If $\frac{dx}{dy}=\frac{1+x-y^2}{y},x\left(1\right)=1,$ then $5x\left(2\right)$ is equal to:

Given: $\frac{dx}{dy}=\frac{{\displaystyle 1+x-y^2}}{y}$

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

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

Now let $1+x=t\Rightarrow \frac{dx}{dy}=\frac{dt}{dy}$

$\Rightarrow \frac{dt}{dy}-\frac{t}{y}=-y$

$\Rightarrow \mathrm{IF}=e^{\int -\frac{1}{y}dy}$

$\Rightarrow \mathrm{IF}=e^{-\log y}$

$\Rightarrow \mathrm{IF}=e^{\log \frac{1}{y}}$

$\Rightarrow \mathrm{IF}=\frac{1}{y}$

So, the solution of the given differential equation is

$\frac{t}{y}=\int \frac{1}{y}\times \left(-y\right)dy$

$\Rightarrow \frac{\left(1+x\right)}{y}=\int \frac{1}{y}\times \left(-y\right)dy$

$\Rightarrow \frac{{\displaystyle \left(1+x\right)}}{{\displaystyle y}}=-y+c$

Also, $x\left(1\right)=1$

$\Rightarrow \frac{{\displaystyle 2}}{{\displaystyle 1}}=-1+c$

$\Rightarrow c=3$

$\Rightarrow \frac{{\displaystyle \left(1+x\right)}}{{\displaystyle y}}=-y+3$

Putting $y=2$

$\Rightarrow \frac{{\displaystyle \left(1+x\right)}}{{\displaystyle 2}}=-2+3$

$\Rightarrow \left(1+x\right)=2$

$\Rightarrow x=1$

$\Rightarrow 5x=5$