Question

What is the solution of \[\left( {1 + {x^2}} \right)\dfrac{{dy}}{{dx}} = 1\]?
A. \[y = \log \left( {1 + {x^2}} \right) + c\]
B. \[y + \log \left( {1 + {x^2}} \right) + c = 0\]
C. \[y - \log \left( {1 + {x^2}} \right) = c\]
D. \[y = {\tan ^{ - 1}}x + c\]

Answer 1

Hint: First we will separate x and y variables in the given equation. Then we will take integration on both sides of the equation and apply the integration formula to get the desired result.

Formula used:
Integration formula
\[\int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x + c\]
\[\int {dx} = x + c\]
Where c is the integration constant.

Complete step by step solution:
Given differential equation is
\[\left( {1 + {x^2}} \right)\dfrac{{dy}}{{dx}} = 1\]
Separate x and y variables from the differential equation
\[dy = \dfrac{{dx}}{{\left( {1 + {x^2}} \right)}}\]
Integrating both sides of the equation
\[\int {dy} = \int {\dfrac{{dx}}{{\left( {1 + {x^2}} \right)}}} \]
Apply the integration formula
\[y = {\tan ^{ - 1}}x + c\]
Hence option D is the correct option.

Additional information:
We add a constant after performing integration. Integration is also known as antiderivative. When we derivate a constant term that returns zero. By the operation of antiderivative or integration, we cannot obtain the constant term. For this reason, we add an integration constant or arbitrary constant after performing integration.

Note: We can solve the question by using a substitution method.
Assume that \[x = \tan \theta \]\[ \Rightarrow dx = {\sec ^2}\theta d\theta \]
Putting \[x = \tan \theta \] and \[dx = {\sec ^2}\theta d\theta \] in the given differential equation
\[dy = \dfrac{{{{\sec }^2}\theta d\theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\]
\[dy = \dfrac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^2}\theta }}\]
\[dy = d\theta \]
Integrating both sides
\[\int {dy} = \int {d\theta } \]
\[y = \theta + c\]
Now putting \[\theta = {\tan ^{ - 1}}x\]
\[y = {\tan ^{ - 1}}x + c\]