Question

The solution of differential equation $(1-xy+x^2{y}^2)dx=x^{2}dy$ is

[$C$ is constant of integration]

• A. $\tan xy=\log \left|Cx\right|$
• B. $\tan \frac{y}{x}=\tan \left(\log \left|Cx\right|\right)$
• C. $xy=\tan \left(\log \left|Cx\right|\right)$
• D. $\frac{x}{y}=\tan \left(\log |Cx|\right)$

Answer 1

The correct option is C

$xy=\tan \left(\log \left|Cx\right|\right)$

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

Let $xy=v$

$\Rightarrow y=\frac{v}{x}$

$\therefore \frac{dy}{dx}=\frac{x.\frac{dv}{dx}-v}{x^2}$

$\Rightarrow x^{2}dy=xdv-vdx$

So, from equation (1), $(1-v+v^2)dx=xdv-vdx$

$\Rightarrow (1+v^2)dx=xdv$

$\Rightarrow \int \frac{dv}{1+v^2}=\int \frac{dx}{x}$

$\Rightarrow {\tan }^{-1}v=\log \left|x\right|+\log C$

$\Rightarrow {\tan }^{-1}v=\log \left|Cx\right|$

$\Rightarrow v=\tan \left(\log \left|Cx\right|\right)$

$\Rightarrow xy=\tan \left(\log \left|Cx\right|\right)$