First slide

Methods of solving first order first degree differential equations

Question

The slope of the tangent at (x, y) to a curve passing through $(1, \frac{π}{4})$ is given by $\frac{y}{x} - \cos^{2} (\frac{y}{x})$ , then the equation of the curve is

Moderate

A

$y = \tan^{- 1} (\ln (\frac{e}{x}))$
B

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

$y = x \tan^{- 1} (\ln (\frac{e}{x}))$
D

none

Solution

$\begin{array}{l} Given \Rightarrow s l o p e = \frac{d y}{d x} = \frac{y}{x} - \cos^{2} (\frac{y}{x}) \\ Put y = v x then \\ v + x \frac{d v}{d x} = v - \cos^{2} v \\ \Rightarrow x \frac{d v}{d x} = - \cos^{2} v \\ \Rightarrow - \sec^{2} d v = \frac{d x}{x} \\ Integrate, we get - \tan v = \ln x + c \\ \Rightarrow - Tan (\frac{y}{x}) = \ln x + c \end{array}$
$\begin{array}{l} It passes through (1, \frac{π}{4}) then - 1 = 0 + c \Rightarrow c = - 1 \\ ∴ - Tan (\frac{y}{x}) = \log_{e} x - 1 \\ \Rightarrow \tan \frac{y}{x} = 1 - \log_{e}^{x} \Rightarrow \tan \frac{y}{x} = 1 - \log_{e}^{x} = \log (\frac{e}{x}) \\ \Rightarrow y = x \tan^{- 1} (\ln (\frac{e}{x})) \end{array}$

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