{2}$

Given: $\frac{x^{3}dx+y{x}^{2}dy}{\sqrt{x^2+y^2}}=ydx-xdy$
$\frac{x^2(xdx+ydy)}{\sqrt{x^2+y^2}}=ydx-xdy$
We know that -
$\left[d\right(x^2+y^2)=2(xdx+ydy\left)\right]$
$\left[d\right(\sqrt{x^2+y^2})=\frac{xdx+ydy}{\sqrt{x^2+y^2}}]$
$\Rightarrow x^{2}d\left(\sqrt{x^2+y^2}\right)=ydx-xdy$
$\Rightarrow d\left(\sqrt{x^2+y^2}\right)=ydx-xdy$
$\Rightarrow d\left(\sqrt{x^2+y^2}\right)=\frac{y}{x^2}dx-\frac{1}{x}dy=d(\frac{-1}{x}.y)$
$\Rightarrow \sqrt{x^2+y^2}=\frac{-y}{x}+c$
$\because y\left(1\right)=1$
$\Rightarrow \sqrt{2}=-1+c\Rightarrow c=1+\sqrt{2}$
$\sqrt{x^2+y^2}=\frac{-y}{x}+1+\sqrt{2}$