First slide

Methods of solving first order first degree differential equations

Question

$The general solution of \frac{d y}{d x} = 2 y \tan x + \tan^{2} x is$

Moderate

A

${ycos}^{2} x= \frac{x}{2} - \frac{1}{4} sin2x+c$
B

${ysin}^{2} x= \frac{x}{2} - \frac{cos2x}{4} +c$
C

${ycos}^{2} x= \frac{x}{4} - \frac{1}{2} \sin 2 x+c$
D

${ysin}^{2} x= \frac{x}{4} - \frac{1}{2} \cos 2 x+c$

Solution

$\frac{d y}{d x} = 2 y \frac{\sin x}{\cos x} + \frac{\sin^{2} x}{\cos^{2} x}$
$\begin{array}{l} \cos^{2} x \frac{dy}{dx} - ysin 2 x = \sin^{2} x \\ \to d (\cos^{2} x \cdot y) = \sin^{2} x = \frac{1}{2} (1 - \cos 2 x) \\ \int d (\cos^{2} x \cdot y) = \frac{1}{2} \int 1 d x - \int \cos 2 x dx \\ \Rightarrow {ycos}^{2} x = \frac{x}{2} - \frac{1}{4} \sin 2 x + c \end{array}$

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