Given, differential equation
a m p ; ( x cos x ) d y + ( x y sin x + y cos x − 1 ) d x = 0 , 0 l t ; x l t ; π 2 a m p ; ⇒ d y d x + x y sin x + y cos x − 1 x cos x = 0 a m p ; ⇒ d y d x + ( x y sin x + y cos x x cos x ) = 1 x cos x a m p ; ⇒ d y d x + ( tan x + 1 x ) y = 1 x cos x \begin{aligned}& (x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2} \\\\& \Rightarrow \frac{d y}{d x}+\frac{x y \sin x+y \cos x-1}{x \cos x}=0 \\\\& \Rightarrow \frac{d y}{d x}+\left(\frac{x y \sin x+y \cos x}{x \cos x}\right)=\frac{1}{x \cos x} \\\\& \Rightarrow \frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{1}{x \cos x}\end{aligned} am p ; ( x cos x ) d y + ( x y sin x + y cos x − 1 ) d x = 0 , 0 am p ; ⇒ d x d y + x cos x x y sin x + y cos x − 1 = 0 am p ; ⇒ d x d y + ( x cos x x y sin x + y cos x ) = x cos x 1 am p ; ⇒ d x d y + ( tan x + x 1 ) y = x cos x 1 lt ; x lt ; 2 π Which is linear differential equation in the form of
a m p ; d y d x + P y = Q a m p ; ∴ I F = e ∫ ( tan x + 1 x ) d x = e ( log sec x + log x ) = e log ( x sec x ) = x sec x \begin{aligned}& \frac{d y}{d x}+P y=Q \\\\& \therefore \mathrm{IF}=e^{\int\left(\tan x+\frac{1}{x}\right) d x}=e^{(\log \sec x+\log x)}=e^{\log (x \sec x)}=x \sec x\end{aligned} am p ; d x d y + P y = Q am p ; ∴ IF = e ∫ ( t a n x + x 1 ) d x = e ( l o g s e c x + l o g x ) = e l o g ( x s e c x ) = x sec x ∴ \therefore ∴ The general solution of the given differential equation
y ⋅ I F = ∫ ( Q × I F ) d x + c ⇒ y ( x sec x ) = ∫ ( 1 x cos x x sec x ) d x + c ⇒ x y sec x = ∫ sec 2 x d x + c ⇒ x y sec x = tan x + c . . . . . . . . ( i ) \begin{array}{rlrl}y \cdot \mathrm{IF} =\int(Q \times \mathrm{IF}) d x+c \\\\\Rightarrow y(x \sec x) =\int\left(\frac{1}{x \cos x} x \sec x\right) d x+c \\\\\Rightarrow x y \sec x =\int \sec ^2 x d x+c \\\\\Rightarrow x y \sec x =\tan x+c ........(i)\end{array} y ⋅ IF = ∫ ( Q × IF ) d x + c ⇒ y ( x sec x ) = ∫ ( x c o s x 1 x sec x ) d x + c ⇒ x y sec x = ∫ sec 2 x d x + c ⇒ x y sec x = tan x + c ........ ( i ) Since,
π 3 y ( π 3 ) = 3 ⇒ y ( π 3 ) = 3 3 π \frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow y\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{\pi} 3 π y ( 3 π ) = 3 ⇒ y ( 3 π ) = π 3 3 a m p ; ∴ π 3 ( 3 3 π ) sec ( π 3 ) = tan ( π 3 ) + c a m p ; ⇒ 3 ( 2 ) = 3 + c a m p ; ⇒ a m p ; c = 3 \begin{aligned}& \therefore \frac{\pi}{3}\left(\frac{3 \sqrt{3}}{\pi}\right) \sec \left(\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{3}\right)+c \\\\& \Rightarrow \sqrt{3}(2)=\sqrt{3}+c \\\\& \begin{aligned}\Rightarrow & c=\sqrt{3}\end{aligned} \end{aligned} am p ; ∴ 3 π ( π 3 3 ) sec ( 3 π ) = tan ( 3 π ) + c am p ; ⇒ 3 ( 2 ) = 3 + c am p ; ⇒ am p ; c = 3 On putting the value of
c c c in Eq. (i), we get
a m p ; x y sec x = tan x + 3 a m p ; ⇒ y = 1 x ( sin x + 3 cos x ) a m p ; y ′ = 1 x ( cos x − 3 sin x ) − 1 x 2 ( sin x + 3 cos x ) \begin{aligned}&x y \sec x =\tan x+\sqrt{3} \\\\&\Rightarrow y =\frac{1}{x}(\sin x+\sqrt{3} \cos x) \\\\& y^{\prime} =\frac{1}{x}(\cos x-\sqrt{3} \sin x)-\frac{1}{x^2}(\sin x+\sqrt{3} \cos x)\end{aligned} am p ; x y sec x = tan x + 3 am p ; ⇒ y = x 1 ( sin x + 3 cos x ) am p ; y ′ = x 1 ( cos x − 3 sin x ) − x 2 1 ( sin x + 3 cos x ) a m p ; and y ′ ′ = 1 x ( − sin x − 3 cos x ) − 1 x 2 ( cos x − 3 sin x ) − 1 x 2 ( cos x − 3 sin x ) a m p ; − 2 x 3 ( cos x − 3 sin x ) a m p ; ∴ ∣ π 6 y ′ ′ ( π 6 ) + 2 y ′ ( π 6 ) ∣ = ∣ − 2 ∣ = 2 a m p ; \begin{aligned} & \text { and } y^{\prime \prime}=\frac{1}{x}(-\sin x-\sqrt{3} \cos x)-\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) -\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) \\ & -\frac{2}{x^3}(\cos x-\sqrt{3} \sin x) \\\\ & \therefore\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|=|-2|=2 \\\\ & \end{aligned} am p ; and y ′′ = x 1 ( − sin x − 3 cos x ) − x 2 1 ( cos x − 3 sin x ) − x 2 1 ( cos x − 3 sin x ) am p ; − x 3 2 ( cos x − 3 sin x ) am p ; ∴ 6 π y ′′ ( 6 π ) + 2 y ′ ( 6 π ) = ∣ − 2∣ = 2 am p ;