Solving linear ODE using separation of variables

Calculus Level 2

Find the general solution of the following linear differential equation: y ′ + y tan ⁡ x = sin ⁡ 2 x {y}^{\prime} + y\tan x = \sin 2x

Use c c as the constant of integration.

y ( x ) = c tan ⁡ x − 2 sin ⁡ x y(x)=c \tan x - 2 \sin x y ( x ) = c cos ⁡ x − 2 cos ⁡ 2 x y(x)=c \cos x - 2 {\cos}^{2} x y ( x ) = c sin ⁡ x − tan ⁡ 2 x y(x)=c \sin x - {\tan}^{2} x y ( x ) = c tan ⁡ x − 2 sin ⁡ 2 x y(x)=c \tan x - 2 {\sin}^{2} x

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Lawahar Qasid by Lawahar Qasid

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1 solution

I'm not sure if this counts as separable variables, but is how I solved it: y ′ + y tan ⁡ ( x ) = 2 sin ⁡ ( x ) ⋅ cos ⁡ ( x ) ⇒ sec ⁡ ( x ) y ′ + sec ⁡ ( x ) ⋅ tan ⁡ ( x ) y = 2 sin ⁡ ( x ) y'+y\tan(x)=2\sin(x)\cdot \cos(x) \Rightarrow \sec(x)y'+\sec(x)\cdot \tan(x) y=2\sin(x) ⇒ d d x ( sec ⁡ ( x ) y ) = 2 sin ⁡ ( x ) ⇒ sec ⁡ ( x ) y = − 2 cos ⁡ ( x ) + C \Rightarrow \frac{d}{dx}(\sec(x)y)=2\sin(x) \Rightarrow \sec(x)y=-2\cos(x)+C ⇒ y = C cos ⁡ ( x ) − cos ⁡ 2 ( x ) \Rightarrow y= C\cos(x)-\cos^{2}(x)

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