Given differential equation
( sin 2 2 x ) d y d x + ( 8 sin 2 2 x + 2 sin 4 x ) y = 2 e − 4 x ( 2 sin 2 x + cos 2 x ) (\sin^2 2x)\frac{dy}{dx} + (8\sin^2 2x + 2\sin 4x)y = 2e^{-4x}(2\sin 2x + \cos 2x) ( sin 2 2 x ) d x d y + ( 8 sin 2 2 x + 2 sin 4 x ) y = 2 e − 4 x ( 2 sin 2 x + cos 2 x ) for x ∈ ( 0 , π / 2 ) x\in (0,\pi/2) x ∈ ( 0 , π /2 )
y ( π 4 ) = e − π . y\left(\frac{\pi}{4}\right)=e^{-\pi}. y ( 4 π ) = e − π . We need to find y ( π 6 ) y\left(\frac{\pi}{6}\right) y ( 6 π )
Put the equation in linear form
Divide throughout by sin 2 2 x \sin^2 2x sin 2 2 x
d y d x + 8 sin 2 2 x + 2 sin 4 x sin 2 2 x y = 2 e − 4 x ( 2 sin 2 x + cos 2 x ) sin 2 2 x . \frac{dy}{dx} + \frac{8\sin^2 2x + 2\sin 4x}{\sin^2 2x}y
= \frac{2e^{-4x}(2\sin 2x + \cos 2x)}{\sin^2 2x}. d x d y + sin 2 2 x 8 sin 2 2 x + 2 sin 4 x y = sin 2 2 x 2 e − 4 x ( 2 sin 2 x + cos 2 x ) . Now simplify the coefficient of y y y
Using
sin 4 x = 2 sin 2 x cos 2 x , \sin 4x = 2\sin 2x\cos 2x, sin 4 x = 2 sin 2 x cos 2 x , we get
2 sin 4 x sin 2 2 x = 4 sin 2 x cos 2 x sin 2 2 x = 4 cot 2 x . \frac{2\sin 4x}{\sin^2 2x} = \frac{4\sin 2x\cos 2x}{\sin^2 2x} = 4\cot 2x. sin 2 2 x 2 sin 4 x = sin 2 2 x 4 sin 2 x cos 2 x = 4 cot 2 x . Hence,
d y d x + ( 8 + 4 cot 2 x ) y = 2 e − 4 x ( 2 sin 2 x + cos 2 x ) sin 2 2 x . \frac{dy}{dx} + (8 + 4\cot 2x)y
= \frac{2e^{-4x}(2\sin 2x + \cos 2x)}{\sin^2 2x}. d x d y + ( 8 + 4 cot 2 x ) y = sin 2 2 x 2 e − 4 x ( 2 sin 2 x + cos 2 x ) . So the linear differential equation is
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 ) , where
P ( x ) = 8 + 4 cot 2 x . P(x)=8+4\cot 2x. P ( x ) = 8 + 4 cot 2 x .
Find the integrating factor
The integrating factor is
I.F. = e ∫ ( 8 + 4 cot 2 x ) d x . \text{I.F.} = e^{\int (8+4\cot 2x)dx}. I.F. = e ∫ ( 8 + 4 c o t 2 x ) d x . Now,
∫ 8 d x = 8 x , \int 8\,dx = 8x, ∫ 8 d x = 8 x , and
∫ 4 cot 2 x d x = 4 ⋅ 1 2 ln ( sin 2 x ) = 2 ln ( sin 2 x ) , \int 4\cot 2x\,dx = 4\cdot \frac{1}{2}\ln(\sin 2x)=2\ln(\sin 2x), ∫ 4 cot 2 x d x = 4 ⋅ 2 1 ln ( sin 2 x ) = 2 ln ( sin 2 x ) , since
∫ cot ( a x ) d x = 1 a ln ( sin a x ) . \int \cot(ax)dx = \frac{1}{a}\ln(\sin ax). ∫ cot ( a x ) d x = a 1 ln ( sin a x ) . Therefore,
I.F. = e 8 x + 2 ln ( sin 2 x ) = e 8 x sin 2 2 x . \text{I.F.} = e^{8x+2\ln(\sin 2x)} = e^{8x}\sin^2 2x. I.F. = e 8 x + 2 l n ( s i n 2 x ) = e 8 x sin 2 2 x .
Apply the integrating factor
Multiplying the differential equation by e 8 x sin 2 2 x e^{8x}\sin^2 2x e 8 x sin 2 2 x
e 8 x sin 2 2 x d y d x + e 8 x sin 2 2 x ( 8 + 4 cot 2 x ) y = e 8 x sin 2 2 x ⋅ 2 e − 4 x ( 2 sin 2 x + cos 2 x ) sin 2 2 x . e^{8x}\sin^2 2x\frac{dy}{dx} + e^{8x}\sin^2 2x(8+4\cot 2x)y
= e^{8x}\sin^2 2x \cdot \frac{2e^{-4x}(2\sin 2x+\cos 2x)}{\sin^2 2x}. e 8 x sin 2 2 x d x d y + e 8 x sin 2 2 x ( 8 + 4 cot 2 x ) y = e 8 x sin 2 2 x ⋅ sin 2 2 x 2 e − 4 x ( 2 sin 2 x + cos 2 x ) . The left-hand side becomes
d d x ( y e 8 x sin 2 2 x ) . \frac{d}{dx}\left(y e^{8x}\sin^2 2x\right). d x d ( y e 8 x sin 2 2 x ) . The right-hand side simplifies to
2 e 4 x ( 2 sin 2 x + cos 2 x ) . 2e^{4x}(2\sin 2x+\cos 2x). 2 e 4 x ( 2 sin 2 x + cos 2 x ) . So,
d d x ( y e 8 x sin 2 2 x ) = 2 e 4 x ( 2 sin 2 x + cos 2 x ) . \frac{d}{dx}\left(y e^{8x}\sin^2 2x\right)=2e^{4x}(2\sin 2x+\cos 2x). d x d ( y e 8 x sin 2 2 x ) = 2 e 4 x ( 2 sin 2 x + cos 2 x ) .
Recognize the right-hand side as a derivative
Let us check:
d d x ( e 4 x sin 2 x ) = 4 e 4 x sin 2 x + 2 e 4 x cos 2 x = 2 e 4 x ( 2 sin 2 x + cos 2 x ) . \frac{d}{dx}(e^{4x}\sin 2x)=4e^{4x}\sin 2x + 2e^{4x}\cos 2x
= 2e^{4x}(2\sin 2x+\cos 2x). d x d ( e 4 x sin 2 x ) = 4 e 4 x sin 2 x + 2 e 4 x cos 2 x = 2 e 4 x ( 2 sin 2 x + cos 2 x ) . Hence,
d d x ( y e 8 x sin 2 2 x ) = d d x ( e 4 x sin 2 x ) . \frac{d}{dx}\left(y e^{8x}\sin^2 2x\right)=\frac{d}{dx}(e^{4x}\sin 2x). d x d ( y e 8 x sin 2 2 x ) = d x d ( e 4 x sin 2 x ) . Integrating,
y e 8 x sin 2 2 x = e 4 x sin 2 x + C . y e^{8x}\sin^2 2x = e^{4x}\sin 2x + C. y e 8 x sin 2 2 x = e 4 x sin 2 x + C . Therefore,
y = e 4 x sin 2 x + C e 8 x sin 2 2 x . y = \frac{e^{4x}\sin 2x + C}{e^{8x}\sin^2 2x}. y = e 8 x sin 2 2 x e 4 x sin 2 x + C .
Use the initial condition
Given
y ( π 4 ) = e − π . y\left(\frac{\pi}{4}\right)=e^{-\pi}. y ( 4 π ) = e − π . At x = π 4 x=\frac{\pi}{4} x = 4 π
sin 2 x = sin π 2 = 1 , \sin 2x = \sin \frac{\pi}{2}=1, sin 2 x = sin 2 π = 1 , so
e − π ⋅ e 8 ⋅ π / 4 ⋅ 1 2 = e 4 ⋅ π / 4 ⋅ 1 + C . e^{-\pi}\cdot e^{8\cdot \pi/4}\cdot 1^2 = e^{4\cdot \pi/4}\cdot 1 + C. e − π ⋅ e 8 ⋅ π /4 ⋅ 1 2 = e 4 ⋅ π /4 ⋅ 1 + C . That is,
e − π e 2 π = e π + C . e^{-\pi}e^{2\pi}=e^{\pi}+C. e − π e 2 π = e π + C . Since e − π e 2 π = e π e^{-\pi}e^{2\pi}=e^{\pi} e − π e 2 π = e π
e π = e π + C ⟹ C = 0. e^{\pi}=e^{\pi}+C \implies C=0. e π = e π + C ⟹ C = 0. Thus,
y e 8 x sin 2 2 x = e 4 x sin 2 x , y e^{8x}\sin^2 2x = e^{4x}\sin 2x, y e 8 x sin 2 2 x = e 4 x sin 2 x , which gives
y = e 4 x sin 2 x e 8 x sin 2 2 x = e − 4 x sin 2 x . y = \frac{e^{4x}\sin 2x}{e^{8x}\sin^2 2x}
= \frac{e^{-4x}}{\sin 2x}. y = e 8 x sin 2 2 x e 4 x sin 2 x = sin 2 x e − 4 x .
Find y ( π 6 ) y\left(\frac{\pi}{6}\right) y ( 6 π )
Substitute x = π 6 x=\frac{\pi}{6} x = 6 π
y ( π 6 ) = e − 4 π / 6 sin ( π / 3 ) = e − 2 π / 3 3 / 2 = 2 3 e − 2 π / 3 . y\left(\frac{\pi}{6}\right)=\frac{e^{-4\pi/6}}{\sin(\pi/3)}
=\frac{e^{-2\pi/3}}{\sqrt{3}/2}
=\frac{2}{\sqrt{3}}e^{-2\pi/3}. y ( 6 π ) = sin ( π /3 ) e − 4 π /6 = 3 /2 e − 2 π /3 = 3 2 e − 2 π /3 .
Compare with options
This matches:
2 3 e − 2 π / 3 \boxed{\frac{2}{\sqrt{3}}e^{-2\pi/3}} 3 2 e − 2 π /3 So the correct option is A .
Comparison with stored correct answer
Stored correct answer: A
Our derived answer: A
Hence, they agree.