Question

Solution of differential equation $x^2=1+{\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}+\frac{{\left(\frac{x}{y}\right)}^{-2}{\left(\frac{dy}{dx}\right)}^2}{2!}+\frac{{\left(\frac{x}{y}\right)}^{-3}{\left(\frac{dy}{dx}\right)}^3}{3!}+....$ is

• A. $y^2=x^2(\ln x^2-1)+C$
• B. $y=x^2(\ln x-1)+C$
• C. $y2=x(\ln x-1)+C$
• D. $y=x^2{e}^{x^2}+C$

Answer 1

Answer: (A)

$x^2=e^{{\left(\frac{x}{y}\right)}^{-1}\left(\frac{dy}{dx}\right)}$
$\Rightarrow x^2=e^{\left(\frac{y}{x}\right)\left(\frac{dy}{dx}\right)}$
$\Rightarrow x^2=\frac{y}{x}\frac{dy}{dx}$
or $\int x\ln x^{2}dx=\int ydy$
Put $x^2=t\Rightarrow 2xdx=dt$
$\therefore \frac{1}{2}\int \ln tdt=\frac{y^2}{2}C+t\ln$
or $t-t=y^2$
or $y^2=x^2\left(\ln x^2-1\right)+C$