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

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

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

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

None of these

Solution

$We have \frac{d y}{d x} = \frac{y}{x} - \cos^{2} (\frac{y}{x})$
$Putting y = v x, so that \frac{d y}{d x} = v + x \frac{d v}{d x}, we get$
$v + x \frac{d v}{d x} = v - \cos^{2} v \Rightarrow \frac{d v}{\cos^{2} v} = - \frac{d x}{x}$
$\Rightarrow \sec^{2} v d v = - \frac{1}{x} d x$
On integration, we get
$\tan v = - \log x + \log C \Rightarrow \tan (\frac{y}{x}) = - \log x + \log C$
$This passes through (1, π / 4), therefore 1 = \log C$
$So, \tan (\frac{y}{x}) = - \log x + 1 \Rightarrow \tan (\frac{y}{x}) = - \log x + \log e$
$\Rightarrow y = x \tan^{- 1} (\log (\frac{e}{x}))$

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