Consider the differential equation $\dfrac{dy}{dx} + \dfrac{3}{\cos^2 x}\,y = \dfrac{1}{\cos^2 x}$, for $x\in\!\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)$. Given $y\!\left(\dfrac{\pi}{4}\right)=\dfrac{4}{3}$, find $y\!\left(-\dfrac{\pi}{4}\right)$.
This is a linear ODE with $P = 3\sec^2 x$ and $Q = \sec^2 x$.
Integrating factor: $e^{\int 3\sec^2 x\,dx} = e^{3\tan x}$
$\frac{d}{dx}(ye^{3\tan x}) = \sec^2 x \cdot e^{3\tan x}$
Integrate: $ye^{3\tan x} = \frac{1}{3}e^{3\tan x} + C$
At $x=\pi/4$, $\tan(\pi/4)=1$, $y=4/3$: $\frac{4}{3}e^3 = \frac{1}{3}e^3 + C \Rightarrow C = e^3$
So $y = \frac{1}{3} + e^3 \cdot e^{-3\tan x}$
At $x=-\pi/4$, $\tan(-\pi/4)=-1$: $y = \frac{1}{3} + e^3 \cdot e^{3} = \frac{1}{3} + e^3$... wait: $e^{-3(-1)}=e^3$, so $y = \frac{1}{3}+e^3 \cdot e^3 = \frac{1}{3}+e^6$.
Correct: $y(-\pi/4) = \mathbf{\frac{1}{3}+e^6}$. So correct option index is 0.
Sign in to join the conversation and share your thoughts.
Log In to Comment