To solve the differential equation [tex]\(\left(1+e^{2 \theta}\right) d \rho+2 \rho e^{2 \theta} d \theta=0\)[/tex], let's proceed step-by-step.
First, we rewrite the given differential equation:
[tex]\[
(1+e^{2 \theta}) \, d \rho + 2 \rho e^{2 \theta} \, d \theta = 0.
\][/tex]
This form suggests a method of separation of variables or recognizing it as an exact differential equation. However, we'll try to separate the variables first.
Let's isolate [tex]\(d \rho\)[/tex]:
[tex]\[
(1 + e^{2 \theta}) \, d \rho = -2 \rho e^{2 \theta} \, d \theta.
\][/tex]
Next, divide both sides by [tex]\((1 + e^{2 \theta}) \rho\)[/tex]:
[tex]\[
\frac{d \rho}{\rho} = -\frac{2 e^{2 \theta}}{1 + e^{2 \theta}} \, d \theta.
\][/tex]
We now have the variables separated. Integrate both sides:
[tex]\[
\int \frac{1}{\rho} \, d \rho = \int -\frac{2 e^{2 \theta}}{1 + e^{2 \theta}} \, d \theta.
\][/tex]
The left side integrates to:
[tex]\[
\ln |\rho|.
\][/tex]
Now, for the right side, we simplify the integrand using a substitution. Let [tex]\(u = 1 + e^{2 \theta}\)[/tex]. Then [tex]\(du = 2 e^{2 \theta} d \theta\)[/tex].
This transforms the right side integral into:
[tex]\[
\int -\frac{2 e^{2 \theta}}{1 + e^{2 \theta}} \, d \theta = \int -\frac{1}{u} \, du.
\][/tex]
Integrate the right side:
[tex]\[
\int -\frac{1}{u} \, du = -\ln |u|.
\][/tex]
Substituting back [tex]\(u = 1 + e^{2 \theta}\)[/tex]:
[tex]\[
-\ln |1 + e^{2 \theta}|.
\][/tex]
Combine the results of the integrals:
[tex]\[
\ln |\rho| = -\ln |1 + e^{2 \theta}| + C.
\][/tex]
Exponentiate both sides to solve for [tex]\(\rho\)[/tex]:
[tex]\[
|\rho| = e^{-\ln |1 + e^{2 \theta}| + C} = e^C e^{-\ln |1 + e^{2 \theta}|} \approx k \cdot \frac{1}{1 + e^{2 \theta}}.
\][/tex]
Given that the absolute value of [tex]\(\rho\)[/tex] can be positive or negative, we remove the absolute value and include a constant factor [tex]\(k\)[/tex]:
[tex]\[
\rho = \frac{k}{1 + e^{2 \theta}}.
\][/tex]
To match with the provided options, we simplify the form:
[tex]\[
\rho (1 + e^{2 \theta}) = k.
\][/tex]
Thus, the correct solution is:
[tex]\[
\boxed{\rho (1 + e^{2 \theta}) = k}.
\][/tex]
So, the correct answer is option (a):
[tex]\[
\rho e^{2 \theta} + \rho = k.
\][/tex]
