Answer :
To solve the given differential equation [tex]\(\cot(x) \, dx + \frac{e^y}{e^y + 1} \, dy = 0\)[/tex], let's follow these steps:
1. Rearrange the given equation:
[tex]\[ \cot(x) \, dx + \frac{e^y}{e^y + 1} \, dy = 0 \][/tex]
2. Separate the variables:
[tex]\[ \cot(x) \, dx = -\frac{e^y}{e^y + 1} \, dy \][/tex]
3. Integrate both sides:
[tex]\[ \int \cot(x) \, dx = - \int \frac{e^y}{e^y + 1} \, dy \][/tex]
4. Compute the integral on the left-hand side:
The integral of [tex]\(\cot(x)\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ \int \cot(x) \, dx = \ln|\sin(x)| \][/tex]
5. Compute the integral on the right-hand side:
To integrate [tex]\(- \int \frac{e^y}{e^y + 1} \, dy\)[/tex], we can use the substitution [tex]\(u = e^y + 1\)[/tex]. Notice that [tex]\(du = e^y \, dy\)[/tex], thus:
[tex]\[ \int \frac{e^y}{e^y + 1} \, dy = \int \frac{1}{u} \, du = \ln|u| \][/tex]
Substituting [tex]\(u\)[/tex] back, we get:
[tex]\[ - \int \frac{e^y}{e^y + 1} \, dy = - \ln|e^y + 1| \][/tex]
6. Combine the results of both integrals:
[tex]\[ \ln|\sin(x)| = - \ln|e^y + 1| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
7. Simplify the equation:
Combine the logarithmic terms on one side:
[tex]\[ \ln|\sin(x)| + \ln|e^y + 1| = C \][/tex]
Using the property of logarithms [tex]\(\ln(a) + \ln(b) = \ln(ab)\)[/tex]:
[tex]\[ \ln|\sin(x) \cdot (e^y + 1)| = C \][/tex]
Exponentiating both sides to eliminate the logarithm, we get:
[tex]\[ |\sin(x) \cdot (e^y + 1)| = e^C \][/tex]
Let [tex]\(e^C = C'\)[/tex], where [tex]\(C'\)[/tex] is a new constant. Thus:
[tex]\[ \sin(x) \cdot (e^y + 1) = C' \][/tex]
8. Final result:
[tex]\[ C' = \sin(x) \cdot (e^y + 1) \][/tex]
1. Rearrange the given equation:
[tex]\[ \cot(x) \, dx + \frac{e^y}{e^y + 1} \, dy = 0 \][/tex]
2. Separate the variables:
[tex]\[ \cot(x) \, dx = -\frac{e^y}{e^y + 1} \, dy \][/tex]
3. Integrate both sides:
[tex]\[ \int \cot(x) \, dx = - \int \frac{e^y}{e^y + 1} \, dy \][/tex]
4. Compute the integral on the left-hand side:
The integral of [tex]\(\cot(x)\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ \int \cot(x) \, dx = \ln|\sin(x)| \][/tex]
5. Compute the integral on the right-hand side:
To integrate [tex]\(- \int \frac{e^y}{e^y + 1} \, dy\)[/tex], we can use the substitution [tex]\(u = e^y + 1\)[/tex]. Notice that [tex]\(du = e^y \, dy\)[/tex], thus:
[tex]\[ \int \frac{e^y}{e^y + 1} \, dy = \int \frac{1}{u} \, du = \ln|u| \][/tex]
Substituting [tex]\(u\)[/tex] back, we get:
[tex]\[ - \int \frac{e^y}{e^y + 1} \, dy = - \ln|e^y + 1| \][/tex]
6. Combine the results of both integrals:
[tex]\[ \ln|\sin(x)| = - \ln|e^y + 1| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
7. Simplify the equation:
Combine the logarithmic terms on one side:
[tex]\[ \ln|\sin(x)| + \ln|e^y + 1| = C \][/tex]
Using the property of logarithms [tex]\(\ln(a) + \ln(b) = \ln(ab)\)[/tex]:
[tex]\[ \ln|\sin(x) \cdot (e^y + 1)| = C \][/tex]
Exponentiating both sides to eliminate the logarithm, we get:
[tex]\[ |\sin(x) \cdot (e^y + 1)| = e^C \][/tex]
Let [tex]\(e^C = C'\)[/tex], where [tex]\(C'\)[/tex] is a new constant. Thus:
[tex]\[ \sin(x) \cdot (e^y + 1) = C' \][/tex]
8. Final result:
[tex]\[ C' = \sin(x) \cdot (e^y + 1) \][/tex]
