The solution of the differential equation

(1 + y²)dx = (tan^-1 y - x)dy is

Concept:

The standard form of a first-order linear differential equation is,

Form 1: 

\(\frac{{dy}}{{dx}} + Py = Q\)

Where P and Q are the functions of x.

Integrating factor, \(IF = {e^{\smallint Pdx}}\)

Now, the solution for the above differential equation is,

\(y\left( {IF} \right) = \smallint IF.Qdx+C\)

Form 2: 

\(\frac{{dx}}{{dy}} + Px = Q\)

Where P and Q are the functions of y.

Integrating factor, \(IF = {e^{\smallint Pdy}}\)

Now, the solution for the above differential equation is,

\(x\left( {IF} \right) = \smallint IF.Qdy+C\)

Calculation:

The given differential equation is,

(1 + y²) dx = (tan^-1 y - x) dy

\( \Rightarrow \frac{{dx}}{{dy}} = \frac{1}{{1 + {y^2}}}\left( {{{\tan }^{ - 1}}y - x} \right)\)

\( \Rightarrow \frac{{dx}}{{dy}} + \frac{x}{{1 + {y^2}}} = \frac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}}\)

Now, it is in the standard form of a first-order linear differential equation.

Here, \(P = \frac{1}{{1 + {y^2}}},Q = \frac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}}\)

Integrating factor:

\( = {e^{\smallint \frac{1}{{1 + {y^2}}}dy}} = {e^{{{\tan }^{ - 1}}y}}\)

Now, the solution of the given differential equation is

\(x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \smallint {e^{{{\tan }^{ - 1}}y}}\frac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}}dy + C\)

Let \({\tan ^{ - 1}}y = t\)

\( \Rightarrow \frac{1}{{1 + {y^2}}}dy = dt\)

Now, the solution becomes

\(x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \smallint t{e^t}dt + C\)

\( \Rightarrow x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = t{e^t} - e^t + C\)

\( \Rightarrow x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {e^t}(t-1) + C\)

\( \Rightarrow x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {e^{\tan^{-1}y}}(\tan^{-1}y-1) + C\)

\( \Rightarrow x = {\tan ^{ - 1}}y - 1 + C{e^{ - {{\tan }^{ - 1}}y}}\)