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JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

The general solution of the differential equation $(tan^{- 1} y - x) d y = (1 + y^{2}) d x$ is

Choose the correct answer:

Explanation

1. Rearrange the Equation

Rewrite the equation to isolate $\frac{dx}{dy}$ :

$\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1 + y^2}$

$\frac{dx}{dy} = \frac{\tan^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$

Now, move the term containing $x$ to the left side:

$\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2}$

2. Identify the Linear Form

This is a first-order linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$ , where:

$P(y) = \frac{1}{1 + y^2}$

$Q(y) = \frac{\tan^{-1} y}{1 + y^2}$

3. Find the Integrating Factor (I.F.)

$I.F. = e^{\int P(y) \, dy} = e^{\int \frac{1}{1 + y^2} \, dy}$

$I.F. = e^{\tan^{-1} y}$

4. General Solution Formula

The general solution is given by:

$x \cdot (I.F.) = \int Q(y) \cdot (I.F.) \, dy + C$

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

5. Evaluate the Integral

To solve $\int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} \, dy$ , use substitution:

Let $t = \tan^{-1} y$ , then $dt = \frac{1}{1 + y^2} dy$ .

The integral becomes:

$\int t e^t \, dt$

Using integration by parts ( $\int u dv = uv - \int v du$ ):

$u = t \implies du = dt$

$dv = e^t dt \implies v = e^t$

$\int t e^t \, dt = t e^t - \int e^t \, dt = t e^t - e^t = (t - 1)e^t$

Substitute back $t = \tan^{-1} y$ :

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

6. Final General Solution

Substitute the integral back into the solution equation:

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

Divide the entire equation by $e^{\tan^{-1} y}$ (or multiply by $e^{-\tan^{-1} y}$ ):

$x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}$

Result:

The general solution is:

$x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}$

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