Question

Find the equation of the curve with D.E. $(1+y^2)dx=xydy$, and passing through $(1,0)$.

• A. $x^2-y^2=1$
• B. $4x^2-y^2=4$
• C. $x^2+y^2=1$
• D. $4x^2+y^2=4$

Answer 1

The correct option is A $x^2-y^2=1$

Given, $(1+y^2)dx=xydy$

$\Rightarrow 2\frac{dx}{x}=\frac{2y}{1+y^2}dy$

Let $t=1+y^2$

$\Rightarrow dt=2ydy$

$\Rightarrow \int \frac{2}{x}dx=\int \frac{dt}{t}+\log c$

Therefore, $\log x^2=\log ct$

$\Rightarrow x^2=c(1+y^2)$

Put $(x,y)=(1,0)$

$\Rightarrow 1=c(1+0)$

$\Rightarrow c=1$

$\Rightarrow x^2-y^2=1$