$\text { Find the solution of the differential equation } \frac { d r } { d t } = \frac { \sec ^ { 2 } t } { \tan t + 1 } \text { which passes through the }$ point $( \pi , 5 )$
A) $r = \ln | \tan t + 1 | + 5$
B) $r = 2 \ln | \sec t + 1 |$
C) $r = \ln | \tan t + 1 |$
D) $r = 5 \ln | \tan t + 1 |$
E) $r = 2 \ln | \sin t | + 5$

Question

Quizplus · Accepted Answer

We start by separating the variables and integrating both sides to get:

$$\int \frac{dr}{\sec^2t + \tan t} = \int dt$$

Using the identity $\sec^2t = 1+\tan^2t$, we can rewrite the denominator as $\tan t + 1$, giving:

$$\int \frac{dr}{\tan t + 1} = \int dt$$

To integrate the left side, we use the substitution $u = \tan t + 1$:

$$\int \frac{dr}{\tan t + 1} = \int \frac{du}{u} = \ln |u| + C_1$$

where $C_1$ is the constant of integration.

Substituting back $u = \tan t + 1$, we get:

$$\ln | \tan t + 1 | + C_1$$

Using the initial condition $(\pi, 5)$, we can find $C_1$:

$$5 = \ln | \tan \pi + 1 | + C_1 \Rightarrow C_1 = 5$$

Therefore, the solution is:

$$r = \ln | \tan t + 1 | + 5$$

which is option A.

Point (π,5)( \pi , 5 )(π,5) A) r=ln⁡∣tan⁡t+1∣+5r = \ln | \tan t + 1 | + 5r=ln∣tant+1∣+5

Point $( \pi , 5 )$ A) $r = \ln | \tan t + 1 | + 5$