Question

The solution of the differential equation $xy\dfrac{{dy}}{{dx}} = 1 - {x^2} + {y^2} - {x^2}{y^2}$ is
1. $\left( {1 + {y^2}} \right)\left( {1 + {x^2}} \right) = c{e^{{x^2}}}$
2. $1 + {y^2} = c{x^2}{e^{ - {x^2}}}$
3. ${y^2}\left( {1 + {x^2}} \right) = c{e^{{x^2}}}$
4. $\left( {1 + {y^2}} \right)\left( {1 - {x^2}} \right) = c{e^{{-x^2}}}$

Answer 1

Hint: To solve the given differential equation, we will use factorization method first in left side of equation and we will use cross multiplication method to split function of $x$ variable one side and function of $y$ variable other side. Then, we will imply the integration symbol both sides and will do integration. After that we will simplify the obtained equation to find the required answer.

Complete step by step answer:
Since, the given equation is:
$xy\dfrac{{dy}}{{dx}} = 1 - {x^2} + {y^2} - {x^2}{y^2}$
Now, we will use factorization method and will simplify it as:
$\begin{gathered}
   \Rightarrow xy\dfrac{{dy}}{{dx}} = \left( {1 - {x^2}} \right) + {y^2}\left( {1 - {x^2}} \right) \\
   \Rightarrow xy\dfrac{{dy}}{{dx}} = \left( {1 - {x^2}} \right)\left( {1 + {y^2}} \right) \\
\end{gathered} $
Here, we will do the cross multiplication as:
$ \Rightarrow \dfrac{y}{{\left( {1 + {y^2}} \right)}}dy = \dfrac{{\left( {1 - {x^2}} \right)}}{x}dx$
Now, we will use integration sign each side of above expression as:
$ \Rightarrow \int {\dfrac{y}{{\left( {1 + {y^2}} \right)}}} dy = \int {\dfrac{{\left( {1 - {x^2}} \right)}}{x}} dx$
Here, we will simplify the above equation as:
$ \Rightarrow \int {\dfrac{y}{{\left( {1 + {y^2}} \right)}}} dy = \int {\left( {\dfrac{1}{x} - x} \right)} dx$
Let’s consider, $t = 1 + {y^2}$ and differentiate it. We will get $dy = \dfrac{{dt}}{{2y}}$. So, we will replace these values in the above equation as:
$ \Rightarrow \int {\dfrac{y}{t}} \cdot \dfrac{{dt}}{{2y}} = \int {\left( {\dfrac{1}{x} - x} \right)} dx$
Here, we will cancel out the equal like term as:
$ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{t}} dt = \int {\left( {\dfrac{1}{x} - x} \right)} dx$
Now, we will do the integration and substitute $\log t$ for $\dfrac{1}{t}$, $\log x$ for \[\] and $\dfrac{{{x^2}}}{2}$ for $x$ in the above step as:
$ \Rightarrow \dfrac{1}{2}\log t = \log x - \dfrac{{{x^2}}}{2} + \log c$
Where $\log c$ is a constant.
Here, we will multiply with $2$ each side of the above step as:
$ \Rightarrow 2 \cdot \dfrac{1}{2}\log t = 2 \cdot \log x - 2 \cdot \dfrac{{{x^2}}}{2} + 2 \cdot \log c$
Now, we will cancel out the equal like term and will simplify the above expression as:
\[\begin{gathered}
   \Rightarrow \log t = \log {x^2} - {x^2} + \log c \\
   \Rightarrow \log t = \log {x^2} + \log c - {x^2} \\
\end{gathered} \]
Here, we can write \[\log {x^2} + \log c\] as $\log c{x^2}$ in the above expression as:
\[\begin{gathered}
   \Rightarrow \log t = \log c{x^2} - {x^2} \\
   \Rightarrow \log t - \log c{x^2} = - {x^2} \\
\end{gathered} \]
Now, we will substitute $\log \dfrac{t}{{c{x^2}}}$ for $\log t - \log c{x^2}$ as:
\[ \Rightarrow \log \dfrac{t}{{c{x^2}}} = - {x^2}\]
 By removing $\log $, we can write above equation as:
\[ \Rightarrow \dfrac{t}{{c{x^2}}} = {e^{ - {x^2}}}\]
Now, we will do cross multiplication as:
$ \Rightarrow t = c{x^2}{e^{ - {x^2}}}$
Here, we will substitute \[\] for \[t\] as:
$ \Rightarrow \left( {1 - {y^2}} \right) = c{x^2}{e^{ - {x^2}}}$

So, the correct answer is “Option 2”.

Note: Let’s consider that:
$1 + {y^2} = t$
Now, we will differentiate the above function with respect to $y$ and will get $2y$as:
$ \Rightarrow 2y = \dfrac{{dt}}{{dy}}$
We will find the value of $dy$ from the above step as:
$ \Rightarrow dy = \dfrac{{dt}}{{2y}}$