Question

$(1-xy+x^2{y}^2)dx=x^{2}dy$

• A. $y=\frac{1}{x}tan\left(ln|cy|\right)$
• B. $x=\frac{1}{y}tan\left(ln|cx|\right)$
• C. $x=\frac{1}{y}cot\left(ln|cy|\right)$
• D. $y=\frac{1}{x}cot\left(ln|cx|\right)$

Answer 1

The correct option is B $x=\frac{1}{y}tan\left(ln|cx|\right)$

$xy=t\Rightarrow x\frac{dy}{dx}+y=\frac{dt}{dx}$
$\Rightarrow x\frac{dy}{dx}=\frac{dt}{dx}-\frac{t}{x}$
$\Rightarrow (1-t+t^2)=x\left(\frac{dt}{dx}-\frac{t}{x}\right)$
$\Rightarrow \frac{(1-t+t^2)}{x}+\frac{t}{x}=\frac{dt}{dx}\Rightarrow \frac{1+t^2}{x}=\frac{dt}{dx}$
$\Rightarrow \frac{dt}{1+t^2}=\frac{dx}{x}\Rightarrow ta{n}^{-1}t=ln|x|+lnc$
$\Rightarrow ta{n}^{-1}\left(xy\right)=ln|cx|$
$xy=tanln|xc|$