Question

The solution of \[y\left( xy+1 \right)dx+x\left( 1+xy+{{x}^{2}}{{y}^{2}} \right)dy=0\] is:
\[\left( a \right)2{{x}^{2}}{{y}^{2}}\log y-2xy-1=c{{x}^{2}}{{y}^{2}}\]
\[\left( b \right)2{{x}^{2}}{{y}^{2}}\log y+2xy-1=c{{x}^{2}}{{y}^{2}}\]
\[\left( c \right)2xy\log x+2xy+1=c{{x}^{2}}{{y}^{2}}\]
\[\left( d \right)2xy\log y-2xy-1=c{{x}^{2}}{{y}^{2}}\]

Answer 1

Hint: In this question, we are given a differential equation and we need to solve it. We will solve it by first splitting the terms and then using inspection methods we will find the integrating factor for the equation. We will then separate the variables and take integration on both the sides and evaluate our answer. We will use the following properties.
\[\left( i \right)d\left( xy \right)=xdy+ydx\]
\[\left( ii \right)\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}\]
\[\left( iii \right)\int{\dfrac{1}{x}dx=\ln \left| x \right|+c}\]

Complete step-by-step answer:
Here we are given the differential equation as \[y\left( xy+1 \right)dx+x\left( 1+xy+{{x}^{2}}{{y}^{2}} \right)dy=0.\] Opening up the brackets and separating the terms, we get,
 \[\left( xy \right)ydx+ydx+xdy+\left( xy \right)xdy+{{x}^{3}}{{y}^{2}}dy=0\]
Rearranging the terms, we get,
\[\Rightarrow ydx+xdy+\left( xy \right)ydx+\left( xy \right)ydy+{{\left( xy \right)}^{2}}xdy=0\]
Now, taking xy common from the third and fourth terms, we get,
\[\Rightarrow ydx+xdy+\left( xy \right)\left( ydx+xdy \right)+{{\left( xy \right)}^{2}}xdy=0\]
Now we know that \[xdy+ydx=d\left( xy \right),\] so putting it in the above equation, we get,
\[\Rightarrow d\left( xy \right)+\left( xy \right)d\left( xy \right)+{{\left( xy \right)}^{2}}xdy=0\]
Taking d (xy) common from the first two terms, we get,
\[\Rightarrow d\left( xy \right)\left[ 1+xy \right]+{{\left( xy \right)}^{2}}xdy=0\]
Now, we need only the terms of y in the second term, so that we can integrate with respect to x. Hence, by inspection, we can take the integrating factor as \[{{\left( xy \right)}^{3}}.\]
Dividing by \[{{\left( xy \right)}^{3}}\] on both the sides, we get,
\[\Rightarrow d\left( xy \right)\left[ \dfrac{1+xy}{{{\left( xy \right)}^{3}}} \right]+\dfrac{{{\left( xy \right)}^{2}}xdy}{{{\left( xy \right)}^{3}}}=0\]
\[\Rightarrow d\left( xy \right)\left( \dfrac{1+xy}{{{\left( xy \right)}^{3}}} \right)+\dfrac{xdy}{xy}=0\]
\[\Rightarrow d\left( xy \right)\left( \dfrac{1+xy}{{{\left( xy \right)}^{3}}} \right)+\dfrac{dy}{y}=0\]
\[\Rightarrow \left( \dfrac{1+xy}{{{\left( xy \right)}^{3}}} \right)d\left( xy \right)=-\dfrac{dy}{y}\]
Integrating both the sides, we get,
\[\Rightarrow \int{\dfrac{1+xy}{{{\left( xy \right)}^{3}}}}dy=-\int{\dfrac{dy}{y}}\]
We are integrating with respect on the left side of the equation and we are integrating with respect to y on the right side of the equation.
\[\Rightarrow \int{\dfrac{1}{{{\left( xy \right)}^{3}}}d\left( xy \right)+\int{\dfrac{xy}{{{\left( xy \right)}^{3}}}d\left( xy \right)}=-\int{\dfrac{dy}{y}}}\]
\[\Rightarrow \int{\dfrac{1}{{{\left( xy \right)}^{3}}}d\left( xy \right)+\int{\dfrac{1}{{{\left( xy \right)}^{2}}}d\left( xy \right)}=-\int{\dfrac{1}{y}}dy}\]
\[\Rightarrow \int{{{\left( xy \right)}^{-3}}d\left( xy \right)+\int{{{\left( xy \right)}^{-2}}d\left( xy \right)}=-\int{\dfrac{1}{y}}dy}\]
We know that, \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}\] and \[\int{\dfrac{1}{x}dx=\ln \left| x \right|+c},\] so we get,
\[\Rightarrow \dfrac{{{\left( xy \right)}^{-3+1}}}{-3+1}+{{c}_{1}}+\dfrac{{{\left( xy \right)}^{-2+1}}}{-2+1}+{{c}_{2}}=-\ln \left| y \right|+{{c}_{3}}\]
\[\Rightarrow \dfrac{{{\left( xy \right)}^{-2}}}{-2}+\dfrac{{{\left( xy \right)}^{-1}}}{-1}+\ln \left| y \right|={{c}_{3}}-{{c}_{1}}-{{c}_{2}}\]
\[\Rightarrow \dfrac{-1}{2{{\left( xy \right)}^{3}}}-\dfrac{1}{\left( xy \right)}+\ln \left| y \right|={{c}_{*}}\left[ {{c}_{*}}={{c}_{3}}-{{c}_{1}}-{{c}_{2}} \right]\]
Taking LCM of \[2{{\left( xy \right)}^{2}},\] we get,
\[\Rightarrow \dfrac{-1-2\left( xy \right)+2{{\left( xy \right)}^{2}}\ln \left| y \right|}{2{{\left( xy \right)}^{2}}}={{c}_{*}}\]
Cross multiplying and rearranging, we get,
\[\Rightarrow 2{{\left( xy \right)}^{2}}\ln \left| y \right|-2\left( xy \right)-1=2{{c}_{*}}{{\left( xy \right)}^{2}}\]
\[\Rightarrow 2{{\left( xy \right)}^{2}}\ln \left| y \right|-2\left( xy \right)-1=c{{\left( xy \right)}^{2}}\left[ {{c}_{*}}=2{{c}_{*}} \right]\]
\[\Rightarrow 2{{\left( xy \right)}^{2}}\ln \left| y \right|-2xy-1=c{{x}^{2}}{{y}^{2}}\]

So, the correct answer is “Option (a)”.

Note: While solving any differential equation, we need to do an inspection to find how we can solve it as there are a lot of methods involved. While integrating, students should take care that we have taken (xy) as a simple term and integrate with respect to (xy) only. To avoid confusion, they can replace xy as t on the left side and solve accordingly. Students should take care of the signs. Here we have assumed \[{{c}_{*}}={{c}_{3}}-{{c}_{1}}-{{c}_{2}}\] and \[c=2{{c}_{*}}\] because all are constants and unknown, so we can use any notation to describe the constant.