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**Solve the differential equation: **

(1 + y^{2}) dx = (tan^{−1 }y − x) dy

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#### Solution

We have,

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

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

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

\[\text{Comparing with }\frac{dx}{dy} + Px = Q,\text{ we get}\]

\[P = \frac{1}{1 + y^2} \]

\[Q = \frac{\tan^{- 1} y}{1 + y^2}\]

Now,

\[I . F . = e^{\int\frac{1}{1 + y^2}dy} = e^{\tan^{- 1} y} \]

So, the solution is given by

\[x \times e^{\tan^{- 1} y} = \int\frac{\tan^{- 1} y}{1 + y^2} \times e^{\tan^{- 1} y} dy + C\]

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

Now,

\[I = \int\frac{\tan^{- 1} y}{1 + y^2} \times e^{\tan^{- 1} y} dy\]

\[\text{Putting }t = \tan^{- 1} y,\text{ we get}\]

\[dt = \frac{1}{1 + y^2}dy\]

\[\therefore I = \int\underset{I}{t}\times \underset{II}{e^t} \text{ }dt\]

\[ = t \times \int e^t dt - \int\left( \frac{d t}{d t} \times \int e^t dt \right)dt\]

\[ = t e^t - \int e^t dt\]

\[ = t e^t - e^t \]

\[ \therefore I = \tan^{- 1} y e^{\tan^{- 1} y} - e^{\tan^{- 1} y} \]

\[ = e^{\tan^{- 1} y} \left( \tan^{- 1} y - 1 \right)\]

Putting the value of `I` in (1), we get

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

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