Given differential equation
d y d x + y tan x = 2 x + x 2 tan x , x ∈ ( − π 2 , π 2 ) \frac{dy}{dx}+y\tan x=2x+x^2\tan x, \qquad x\in\left(-\frac\pi2,\frac\pi2\right) d x d y + y tan x = 2 x + x 2 tan x , x ∈ ( − 2 π , 2 π )
with initial condition
y ( 0 ) = 1. y(0)=1. y ( 0 ) = 1.
Solve the linear differential equation
Rewrite in standard form:
y ′ + ( tan x ) y = 2 x + x 2 tan x . y'+(\tan x)y=2x+x^2\tan x. y ′ + ( tan x ) y = 2 x + x 2 tan x .
The integrating factor is
I.F. = e ∫ tan x d x = e − ln ( cos x ) = sec x . \text{I.F.}=e^{\int \tan x\,dx}=e^{-\ln(\cos x)}=\sec x. I.F. = e ∫ t a n x d x = e − l n ( c o s x ) = sec x .
Multiply throughout by the integrating factor
sec x y ′ + y sec x tan x = 2 x sec x + x 2 sec x tan x . \sec x\, y' + y\sec x\tan x = 2x\sec x + x^2\sec x\tan x. sec x y ′ + y sec x tan x = 2 x sec x + x 2 sec x tan x .
Left side becomes:
d d x ( y sec x ) = 2 x sec x + x 2 sec x tan x . \frac{d}{dx}(y\sec x)=2x\sec x+x^2\sec x\tan x. d x d ( y sec x ) = 2 x sec x + x 2 sec x tan x .
Notice that
d d x ( x 2 sec x ) = 2 x sec x + x 2 sec x tan x . \frac{d}{dx}(x^2\sec x)=2x\sec x+x^2\sec x\tan x. d x d ( x 2 sec x ) = 2 x sec x + x 2 sec x tan x .
Hence,
d d x ( y sec x ) = d d x ( x 2 sec x ) . \frac{d}{dx}(y\sec x)=\frac{d}{dx}(x^2\sec x). d x d ( y sec x ) = d x d ( x 2 sec x ) .
Integrating,
y sec x = x 2 sec x + C . y\sec x=x^2\sec x+C. y sec x = x 2 sec x + C .
Therefore,
y = x 2 + C cos x . y=x^2+C\cos x. y = x 2 + C cos x .
Use the initial condition
y ( 0 ) = 0 2 + C cos 0 = C = 1. y(0)=0^2+C\cos 0=C=1. y ( 0 ) = 0 2 + C cos 0 = C = 1.
So,
y = x 2 + cos x . \boxed{y=x^2+\cos x}. y = x 2 + cos x .
Differentiate
y ′ = 2 x − sin x . y'=2x-\sin x. y ′ = 2 x − sin x .
Evaluate at x = ± π 4 x=\pm \frac\pi4 x = ± 4 π
Since
cos π 4 = 1 2 , sin π 4 = 1 2 , sin ( − π 4 ) = − 1 2 , \cos\frac\pi4=\frac{1}{\sqrt2}, \qquad \sin\frac\pi4=\frac{1}{\sqrt2}, \qquad \sin\left(-\frac\pi4\right)=-\frac{1}{\sqrt2}, cos 4 π = 2 1 , sin 4 π = 2 1 , sin ( − 4 π ) = − 2 1 ,
we get
y ( π 4 ) = ( π 4 ) 2 + 1 2 = π 2 16 + 1 2 , y\left(\frac\pi4\right)=\left(\frac\pi4\right)^2+\frac{1}{\sqrt2}=\frac{\pi^2}{16}+\frac{1}{\sqrt2}, y ( 4 π ) = ( 4 π ) 2 + 2 1 = 16 π 2 + 2 1 ,
y ( − π 4 ) = ( π 4 ) 2 + 1 2 = π 2 16 + 1 2 . y\left(-\frac\pi4\right)=\left(\frac\pi4\right)^2+\frac{1}{\sqrt2}=\frac{\pi^2}{16}+\frac{1}{\sqrt2}. y ( − 4 π ) = ( 4 π ) 2 + 2 1 = 16 π 2 + 2 1 .
Hence,
y ( π 4 ) − y ( − π 4 ) = 0 , y\left(\frac\pi4\right)-y\left(-\frac\pi4\right)=0, y ( 4 π ) − y ( − 4 π ) = 0 ,
so A is false .
Also,
y ( π 4 ) + y ( − π 4 ) = 2 ( π 2 16 + 1 2 ) = π 2 8 + 2 , y\left(\frac\pi4\right)+y\left(-\frac\pi4\right)=2\left(\frac{\pi^2}{16}+\frac{1}{\sqrt2}\right)=\frac{\pi^2}{8}+\sqrt2, y ( 4 π ) + y ( − 4 π ) = 2 ( 16 π 2 + 2 1 ) = 8 π 2 + 2 ,
which is not
π 2 2 + 2. \frac{\pi^2}{2}+2. 2 π 2 + 2.
So C is false .
Evaluate derivatives
y ′ ( π 4 ) = 2 ⋅ π 4 − 1 2 = π 2 − 1 2 , y'\left(\frac\pi4\right)=2\cdot\frac\pi4-\frac{1}{\sqrt2}=\frac\pi2-\frac{1}{\sqrt2}, y ′ ( 4 π ) = 2 ⋅ 4 π − 2 1 = 2 π − 2 1 ,
y ′ ( − π 4 ) = 2 ⋅ ( − π 4 ) − ( − 1 2 ) = − π 2 + 1 2 . y'\left(-\frac\pi4\right)=2\cdot\left(-\frac\pi4\right)-\left(-\frac{1}{\sqrt2}\right)=-\frac\pi2+\frac{1}{\sqrt2}. y ′ ( − 4 π ) = 2 ⋅ ( − 4 π ) − ( − 2 1 ) = − 2 π + 2 1 .
Therefore,
y ′ ( π 4 ) − y ′ ( − π 4 ) = ( π 2 − 1 2 ) − ( − π 2 + 1 2 ) y'\left(\frac\pi4\right)-y'\left(-\frac\pi4\right)=\left(\frac\pi2-\frac{1}{\sqrt2}\right)-\left(-\frac\pi2+\frac{1}{\sqrt2}\right) y ′ ( 4 π ) − y ′ ( − 4 π ) = ( 2 π − 2 1 ) − ( − 2 π + 2 1 )
= π − 2 2 = π − 2 . =\pi-\frac{2}{\sqrt2}=\pi-\sqrt2. = π − 2 2 = π − 2 .
So B is true .
Next,
y ′ ( π 4 ) + y ′ ( − π 4 ) = ( π 2 − 1 2 ) + ( − π 2 + 1 2 ) = 0. y'\left(\frac\pi4\right)+y'\left(-\frac\pi4\right)=\left(\frac\pi2-\frac{1}{\sqrt2}\right)+\left(-\frac\pi2+\frac{1}{\sqrt2}\right)=0. y ′ ( 4 π ) + y ′ ( − 4 π ) = ( 2 π − 2 1 ) + ( − 2 π + 2 1 ) = 0.
So D is false .
Conclusion
The only correct option is
B . \boxed{\text{B}}. B .