Given differential equation
We need to solve
d y d x + 1 x 2 − 1 y = ( x − 1 x + 1 ) 1 / 2 , x > 1 \frac{dy}{dx}+\frac{1}{x^2-1}y=\left(\frac{x-1}{x+1}\right)^{1/2}, \qquad x>1 d x d y + x 2 − 1 1 y = ( x + 1 x − 1 ) 1/2 , x > 1 with condition
y ( 2 ) = 1 3 = 1 3 . y(2)=\sqrt{\frac13}=\frac{1}{\sqrt3}. y ( 2 ) = 3 1 = 3 1 . This is a linear differential equation of the 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 ) , where
P ( x ) = 1 x 2 − 1 = 1 ( x − 1 ) ( x + 1 ) , Q ( x ) = x − 1 x + 1 . P(x)=\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)},
\qquad
Q(x)=\sqrt{\frac{x-1}{x+1}}. P ( x ) = x 2 − 1 1 = ( x − 1 ) ( x + 1 ) 1 , Q ( x ) = x + 1 x − 1 .
Find the integrating factor
We compute
∫ P ( x ) d x = ∫ 1 x 2 − 1 d x . \int P(x)\,dx=\int \frac{1}{x^2-1}\,dx. ∫ P ( x ) d x = ∫ x 2 − 1 1 d x . Using partial fractions,
1 x 2 − 1 = 1 ( x − 1 ) ( x + 1 ) = 1 2 ( 1 x − 1 − 1 x + 1 ) . \frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}=\frac12\left(\frac{1}{x-1}-\frac{1}{x+1}\right). x 2 − 1 1 = ( x − 1 ) ( x + 1 ) 1 = 2 1 ( x − 1 1 − x + 1 1 ) . Hence,
∫ 1 x 2 − 1 d x = 1 2 ∫ ( 1 x − 1 − 1 x + 1 ) d x = 1 2 ln ( x − 1 x + 1 ) . \int \frac{1}{x^2-1}\,dx
=\frac12\int\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx
=\frac12\ln\left(\frac{x-1}{x+1}\right). ∫ x 2 − 1 1 d x = 2 1 ∫ ( x − 1 1 − x + 1 1 ) d x = 2 1 ln ( x + 1 x − 1 ) . Since x > 1 x>1 x > 1
Therefore the integrating factor is
I . F . = e ∫ P ( x ) d x = e 1 2 ln ( x − 1 x + 1 ) = ( x − 1 x + 1 ) 1 / 2 . I.F.=e^{\int P(x)dx}=e^{\frac12\ln\left(\frac{x-1}{x+1}\right)}
=\left(\frac{x-1}{x+1}\right)^{1/2}. I . F . = e ∫ P ( x ) d x = e 2 1 l n ( x + 1 x − 1 ) = ( x + 1 x − 1 ) 1/2 .
Multiply the equation by the integrating factor
Multiplying throughout,
( x − 1 x + 1 ) 1 / 2 d y d x + 1 x 2 − 1 ( x − 1 x + 1 ) 1 / 2 y = ( x − 1 x + 1 ) . \left(\frac{x-1}{x+1}\right)^{1/2}\frac{dy}{dx}
+\frac{1}{x^2-1}\left(\frac{x-1}{x+1}\right)^{1/2}y
=\left(\frac{x-1}{x+1}\right). ( x + 1 x − 1 ) 1/2 d x d y + x 2 − 1 1 ( x + 1 x − 1 ) 1/2 y = ( x + 1 x − 1 ) . The left side becomes
d d x [ y ( x − 1 x + 1 ) 1 / 2 ] . \frac{d}{dx}\left[y\left(\frac{x-1}{x+1}\right)^{1/2}\right]. d x d [ y ( x + 1 x − 1 ) 1/2 ] . So,
d d x [ y ( x − 1 x + 1 ) 1 / 2 ] = x − 1 x + 1 . \frac{d}{dx}\left[y\left(\frac{x-1}{x+1}\right)^{1/2}\right]=\frac{x-1}{x+1}. d x d [ y ( x + 1 x − 1 ) 1/2 ] = x + 1 x − 1 .
Integrate both sides
y ( x − 1 x + 1 ) 1 / 2 = ∫ x − 1 x + 1 d x + C . y\left(\frac{x-1}{x+1}\right)^{1/2}=\int \frac{x-1}{x+1}\,dx + C. y ( x + 1 x − 1 ) 1/2 = ∫ x + 1 x − 1 d x + C . Now,
x − 1 x + 1 = 1 − 2 x + 1 . \frac{x-1}{x+1}=1-\frac{2}{x+1}. x + 1 x − 1 = 1 − x + 1 2 . Thus,
∫ x − 1 x + 1 d x = ∫ ( 1 − 2 x + 1 ) d x = x − 2 ln ( x + 1 ) . \int \frac{x-1}{x+1}\,dx=\int \left(1-\frac{2}{x+1}\right)dx
=x-2\ln(x+1). ∫ x + 1 x − 1 d x = ∫ ( 1 − x + 1 2 ) d x = x − 2 ln ( x + 1 ) . Hence,
y ( x − 1 x + 1 ) 1 / 2 = x − 2 ln ( x + 1 ) + C . y\left(\frac{x-1}{x+1}\right)^{1/2}=x-2\ln(x+1)+C. y ( x + 1 x − 1 ) 1/2 = x − 2 ln ( x + 1 ) + C . So,
y = ( x + 1 x − 1 ) 1 / 2 ( x − 2 ln ( x + 1 ) + C ) . y=\left(\frac{x+1}{x-1}\right)^{1/2}\bigl(x-2\ln(x+1)+C\bigr). y = ( x − 1 x + 1 ) 1/2 ( x − 2 ln ( x + 1 ) + C ) .
Use the initial condition
Given y ( 2 ) = 1 3 y(2)=\frac{1}{\sqrt3} y ( 2 ) = 3 1
Substitute x = 2 x=2 x = 2
1 3 = ( 3 1 ) 1 / 2 ( 2 − 2 ln 3 + C ) = 3 ( 2 − 2 ln 3 + C ) . \frac{1}{\sqrt3}=\left(\frac{3}{1}\right)^{1/2}\bigl(2-2\ln 3 + C\bigr)
=\sqrt3\,(2-2\ln 3 + C). 3 1 = ( 1 3 ) 1/2 ( 2 − 2 ln 3 + C ) = 3 ( 2 − 2 ln 3 + C ) . Therefore,
1 3 = 2 − 2 ln 3 + C . \frac{1}{3}=2-2\ln 3 + C. 3 1 = 2 − 2 ln 3 + C . So,
C = 1 3 − 2 + 2 ln 3 = 2 ln 3 − 5 3 . C=\frac13-2+2\ln 3=2\ln 3-\frac53. C = 3 1 − 2 + 2 ln 3 = 2 ln 3 − 3 5 . Thus,
y = ( x + 1 x − 1 ) 1 / 2 ( x − 2 ln ( x + 1 ) + 2 ln 3 − 5 3 ) . y=\left(\frac{x+1}{x-1}\right)^{1/2}
\left(x-2\ln(x+1)+2\ln 3-\frac53\right). y = ( x − 1 x + 1 ) 1/2 ( x − 2 ln ( x + 1 ) + 2 ln 3 − 3 5 ) .
Find y ( 8 ) y(8) y ( 8 )
At x = 8 x=8 x = 8
y ( 8 ) = ( 9 7 ) 1 / 2 ( 8 − 2 ln 9 + 2 ln 3 − 5 3 ) . y(8)=\left(\frac{9}{7}\right)^{1/2}
\left(8-2\ln 9+2\ln 3-\frac53\right). y ( 8 ) = ( 7 9 ) 1/2 ( 8 − 2 ln 9 + 2 ln 3 − 3 5 ) . Since ln 9 = 2 ln 3 \ln 9=2\ln 3 ln 9 = 2 ln 3
− 2 ln 9 + 2 ln 3 = − 4 ln 3 + 2 ln 3 = − 2 ln 3. -2\ln 9+2\ln 3=-4\ln 3+2\ln 3=-2\ln 3. − 2 ln 9 + 2 ln 3 = − 4 ln 3 + 2 ln 3 = − 2 ln 3. Also,
8 − 5 3 = 24 − 5 3 = 19 3 . 8-\frac53=\frac{24-5}{3}=\frac{19}{3}. 8 − 3 5 = 3 24 − 5 = 3 19 . Hence,
y ( 8 ) = 3 7 ( 19 3 − 2 ln 3 ) . y(8)=\frac{3}{\sqrt7}\left(\frac{19}{3}-2\ln 3\right). y ( 8 ) = 7 3 ( 3 19 − 2 ln 3 ) . Therefore,
7 y ( 8 ) = 3 ( 19 3 − 2 ln 3 ) = 19 − 6 ln 3. \sqrt7\,y(8)=3\left(\frac{19}{3}-2\ln 3\right)=19-6\ln 3. 7 y ( 8 ) = 3 ( 3 19 − 2 ln 3 ) = 19 − 6 ln 3.
Compare with options
Thus,
7 y ( 8 ) = 19 − 6 log e 3. \sqrt7\,y(8)=19-6\log_e 3. 7 y ( 8 ) = 19 − 6 log e 3. This matches Option D .
Comparison with stored correct answer
Stored correct answer: D
Our derived answer: D
So they agree.