Question

Find the particular solution of the following differential equation; $\frac{dx}{dy}=1+x^2+y^2+x^2{y}^2$ given that $y=1$ when $x=0.$

Answer 1

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

$\therefore \frac{dx}{dy}=(1+x^2)(1+y^2)$

$\therefore \frac{dx}{1+x^2}=(1+y^2)dy$

Integrating both sides, we get

${\tan }^{-1}x=y+\frac{y^3}{3}+c$,

where $c$ is a cons\tan t of integration, whose value can be found out by substituting the given values.

Substituting $y=1$ when $x=0$, we get

${\tan }^{-1}0=1+\frac{1}{3}+c$

$\therefore 0=c+\frac{4}{3}$

$\Rightarrow c=\frac{-4}{3}$

Thus, the solution can be written as ${\tan }^{-1}x=y+\frac{y^3}{3}-\frac{4}{3}$

i.e. $3{\tan }^{-1}x=3y+y^3-4.$