Question

Find the equation of the curve which passes through the point $(\alpha ,\alpha )$ and satisfies the differential equation $y-x\frac{dy}{dx}=\alpha (y^2+x^2\frac{dy}{dx}).$

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

Answer 1

Answer: (C)

$\frac{(1-{\alpha }^2)y}{(1-\alpha y)}=\frac{(1+{\alpha }^2)x}{(1+\alpha x)}$

The equation can be re-written as
$y(1-\alpha y)=\frac{dy}{dx}x(1+\alpha x)$
$\therefore \frac{dy}{y(1-\alpha y)}=\frac{dx}{x(1+\alpha x)}$
or $(\frac{1}{y}+\frac{\alpha }{1-\alpha y})dy=(\frac{1}{x}-\frac{\alpha }{1+\alpha x})dx$
Intergrating
$\log y-\log (1-\alpha y)=\log x-\log (1+\alpha x)+\log k$
$\therefore \frac{y}{1-\alpha y}=k.\frac{x}{1+\alpha x}$ It passes through $(\alpha ,\alpha )$ $\therefore \frac{\alpha }{1-{\alpha }^2}=k.\frac{\alpha }{1+{\alpha }^2}$ $\therefore k=\frac{1+{\alpha }^2}{1-{\alpha }^2}$
$\therefore \frac{(1-{\alpha }^2)y}{(1-\alpha y)}=\frac{(1+{\alpha }^2)x}{(1+\alpha x)}$