First slide

Methods of solving first order first degree differential equations

Question

The solution of the differential equation $(x - y a r c \tan \frac{y}{x}) d x + (x a r c T a n \frac{y}{x}) d y = 0$ is of the form $\log |\frac{x^{4}}{x^{2} + y^{2}}| + k \tan^{- 1} \frac{y}{x}$ = c then the value of k is

Moderate

A

$\frac{x}{y}$
B

$\frac{y}{x}$
C

$\frac{2 x}{y}$
D

$\frac{2 y}{x}$

Solution

The given equation can rewritten as
$\begin{array}{l} \frac{d y}{d x} = \frac{y \tan^{- 1} \frac{y}{x} - x}{x \tan^{- 1} \frac{y}{x}} = \frac{y}{x} - \frac{1}{\tan^{- 1} y / x} \\ v + x \frac{d v}{d x} = v - \frac{1}{\tan^{- 1} v} (on substituting y = v x) \\ \Rightarrow \int \tan^{- 1} v d v = - \int \frac{d x}{x} \\ v \tan^{- 1} v - \int v \frac{1}{1 + v^{2}} d v + \ln x = c (b y p a r t s) \\ \Rightarrow \ln x - \frac{1}{2} \int \frac{2 v}{1 + v^{2}} d v + \frac{y}{x} \tan^{- 1} (\frac{y}{x}) = c \\ \Rightarrow 2 \ln x - \ln |1 + v^{2}| + \frac{2 y}{x} \tan^{- 1} (\frac{y}{x}) = c \end{array}$
$\begin{array}{l} \Rightarrow 2 \ln x - \ln |\frac{x^{2} + y^{2}}{x^{2}}| + \frac{2 y}{x} \tan^{- 1} (\frac{y}{x}) = c \\ \Rightarrow \ln |\frac{x^{4}}{x^{2} + y^{2}}| + \frac{2 y}{x} \tan^{- 1} (\frac{y}{x}) = c \end{array}$

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