Solve $\frac { d y } { d t } = y ^ { 2 } + 1$ if y = 1 when t = 0.
A) $y = \arctan \left( t + \frac { \pi } { 4 } \right)$
B) $y = \arctan \left( t - \frac { \pi } { 4 } \right)$
C) $y = \tan \left( t + \frac { \pi } { 4 } \right)$
D) $y = \tan \left( t - \frac { \pi } { 4 } \right)$

Question

Quizplus · Accepted Answer

This differential equation can be solved using separation of variables. Rearranging, we get $\frac{dy}{y^2+1} = dt$. Integrating both sides, we obtain $\arctan(y) = t + C$, where $C$ is the constant of integration. Given $y(0) = 1$, we find $C = \frac{\pi}{4}$. Therefore, $y = \tan(t + \frac{\pi}{4})$, making option C correct.

Solve dydt=y2+1\frac { d y } { d t } = y ^ { 2 } + 1dtdy​=y2+1

Solve $\frac { d y } { d t } = y ^ { 2 } + 1$