Question

$(x^2+y^2)dy=xydx$. If $y\left(x_0\right)=1$, then the value of $x_0$ is equal to :

• A. $\sqrt{2}c$
• B. $\sqrt{e^2-\frac{1}{2}}$
• C. $\sqrt{\frac{e^2-1}{2}}$
• D. $\sqrt{e^2+\frac{1}{2}}$

Answer 1

The correct option is A $\sqrt{2}c$

$(x^2+y^2)dy=xydx$
$\frac{dy}{dx}=\frac{xy}{x^2+y^2}$
$\frac{dy}{dx}=\frac{\frac{xy}{x^2}}{\frac{x^2+y^2}{x^2}}=\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}$
Let $\frac{y}{x}=z$
$y=zx$
$\frac{dy}{dx}=x\frac{dz}{dx}+z$
$x\frac{dz}{dx}+z=\frac{z}{1+z^2}$
$x\frac{dz}{dx}=\frac{z}{1+z^2}-z$
$x\frac{dz}{dx}=\frac{-z^3}{1+z^2}$
$\frac{(1+z^2)}{z^3}dz=-\frac{dx}{x}$
$\int (\frac{1}{z^3}+\frac{1}{z})dz=\int -\frac{1}{x}dx$
$\frac{z^{-2}}{-z}+\ln z=-\ln x+c$
$-\frac{1}{2z^2}+\ln xz=c$
$-\frac{x^2}{2y^2}+\ln y=c$
$x=x_0,y=1$
$-\frac{x_0^2}{2}+\ln 1=-c$
$-\frac{x_0^2}{2}=-c;\frac{x_0^2}{2}=c$
$x_0=\sqrt{2}c$