Question

What is the differential equation \[\left( {1 + {x^2}} \right)\dfrac{{dy}}{{dx}} = x\left( {1 + {y^2}} \right)\]?
A. \[2{\tan ^{ - 1}}y = \log \left( {1 + {x^2}} \right) + c\]
B. \[{\tan ^{ - 1}}y = \log \left( {1 + {x^2}} \right) + c\]
C. \[2{\tan ^{ - 1}}y + \log \left( {1 + {x^2}} \right) = + c\]
D. None of these

Answer 1

Hint First we will separate the variable of the differential equation. Then apply the integration formula to find 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
\[\left( {1 + {x^2}} \right)\dfrac{{dy}}{{dx}} = x\left( {1 + {y^2}} \right)\]
Separate the variables of the differential equation:
\[\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}} = \dfrac{x}{{\left( {1 + {x^2}} \right)}}dx\]
Taking integration both sides
\[\int {\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}}} = \int {\dfrac{x}{{\left( {1 + {x^2}} \right)}}dx} \] …..(i)
Assume that, \[1 + {x^2} = t\]
Differentiate with respect to x
\[ \Rightarrow 2xdx = dt\]
\[ \Rightarrow xdx = \dfrac{{dt}}{2}\]
Substitute \[xdx = \dfrac{{dt}}{2}\] and \[1 + {x^2} = t\] in equation (i)
\[\int {\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}}} = \int {\dfrac{1}{t} \cdot \dfrac{{dt}}{2}} \]
\[ \Rightarrow 2\int {\dfrac{{dy}}{{\left( {1 + {y^2}} \right)}}} = \int {\dfrac{1}{t}dt} \]
Applying integration formula
\[ \Rightarrow 2{\tan ^{ - 1}}y = \log t + c\]
Putting \[1 + {x^2} = t\] in the above equation:
\[2{\tan ^{ - 1}}y = \log \left( {1 + {x^2}} \right) + c\]
Hence option A is the correct option.

Additional information
We add c after integration. c is an arbitrary constant or integration constant. The solution of an integration is known as general value when the value of c is unknown.
When we have the value of c, then the solution is known as particular solution.

Note: Students often do common mistake to integrating \[\int {\dfrac{x}{{\left( {1 + {x^2}} \right)}}dx} \]. When they used substitution method, forgot to multiply \[\dfrac{1}{2}\]. The correct substitution is \[\int {\dfrac{1}{t} \cdot \dfrac{{dt}}{2}} \] where \[1 + {x^2} = t\].