Solve the differential equation with the given initial condition.
-$y ^ { \prime } = \tan t \sec ^ { 2 } t ; y ( 0 ) = 1$
A) y = ln $| \tan \mathrm { t } |$ + 1
B) y = $\frac { \tan ^ { 2 } \mathrm { t } } { 2 }$ + 1
C) y = tan t + 1
D) y = $\frac { \sec ^ { 2 } t } { 3 }$ + $\frac { 2 } { 3 }$

Question

Solve the differential equation with the given initial condition. 
-$y ^ { \prime } = \tan t \sec ^ { 2 } t ; y ( 0 ) = 1$
A) y = ln $| \tan \mathrm { t } |$ + 1
B) y = $\frac { \tan ^ { 2 } \mathrm { t } } { 2 }$ + 1
C) y = tan t + 1
D) y = $\frac { \sec ^ { 2 } t } { 3 }$ + $\frac { 2 } { 3 }$

Quizplus · Accepted Answer

To solve the differential equation $y' = \tan t \sec^2 t$, we integrate the right-hand side with respect to $t$. The integral of $\tan t \sec^2 t$ is $\frac{\tan^2 t}{2} + C$, where $C$ is the constant of integration. Using the initial condition $y(0) = 1$, we find $C$ by substituting $t = 0$ into the solution, which gives us $C = 1$. Therefore, the solution is $y = \frac{\tan^2 t}{2} + 1$.

Solve the Differential Equation with the Given Initial Condition y′=tan⁡tsec⁡2t;y(0)=1y ^ { \prime } = \tan t \sec ^ { 2 } t ; y ( 0 ) = 1y′=tantsec2t;y(0)=1

Solve the Differential Equation with the Given Initial Condition $y ^ { \prime } = \tan t \sec ^ { 2 } t ; y ( 0 ) = 1$