First slide

Methods of solving first order first degree differential equations

Question

$The solution of the equation \cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x, | x | < \frac{π}{4} when y (π / 6) = \frac{3 \sqrt{3}}{8} is$

Moderate

A

${y=tan2xcos}^{2} x$
B

${y=cot2x.cos}^{2} x$
C

$y= \frac{1}{2} \tan 2 {xcos}^{2} x$
D

$y= \frac{1}{2} \cot 2 {xcos}^{2} x$

Solution

$\begin{array}{l} \frac{d y}{d x} - \frac{\tan 2 x}{\cos^{2} x} y = \cos^{2} x \\ I.E. = e^{- \int \frac{\tan 2 x}{\cos^{2} x} dx} dx = e^{- \int \frac{2 \tan x}{1 - \tan^{2} x} s e c^{2} x d x} \\ p u t \tan x = t, s e c^{2} x d x = d t \\ = e^{\int \frac{- 2 t}{1 - t^{2}} d t} = e^{\log (1 - t^{2})} = 1 - \tan^{2} x \\ Solution of D.E is y (1 - \tan^{2} x) = \int (1 - \tan^{2} x) \cos^{2} x dx \\ y (1 - \frac{\sin^{2} x}{\cos^{2} x}) = \int (1 - \frac{\sin^{2} x}{\cos^{2} x}) \cos^{2} x d x \\ \frac{y \cos 2 x}{\cos^{2} x} = \int \cos 2 x d x \\ \frac{y \cos 2 x}{\cos^{2} x} = \frac{\sin 2 x}{2} + C \\ When x = π / 6, y = \frac{3 \sqrt{3}}{8} \\ \frac{\frac{3 \sqrt{3}}{8} \cos \frac{π}{3}}{\cos^{2} \frac{π}{6}} = \frac{\sin \frac{π}{3}}{2} + C \\ \frac{3 \sqrt{3}}{8} \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{\frac{\sqrt{3}}{2}}{2} + C \\ \Rightarrow C = 0 \\ \frac{y \cos 2 x}{\cos^{2} x} = \frac{\sin 2 x}{2} \\ y = \frac{1}{2} \tan 2 {xcos}^{2} x \end{array}$

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