Solve the initial value problem.
-$$\frac { d y } { d t } = e ^ { - t } \sec ^ { 2 } \left( \pi e ^ { - t } \right) , y ( - \ln 5 ) = \frac { 1 } { \pi }$$
A) $y = \frac { \tan \left( \pi \mathrm { e } ^ { - \mathrm { t } } \right) + 6 } { \pi }$
B) $y = \frac { - e ^ { - t } \cot \left( \pi e ^ { - t } \right) + 0 } { \pi }$
C) $y = \cot \left( \pi e ^ { - t } \right) + 1$
D) $y = \frac { - \tan \left( \pi e ^ { - t } \right) + 1 } { \pi }$

Question

Quizplus · Accepted Answer

First, we can simplify the right-hand side of the differential equation by using the identity $\sec^2 	heta = 1 + 	an^2 	heta$. Then we have:
$$\frac{dy}{dt} = e^{-t} (1 + 	an^2 (\pi e^{-t})) = e^{-t} + e^{-t} 	an^2 (\pi e^{-t})$$
Now we need to integrate both sides with respect to $t$. On the left side, we get:
$$y(t) = \int \frac{dy}{dt} dt = \int e^{-t} + e^{-t} 	an^2 (\pi e^{-t}) dt = -e^{-t} - \frac{1}{\pi} 	an(\pi e^{-t}) + C$$
where $C$ is the constant of integration. To find $C$, we use the initial condition $y(-\ln 5) = \frac{1}{\pi}$:
$$\frac{1}{\pi} = -e^{\ln 5} - \frac{1}{\pi} 	an(\pi e^{\ln 5}) + C$$
Simplifying gives $C = \frac{1}{\pi} + e^{\ln 5} + \frac{1}{\pi} 	an(\pi e^{\ln 5})$. Plugging this back into the equation for $y(t)$, we get:
$$y(t) = -e^{-t} - \frac{1}{\pi} 	an(\pi e^{-t}) + \left(\frac{1}{\pi} + e^{\ln 5} + \frac{1}{\pi} 	an(\pi e^{\ln 5})ight)$$
Simplifying further, we get:
$$y(t) = \frac{1}{\pi} - e^{-t} - \frac{1}{\pi} 	an(\pi e^{-t}) + \frac{1}{\pi} 	an(\pi e^{\ln 5})$$
Using the fact that $e^{\ln 5} = 5$, we can simplify the last term to $\frac{1}{\pi} 	an(\pi 5)$. Finally, we can simplify the second term by using the identity $	an(\pi - 	heta) = -	an(	heta)$:
$$y(t) = \frac{1}{\pi} - e^{-t} + \frac{1}{\pi} 	an(\pi e^{-t} - \pi)$$
Since $	an(\pi - 	heta) = -	an(	heta)$, this is equivalent to:
$$y(t) = \frac{1}{\pi} - e^{-t} - \frac{1}{\pi} 	an(\pi e^{-t})$$
which matches the form of choice D. Therefore, the answer is D.

Solve the Initial Value Problem dydt=e−tsec⁡2(πe−t),y(−ln⁡5)=1π\frac { d y } { d t } = e ^ { - t } \sec ^ { 2 } \left( \pi e ^ { - t } \right) , y ( - \ln 5 ) = \frac { 1 } { \pi }dtdy​=e−tsec2(πe−t),y(−ln5)=π1​

Solve the Initial Value Problem $\frac { d y } { d t } = e ^ { - t } \sec ^ { 2 } \left( \pi e ^ { - t } \right) , y ( - \ln 5 ) = \frac { 1 } { \pi }$