What is the solution of the differential equation \[\dfrac{{dy}}{{dx}} = {\left( {x + y} \right)^2}\]?
A. \[x + y + \tan \left( {x + c} \right) = 0\]
B. \[x - y + \tan \left( {x + c} \right) = 0\]
C. \[x + y - \tan \left( {x + c} \right) = 0\]
D. None of these

Hint: The given equation is a homogeneous equation. Thus to solve the given equation we will use the substitution method. We will substitute the value of x + y as t and then simplify it to get the required solution.

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

Complete step by step solution:
Given differential equation is
\[\dfrac{{dy}}{{dx}} = {\left( {x + y} \right)^2}\]
Since the degree of each term of the right side expression is 2, it is a homogeneous equation.
Assume that, \[x + y = t\]
Differentiate both sides with respect to x:
\[1 + \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} - 1\]
Putting \[x + y = t\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} - 1\] in the given differential equation:
\[\dfrac{{dt}}{{dx}} - 1 = {t^2}\]
Separate the variables of the above equation:
\[ \Rightarrow \dfrac{{dt}}{{dx}} = {t^2} + 1\]
\[ \Rightarrow \dfrac{{dt}}{{{t^2} + 1}} = dx\]
Taking integration on both sides:
\[ \Rightarrow \int {\dfrac{{dt}}{{{t^2} + 1}}} = \int {dx} \]
Now applying integration formula
\[ \Rightarrow {\tan ^{ - 1}}t = x + c\]
Substitute the value of t in above equation:
\[ \Rightarrow {\tan ^{ - 1}}\left( {x + y} \right) = x + c\]
Taking tangent on both sides of the equation:
\[ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {x + y} \right)} \right] = \tan \left( {x + c} \right)\]
Applying the formula \[{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta \] on the left side expression:
\[ \Rightarrow x + y = \tan \left( {x + c} \right)\]
\[ \Rightarrow x + y - \tan \left( {x + c} \right) = 0\]
Hence option C is the correct option.

Note: Students often forget to add c after integration. c is the integration constant. Integration is also known as antiderivative. We are unable to get any constant term after performing integration. Thus we add c after performing integration.