Equation of curve through point(1,0) which satisfies the differential equation $(1 + y^{2}) d x - x y d y = 0$ , is

x^{2} + y^{2} = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} - y^{2} = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 x^{2} + y^{2} = 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} - y^{2} = 1$
We have $\frac{d x}{x} = \frac{y d y}{1 + y^{2}}$
Integrating, we get $l o g | x | = \frac{1}{2} l o g (1 + y^{2}) + l o g c$
or $| x | = c \sqrt{(1 + y^{2})}$
But it passes through (1, 0), so we get c =1
$∴$ Solution is $x^{2} = y^{2} + 1$ or $x^{2} - y^{2} = 1$ .

Suggest Corrections

Similar questions

\frac{x^{2}}{4} + y^{2} = 1

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

^{'} a^{'}

\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0

y {(\frac{d y}{d x})}^{2} + (x - y) \frac{d y}{d x} - x = 0

\frac{2 y}{x}

x y \frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}} (1 + x + x^{2})

Join BYJU'S Learning Program