Simplify the differential equation
Given
d y d x = tan x + y sin x ( sec x − sin x tan x ) , x ∈ ( 0 , π 2 ) \frac{dy}{dx}=\frac{\tan x+y}{\sin x(\sec x-\sin x\tan x)}, \qquad x\in\left(0,\frac\pi2\right) d x d y = sin x ( sec x − sin x tan x ) tan x + y , x ∈ ( 0 , 2 π ) First simplify the denominator:
sec x − sin x tan x = 1 cos x − sin x ⋅ sin x cos x = 1 − sin 2 x cos x = cos 2 x cos x = cos x \sec x-\sin x\tan x=\frac1{\cos x}-\sin x\cdot\frac{\sin x}{\cos x}
=\frac{1-\sin^2 x}{\cos x}=\frac{\cos^2 x}{\cos x}=\cos x sec x − sin x tan x = cos x 1 − sin x ⋅ cos x sin x = cos x 1 − sin 2 x = cos x cos 2 x = cos x So,
sin x ( sec x − sin x tan x ) = sin x cos x \sin x(\sec x-\sin x\tan x)=\sin x\cos x sin x ( sec x − sin x tan x ) = sin x cos x Hence the DE becomes
d y d x = tan x + y sin x cos x \frac{dy}{dx}=\frac{\tan x+y}{\sin x\cos x} d x d y = sin x cos x tan x + y Now,
tan x sin x cos x = sin x / cos x sin x cos x = 1 cos 2 x = sec 2 x \frac{\tan x}{\sin x\cos x}=\frac{\sin x/\cos x}{\sin x\cos x}=\frac1{\cos^2 x}=\sec^2 x sin x cos x tan x = sin x cos x sin x / cos x = cos 2 x 1 = sec 2 x Therefore,
d y d x = sec 2 x + y sin x cos x \frac{dy}{dx}=\sec^2 x+\frac{y}{\sin x\cos x} d x d y = sec 2 x + sin x cos x y or
d y d x − 1 sin x cos x y = sec 2 x \frac{dy}{dx}-\frac{1}{\sin x\cos x}y=\sec^2 x d x d y − sin x cos x 1 y = sec 2 x This is a linear differential equation.
Write in standard linear form
d y d x + P ( x ) y = Q ( x ) \frac{dy}{dx}+P(x)y=Q(x) d x d y + P ( x ) y = Q ( x ) with
P ( x ) = − 1 sin x cos x , Q ( x ) = sec 2 x P(x)=-\frac{1}{\sin x\cos x}, \qquad Q(x)=\sec^2 x P ( x ) = − sin x cos x 1 , Q ( x ) = sec 2 x The integrating factor is
I.F. = e ∫ P ( x ) d x = e − ∫ d x sin x cos x \text{I.F.}=e^{\int P(x)\,dx}=e^{-\int \frac{dx}{\sin x\cos x}} I.F. = e ∫ P ( x ) d x = e − ∫ s i n x c o s x d x Now,
1 sin x cos x = tan x + cot x \frac{1}{\sin x\cos x}=\tan x+\cot x sin x cos x 1 = tan x + cot x So,
∫ d x sin x cos x = ∫ ( tan x + cot x ) d x \int \frac{dx}{\sin x\cos x}=\int (\tan x+\cot x)\,dx ∫ sin x cos x d x = ∫ ( tan x + cot x ) d x = ∫ tan x d x + ∫ cot x d x =\int \tan x\,dx+\int \cot x\,dx = ∫ tan x d x + ∫ cot x d x = − ln ( cos x ) + ln ( sin x ) = ln ( tan x ) =-\ln(\cos x)+\ln(\sin x)=\ln(\tan x) = − ln ( cos x ) + ln ( sin x ) = ln ( tan x ) Thus,
I.F. = e − ln ( tan x ) = 1 tan x = cot x \text{I.F.}=e^{-\ln(\tan x)}=\frac{1}{\tan x}=\cot x I.F. = e − l n ( t a n x ) = tan x 1 = cot x
Multiply the DE by the integrating factor
Multiplying by cot x \cot x cot x
cot x d y d x − cot x ⋅ 1 sin x cos x y = cot x sec 2 x \cot x\frac{dy}{dx}-\cot x\cdot\frac{1}{\sin x\cos x}y=\cot x\sec^2 x cot x d x d y − cot x ⋅ sin x cos x 1 y = cot x sec 2 x The left side becomes
d d x ( y cot x ) \frac{d}{dx}(y\cot x) d x d ( y cot x ) And the right side simplifies as
cot x sec 2 x = cos x sin x ⋅ 1 cos 2 x = 1 sin x cos x \cot x\sec^2 x=\frac{\cos x}{\sin x}\cdot\frac{1}{\cos^2 x}=\frac{1}{\sin x\cos x} cot x sec 2 x = sin x cos x ⋅ cos 2 x 1 = sin x cos x 1 So,
d d x ( y cot x ) = 1 sin x cos x \frac{d}{dx}(y\cot x)=\frac{1}{\sin x\cos x} d x d ( y cot x ) = sin x cos x 1 Integrating,
y cot x = ∫ d x sin x cos x + C = ln ( tan x ) + C y\cot x=\int \frac{dx}{\sin x\cos x}+C=\ln(\tan x)+C y cot x = ∫ sin x cos x d x + C = ln ( tan x ) + C Hence,
y = tan x [ ln ( tan x ) + C ] y=\tan x\,[\ln(\tan x)+C] y = tan x [ ln ( tan x ) + C ]
Use the initial condition
Given
y ( π 4 ) = 2 y\left(\frac\pi4\right)=2 y ( 4 π ) = 2 Since
tan π 4 = 1 , ln 1 = 0 \tan\frac\pi4=1, \qquad \ln 1=0 tan 4 π = 1 , ln 1 = 0 we get
2 = 1 ⋅ ( 0 + C ) 2=1\cdot(0+C) 2 = 1 ⋅ ( 0 + C ) so
C = 2 C=2 C = 2 Therefore the solution is
y = tan x ( ln ( tan x ) + 2 ) y=\tan x\,(\ln(\tan x)+2) y = tan x ( ln ( tan x ) + 2 )
Find y ( π 3 ) y\left(\frac\pi3\right) y ( 3 π )
Now,
tan π 3 = 3 \tan\frac\pi3=\sqrt3 tan 3 π = 3 Thus,
y ( π 3 ) = 3 ( ln ( 3 ) + 2 ) y\left(\frac\pi3\right)=\sqrt3\left(\ln(\sqrt3)+2\right) y ( 3 π ) = 3 ( ln ( 3 ) + 2 ) So,
y ( π 3 ) = 3 ( 2 + ln 3 ) \boxed{y\left(\frac\pi3\right)=\sqrt3\left(2+\ln\sqrt3\right)} y ( 3 π ) = 3 ( 2 + ln 3 )
Evaluate the options
A: 3 ( 2 + ln 3 ) \sqrt3(2+\ln 3) 3 ( 2 + ln 3 ) B: 3 ( 1 + 2 ln 3 ) \sqrt3(1+2\ln 3) 3 ( 1 + 2 ln 3 ) C: 3 ( 2 + ln 3 ) \sqrt3(2+\ln\sqrt3) 3 ( 2 + ln 3 ) D: 3 2 ( 2 + ln 3 ) \dfrac{\sqrt3}{2}(2+\ln 3) 2 3 ( 2 + ln 3 ) ln 3 = 1 2 ln 3 \ln\sqrt3=\frac12\ln 3 ln 3 = 2 1 ln 3 3 ( 2 + ln 3 ) = 3 ( 2 + 1 2 ln 3 ) \sqrt3\left(2+\ln\sqrt3\right)=\sqrt3\left(2+\frac12\ln 3\right) 3 ( 2 + ln 3 ) = 3 ( 2 + 2 1 ln 3 )
Hence the correct option is C .