Question

Find the solution of differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{{1 + {y^2}}}{{1 + {x^2}}}\].
A. \[1 + xy + c\left( {y + x} \right) = 0\]
B. \[x + y = c\left( {1 - xy} \right)\]
C. \[y - x = c\left( {1 + xy} \right)\]
D. \[1 + xy = c\left( {x + y} \right)\]

Answer 1

Hint: To find the solution of the differential equation use the variable separation method. There are x and y variables in the given differential equation. Separate x containing terms and y containing terms. Then integrate the equation to find a general solution.

Formula used:
\[\begin{array}{l}\int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x\\{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - yx}}} \right)\end{array}\]

Complete step by step solution:
The given differential equation is \[\dfrac{{dy}}{{dx}} = \dfrac{{1 + {y^2}}}{{1 + {x^2}}}\].
First multiply by \[(1 + {y^2})dx\] on both sides of the equation for variable separation.
\[\dfrac{1}{{1 + {y^2}}}dy = \dfrac{1}{{1 + {x^2}}}dx\]
Now integrate the equation.
\[\begin{array}{l}\int {\dfrac{1}{{1 + {y^2}}}dy} = \int {\dfrac{1}{{1 + {x^2}}}dx} \\{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}c\end{array}\]
Simplify the equation as follows.
\[{\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + c}}{{1 - cx}}} \right)\]
Further simplify the equation by eliminating \[{\tan ^{ - 1}}\]
\[y = \dfrac{{x + c}}{{1 - cx}}\]
Now multiply both sides by \[\left( {1 - cx} \right)\]
\[y\left( {1 - cx} \right) = x + c\]
Simplify further to get the general solution.
\[\begin{array}{l}y - cxy = x + c\\y - x = c + cxy\\y - x = c\left( {1 + xy} \right)\end{array}\]
So, the general solution is \[y - x = c\left( {1 + xy} \right)\].
Hence Option C is the correct answer.

Note:The common mistake happens while solving this question is writing \[{\tan ^{ - 1}}\left( {x + c} \right)\] from \[{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}c\], Which is wrong. Also while integrating \[\dfrac{1}{{1 + {x^2}}}\] use the formula of \[{\tan ^{ - 1}}x\].