Question

What is the solution of the differential equation \[xy\dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + {y^2}} \right)\left( {1 + x + {x^2}} \right)}}{{\left( {1 + {x^2}} \right)}}\]?
A. \[\dfrac{1}{2}\log \left( {1 + {y^2}} \right) = \log x - {\tan ^{ - 1}}x + c\]
B. \[\dfrac{1}{2}\log \left( {1 + {y^2}} \right) = \log x + {\tan ^{ - 1}}x + c\]
C. \[\log \left( {1 + {y^2}} \right) = \log x - {\tan ^{ - 1}}x + c\]
D. \[\log \left( {1 + {y^2}} \right) = \log x + {\tan ^{ - 1}}x + c\]

Answer 1

Hint: First we separate the variables of the given differential equation. Then simplify the equation after that we use the substitution method to apply the integration formula. Then solve the differential equation to get the result.

Formula used:
Integration formula
\[\int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x + c\]
\[\int {\dfrac{1}{x}dx} = \log x + c\]

Complete step by step solution:
Given differential equation is
\[xy\dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + {y^2}} \right)\left( {1 + x + {x^2}} \right)}}{{\left( {1 + {x^2}} \right)}}\]
Separate the variables of the differential equation
\[ \Rightarrow y\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}} = \dfrac{{\left( {1 + x + {x^2}} \right)}}{{x\left( {1 + {x^2}} \right)}}dx\]
Break as a sum of two terms of the left side expression:
\[ \Rightarrow y\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}} = \left[ {\dfrac{{\left( {1 + {x^2}} \right)}}{{x\left( {1 + {x^2}} \right)}} + \dfrac{x}{{x\left( {1 + {x^2}} \right)}}} \right]dx\]
Cancel out same terms from the right-side expression
\[ \Rightarrow y\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}} = \left[ {\dfrac{1}{x} + \dfrac{1}{{\left( {1 + {x^2}} \right)}}} \right]dx\] …..(i)
Assume that, \[1 + {y^2} = z\]
Differentiate the above equation:
\[2ydy = dz\]
\[ \Rightarrow ydy = \dfrac{1}{2}dz\]
Substitute \[1 + {y^2} = z\] and \[ydy = \dfrac{1}{2}dz\] in equation (i)
\[ \Rightarrow \dfrac{{dz}}{{2z}} = \dfrac{1}{x}dx + \dfrac{1}{{\left( {1 + {x^2}} \right)}}dx\]
Integrating both sides:
\[ \Rightarrow \int {\dfrac{{dz}}{{2z}}} = \int {\dfrac{1}{x}dx} + \int {\dfrac{1}{{\left( {1 + {x^2}} \right)}}dx} \]
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{dz}}{z}} = \int {\dfrac{1}{x}dx} + \int {\dfrac{1}{{\left( {1 + {x^2}} \right)}}dx} \]
Apply the integration formula
\[ \Rightarrow \dfrac{1}{2}\log z = \log x + {\tan ^{ - 1}}x + c\]
Substitute \[1 + {y^2} = z\] in the above equation
\[ \Rightarrow \dfrac{1}{2}\log \left( {1 + {y^2}} \right) = \log x + {\tan ^{ - 1}}x + c\]
Hence option B is the correct option.

Note: Students often forget to substitute the value of z. Then get \[\dfrac{1}{2}\log z = \log x + {\tan ^{ - 1}}x + c\] which is incorrect. The correct answer is \[\dfrac{1}{2}\log \left( {1 + {y^2}} \right) = \log x + {\tan ^{ - 1}}x + c\].