First slide

Methods of solving first order first degree differential equations

Question

The solution to the differential equation $\sin (x \frac{dy}{dx}) \cos y = \frac{dy}{dx} + \sin ycos (x \frac{dy}{dx})$ is

Difficult

A

y =0
B

${cx}^{2} - y = \sin^{- 1} x$
C

$cx - y = \sin^{- 1} c$
D

$y = \sqrt{x^{2} - 1} - \sin^{- 1} \frac{\sqrt{x^{2} - 1}}{x}$

Solution

Given equation is $\sin (x \frac{dy}{dx} - y) = \frac{dy}{dx}$
$\Rightarrow y = x \frac{dy}{dx} - \sin^{- 1} (\frac{dy}{dx})$
Again differentiating we get
$\begin{array}{l} \frac{dy}{dx} = \frac{dy}{dx} + x \frac{d^{2} y}{{dx}^{2}} - \frac{1}{\sqrt{1 - {(\frac{dy}{dx})}^{2}}} \frac{d^{2} y}{{dx}^{2}} \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 0 or x = \frac{1}{\sqrt{1 - {(\frac{dy}{dx})}^{2}}} \\ \Rightarrow \frac{dy}{dx} = c or {(\frac{dy}{dx})}^{2} = 1 - \frac{1}{x^{2}} \end{array}$
using, $\frac{dy}{dx} = c in g i v e n e q u a t i o n we get y = cx - \sin^{- 1} c$
Also for particular value of c = 0,y = 0 is also a solution.
Finally using ${(\frac{dy}{dx})}^{2} = 1 - \frac{1}{x^{2}}$ in (i) we get
$i.e., y = \sqrt{x^{2} - 1} - \sin^{- 1} (\frac{\sqrt{x^{2} - 1}}{x})$

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