Given cos x ( 3 sin x + cos x + 3 ) d y = ( 1 + y s i n x ( 3 sin x + cos x + 3 ) ) d x \cos x ( 3\sin x + \cos x + 3 ) dy = ( 1 + ysinx ( 3\sin x + \cos x + 3 ) ) dx cos x ( 3 sin x + cos x + 3 ) d y = ( 1 + ys in x ( 3 sin x + cos x + 3 )) d x
⇒ d y d x = ( 1 + y s i n x ( 3 sin x + cos x + 3 ) ) cos x ( 3 sin x + cos x + 3 ) ⇒ \dfrac{dy}{dx} = \dfrac{( 1 + ysinx ( 3\sin x + \cos x + 3 ) )}{\cos x ( 3\sin x + \cos x + 3 )} ⇒ d x d y = cos x ( 3 sin x + cos x + 3 ) ( 1 + ys in x ( 3 sin x + cos x + 3 ))
⇒ d y d x = 1 cos x ( 3 sin x + cos x + 3 ) + y s i n x ( 3 sin x + cos x + 3 ) ) cos x ( 3 sin x + cos x + 3 ) ⇒ \dfrac{dy}{dx} = \dfrac{1}{\cos x ( 3\sin x + \cos x + 3 )} + \dfrac{ysinx ( 3\sin x + \cos x + 3 ) )}{\cos x ( 3\sin x + \cos x + 3 )} ⇒ d x d y = cos x ( 3 sin x + cos x + 3 ) 1 + cos x ( 3 sin x + cos x + 3 ) ys in x ( 3 sin x + cos x + 3 ))
⇒ d y d x − ( tan x ) y = 1 ( 3 sin x + cos x + 3 ) cos x ⇒ \dfrac{dy}{dx} - ( \tan x ) y = \dfrac{1}{( 3\sin x + \cos x + 3 ) \cos x} ⇒ d x d y − ( tan x ) y = ( 3 sin x + cos x + 3 ) cos x 1
This is a linear differential equation of the type d y d x + P y = Q , \dfrac{dy}{dx} + Py = Q, d x d y + P y = Q , P = − tan x P = - \tan x P = − tan x Q = 1 cos x ( 3 sin x + cos x + 3 ) . Q = \dfrac{1}{\cos x\left(3\sin x + \cos x + 3\right)} . Q = cos x ( 3 sin x + cos x + 3 ) 1 .
Now, the integrating factor I . F . = e ∫ P d x = e ∫ − t a n x d x I . F . = {e}^{∫ Pdx} = {e}^{∫ - tanxdx} I . F . = e ∫ P d x = e ∫ − t an x d x
= e ln ∣ cos x ∣ = ∣ cos x ∣ = {e}^{\ln \left|\cos x\right|} = \left|\cos x\right| = e l n ∣ c o s x ∣ = ∣ cos x ∣
= cos x ∀ 0 ≤ x ≤ π 2 . = \cos x \forall 0 \le x \le \dfrac{\pi}{2} . = cos x ∀0 ≤ x ≤ 2 π .
The solution of the given differential equation is y ( I . F . ) = ∫ Q ( I . F . ) d x + C y\left(I . F .\right) = ∫ Q\left(I . F .\right)dx + C y ( I . F . ) = ∫ Q ( I . F . ) d x + C
⇒ y ( cos x ) = ∫ ( cos x ) ⋅ 1 cos x ( 3 sin x + cos x + 3 ) d x + C ⇒ y\left(\cos x\right) = ∫ ( \cos x ) \cdot \dfrac{1}{\cos x ( 3\sin x + \cos x + 3 )}dx + C ⇒ y ( cos x ) = ∫ ( cos x ) ⋅ cos x ( 3 sin x + cos x + 3 ) 1 d x + C
⇒ y ( cos x ) = ∫ d x 3 sin x + cos x + 3 d x + C ⇒ y\left(\cos x\right) = ∫ \dfrac{dx}{3\sin x + \cos x + 3}dx + C ⇒ y ( cos x ) = ∫ 3 sin x + cos x + 3 d x d x + C
Using sin 2 θ = 2 t a n θ 1 + tan 2 θ and cos 2 θ = 1 − tan 2 θ 1 + tan 2 θ \sin 2\theta = \dfrac{2tan\theta}{1 + \tan^{2}\theta} \text{ and } \cos 2\theta = \dfrac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} sin 2 θ = 1 + tan 2 θ 2 t an θ and cos 2 θ = 1 + tan 2 θ 1 − tan 2 θ
⇒ y ( cos x ) = ∫ 1 3 × 2 t a n x 2 1 + tan 2 x 2 + 1 − tan 2 x 2 1 + tan 2 x 2 + 3 d x + C ⇒ y\left(\cos x\right) = ∫ \dfrac{1}{3 \times \dfrac{2tan\dfrac{x}{2}}{1 + \tan^{2}\dfrac{x}{2}} + \dfrac{1 - \tan^{2}\dfrac{x}{2}}{1 + \tan^{2}\dfrac{x}{2}} + 3}dx + C ⇒ y ( cos x ) = ∫ 3 × 1 + tan 2 2 x 2 t an 2 x + 1 + tan 2 2 x 1 − tan 2 2 x + 3 1 d x + C
⇒ y ( cos x ) = ∫ 1 + tan 2 x 2 6 t a n x 2 + 1 − tan 2 x 2 + 3 + 3 tan 2 x 2 d x + C ⇒ y\left(\cos x\right) = ∫ \dfrac{1 + \tan^{2}\dfrac{x}{2}}{6tan\dfrac{x}{2} + 1 - \tan^{2}\dfrac{x}{2} + 3 + 3\tan^{2}\dfrac{x}{2}}dx + C ⇒ y ( cos x ) = ∫ 6 t an 2 x + 1 − tan 2 2 x + 3 + 3 tan 2 2 x 1 + tan 2 2 x d x + C
⇒ y ( cos x ) = ∫ ( sec 2 x 2 ) 2 tan 2 x 2 + 6 t a n x 2 + 4 d x + C ⇒ y\left(\cos x\right) = ∫ \dfrac{\left(\sec^{2}\dfrac{x}{2}\right)}{2\tan^{2}\dfrac{x}{2} + 6tan\dfrac{x}{2} + 4}dx + C ⇒ y ( cos x ) = ∫ 2 tan 2 2 x + 6 t an 2 x + 4 ( sec 2 2 x ) d x + C
Let I 1 = ∫ ( sec 2 x 2 ) 2 ( tan 2 x 2 + 3 t a n x 2 + 2 ) d x + C {I}_{1} = ∫ \dfrac{\left(\sec^{2}\dfrac{x}{2}\right)}{2\left(\tan^{2}\dfrac{x}{2} + 3tan\dfrac{x}{2} + 2\right)}dx + C I 1 = ∫ 2 ( tan 2 2 x + 3 t an 2 x + 2 ) ( sec 2 2 x ) d x + C
Put t a n x 2 = t ⇒ 1 2 sec 2 x 2 d x = d t tan\dfrac{x}{2} = t ⇒ \dfrac{1}{2}\sec^{2}\dfrac{x}{2}dx = dt t an 2 x = t ⇒ 2 1 sec 2 2 x d x = d t
⇒ I 1 = ∫ d t t 3 + 3 t + 2 = ∫ d t ( t + 2 ) ( t + 1 ) ⇒ {I}_{1} = ∫ \dfrac{dt}{{t}^{3} + 3t + 2} = ∫ \dfrac{dt}{\left(t + 2\right) ( t + 1 )} ⇒ I 1 = ∫ t 3 + 3 t + 2 d t = ∫ ( t + 2 ) ( t + 1 ) d t
⇒ I 1 = ∫ ( 1 t + 1 − 1 t + 2 ) d t ⇒ {I}_{1} = ∫ \left(\dfrac{1}{t + 1} - \dfrac{1}{t + 2}\right)dt ⇒ I 1 = ∫ ( t + 1 1 − t + 2 1 ) d t
⇒ I 1 = log e ( t + 1 ) − log e ( t + 2 ) ⇒ {I}_{1} = \log_{e}\left(t + 1\right) - \log_{e}\left(t + 2\right) ⇒ I 1 = log e ( t + 1 ) − log e ( t + 2 )
⇒ I 1 = log e ( t + 1 t + 2 ) ⇒ {I}_{1} = \log_{e}\left(\dfrac{t + 1}{t + 2}\right) ⇒ I 1 = log e ( t + 2 t + 1 )
⇒ I 1 = log e ∣ ( t a n x 2 + 1 t a n x 2 + 2 ) ∣ ⇒ {I}_{1} = \log_{e}\left|\left(\dfrac{tan\dfrac{x}{2} + 1}{tan\dfrac{x}{2} + 2}\right)\right| ⇒ I 1 = log e t an 2 x + 2 t an 2 x + 1
So, the solution is y ( cos x ) = log e ∣ 1 + t a n x 2 2 + t a n x 2 ∣ + C y\left(\cos x\right) = \log_{e}\left|\dfrac{1 + tan\dfrac{x}{2}}{2 + tan\dfrac{x}{2}}\right| + C y ( cos x ) = log e 2 + t an 2 x 1 + t an 2 x + C
⇒ y ( cos x ) = log e ( 1 + t a n x 2 2 + t a n x 2 ) + C ⇒ y\left(\cos x\right) = \log_{e}\left(\dfrac{1 + tan\dfrac{x}{2}}{2 + tan\dfrac{x}{2}}\right) + C ⇒ y ( cos x ) = log e 2 + t an 2 x 1 + t an 2 x + C 0 ≤ x ≤ π 2 . 0 \le x \le \dfrac{\pi}{2} . 0 ≤ x ≤ 2 π .
Now, it is given y ( 0 ) = 0 y ( 0 ) = 0 y ( 0 ) = 0
⇒ 0 = log e ( 1 2 ) + C ⇒ C = log e 2 ⇒ 0 = \log_{e}\left(\dfrac{1}{2}\right) + C ⇒ C = \log_{e}2 ⇒ 0 = log e ( 2 1 ) + C ⇒ C = log e 2
⇒ y ( cos x ) = log e ( 1 + t a n x 2 2 + t a n x 2 ) + log e 2 ⇒ y\left(\cos x\right) = \log_{e}\left(\dfrac{1 + tan\dfrac{x}{2}}{2 + tan\dfrac{x}{2}}\right) + \log_{e}2 ⇒ y ( cos x ) = log e 2 + t an 2 x 1 + t an 2 x + log e 2
Now, for x = π 3 , x = \dfrac{\pi}{3}, x = 3 π , y ( 1 2 ) = log e ( 1 + 1 3 2 + 1 3 ) + log e 2 y\left(\dfrac{1}{2}\right) = \log_{e}\left(\dfrac{1 + \dfrac{1}{\sqrt{3}}}{2 + \dfrac{1}{\sqrt{3}}}\right) + \log_{e}2 y ( 2 1 ) = log e 2 + 3 1 1 + 3 1 + log e 2
⇒ y ⋅ ( 1 2 ) = log e ( 3 + 1 2 3 + 1 ) + log e 2 ⇒ y \cdot \left(\dfrac{1}{2}\right) = \log_{e}\left(\dfrac{\sqrt{3} + 1}{2\sqrt{3} + 1}\right) + \log_{e}2 ⇒ y ⋅ ( 2 1 ) = log e ( 2 3 + 1 3 + 1 ) + log e 2
⇒ y ⋅ ( 1 2 ) = log e ( ( 3 + 1 2 3 + 1 ) × ( 2 3 − 1 2 3 − 1 ) ) + log e 2 ⇒ y \cdot \left(\dfrac{1}{2}\right) = \log_{e}\left(\left(\dfrac{\sqrt{3} + 1}{2\sqrt{3} + 1}\right) \times \left(\dfrac{2\sqrt{3} - 1}{2\sqrt{3} - 1}\right)\right) + \log_{e}2 ⇒ y ⋅ ( 2 1 ) = log e ( ( 2 3 + 1 3 + 1 ) × ( 2 3 − 1 2 3 − 1 ) ) + log e 2
⇒ y ⋅ ( 1 2 ) = log e ( 5 + 3 11 ) + log e 2 ⇒ y \cdot \left(\dfrac{1}{2}\right) = \log_{e}\left(\dfrac{5 + \sqrt{3}}{11}\right) + \log_{e}2 ⇒ y ⋅ ( 2 1 ) = log e ( 11 5 + 3 ) + log e 2
⇒ y = 2 log e ( 2 3 + 10 11 ) . ⇒ y = 2\log_{e}\left(\dfrac{2\sqrt{3} + 10}{11}\right) . ⇒ y = 2 log e ( 11 2 3 + 10 ) .
