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