Question

The solution of primitive integral equation $(x^2+y^2)dy=xydx$ is $y=y\left(x\right).$ If $y\left(1\right)=$1 and $y\left(x_0\right)=e$ then $x_0$ is

• A. $\sqrt{2(e^2-1)}$
• B. $\sqrt{2(e^2+1)}$
• C. $\sqrt{3}e$
• D. $\sqrt{\frac{1}{2}(e^2+1)}$

Answer 1

The correct option is C $\sqrt{3}e$

$(x^2+y^2)dy=xydx$

$(\frac{x}{y}+\frac{y}{x})=dx$ $\frac{x}{y}=p$

$x=py$

$(p+\frac{1}{p})dy=dpy+dyp$ $dx=p^{1}y+y^1{p}^1$

$dy(p+\frac{1}{p}-p)=dpy$

$\int \frac{dy}{y}=\int pdp$

$lny=\frac{p^2}{2}+c$

$lny=\frac{{\left(\frac{x}{y}\right)}^2}{2}+c$

$0=\frac{1}{2}+c$ $\therefore c=\frac{-1}{2}$

If $y=e$

than $1=\frac{{\left(\frac{x}{e}\right)}^2}{2}-\frac{1}{2}$

$3e^2=x^2$

$x=\sqrt{3e}$