The curves satisfying the differential equation ( 1 − x 2 ) y 1 + x y = a x are Options: A . &nbsp Ellipse and hyperbola B . &nbsp Ellipse and parabola C . &nbsp Ellipse and straight line D . &nbsp Circle and parabola

Answer: Option A : A The given equation is linear DE and can be written as d y d x + x 1 − x 2 y = a x 1 − x 2 Its integrating factor is e ∫ x 1 − x 2 d x = e − ( 1 2 ) ln ( 1 − x 2 ) = 1 √ 1 − x 2 [ − 1 < x < 1 ] and if x 2 > 1 then I . F . = 1 √ x 2 − 1 d d x ( y 1 √ 1 − x 2 ) = a x ( 1 − x 2 ) 3 2 = − 1 2 − 2 a x ( 1 − x 2 ) 3 2 ⇒ y 1 √ 1 − x 2 = a √ 1 − x 2 + C ⇒ y = a + C √ 1 − x 2 ⇒ ( y − a ) 2 = C 2 ( 1 − x 2 ) ⇒ ( y − a ) 2 + C 2 x 2 = C 2 Thus, if − 1 < x < 1 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.

The curves satisfying the differential equation (1−x2)y1+xy=ax are

Discussion Forum : Differential Equations

Question -

The curves satisfying the differential equation $(1 - x^{2}) y^{1} + x y = a x$ are

Options:

A . Ellipse and hyperbola

B . Ellipse and parabola

C . Ellipse and straight line

D . Circle and parabola

Answer: Option A
:
A
The given equation is linear DE and can be written as

\frac{d y}{d x} + \frac{x}{1 - x^{2}} y = \frac{a x}{1 - x^{2}}

Its integrating factor is

e^{\int \frac{x}{1 - x^{2}} d x} = e^{- (\frac{1}{2}) ln (1 - x^{2})} = \frac{1}{\sqrt{1 - x^{2}}} [- 1 < x < 1]

and

if x^{2} > 1

then

I . F . = \frac{1}{\sqrt{x^{2} - 1}}

\frac{d}{d x} (y \frac{1}{\sqrt{1 - x^{2}}}) = \frac{a x}{(1 - x^{2})^{\frac{3}{2}}} = - \frac{1}{2} \frac{- 2 a x}{(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

- 1 < x < 1

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.

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