Question

The curves satisfying the differential equation $(1-x^2)y^1+xy=ax$ are

• A. ellipses and hyperbolas
• B. ellipses and parabola
• C. ellipses and straight lines
• D. circles and ellipses

Answer 1

The correct option is A ellipses and hyperbolas

The given equation is linear in and can be written as
$\frac{dy}{dx}+\frac{x}{1-x^2}y=\frac{ax}{1-x^2}$
Its integrating factor is $e^{\int \frac{x}{1+-x^2}dt}=e^{-\frac{1}{2}log(1-x^2)}=\frac{1}{\sqrt{1-x^2}}$ if $-1<x<1$ and if $x^2>1$ then $I.F.=\frac{1}{\sqrt{x^2-1}}$
$\frac{d}{dx}\left(y\frac{1}{\sqrt{1-x^2}}\right)=\frac{ax}{(1-x^2{)}^{\frac{3}{2}}}=-\frac{1}{2}a\frac{-2x}{(1-x^2{)}^{\frac{3}{2}}}\Rightarrow y\frac{1}{\sqrt{1-x^2}}=\frac{a}{\sqrt{1-x^2}}+C\Rightarrow y=a+C\sqrt{1-x^2}\Rightarrow (y-a{)}^2=C^2(1-x^2)\Rightarrow (y-a{)}^2+C^2{x}^2=C^2$
Thus if the given equation represents an ellipse. If $x^2>1$ then the solution is of the form $-(y-a{)}^2+C^2{x}^2=C^2$ which represents a hyperbola.