Step 1: Understanding the Question:
We are asked to find the general solution of a given first-order differential equation. This equation is a linear differential equation.
Step 2: Key Formula or Approach:
The given equation is in the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\).
The method to solve this is:
1. Identify \(P(x)\) and \(Q(x)\).
2. Calculate the Integrating Factor (I.F.) using the formula: \(I.F. = e^{\int P(x) dx}\).
3. The general solution is given by: \(y \times (\text{I.F.}) = \int Q(x) \times (\text{I.F.}) dx + C\).
Step 3: Detailed Explanation:
(i) Identify P(x) and Q(x):
Comparing \(\frac{dy}{dx} + y\cot x = \csc x\) with the standard form, we have:
\(P(x) = \cot x\)
\(Q(x) = \csc x\)
(ii) Calculate the Integrating Factor (I.F.):
\[
I.F. = e^{\int \cot x \, dx}
\]
Since \(\int \cot x \, dx = \ln|\sin x|\),
\[
I.F. = e^{\ln|\sin x|} = \sin x
\]
(iii) Find the general solution:
Using the solution formula \(y \times (\text{I.F.}) = \int Q(x) \times (\text{I.F.}) dx + C\):
\[
y \cdot \sin x = \int (\csc x) \cdot (\sin x) \, dx + C
\]
We know that \(\csc x = \frac{1}{\sin x}\), so the product simplifies:
\[
y \sin x = \int \frac{1}{\sin x} \cdot \sin x \, dx + C
\]
\[
y \sin x = \int 1 \, dx + C
\]
Now, integrate with respect to \(x\):
\[
y \sin x = x + C
\]
Step 4: Final Answer:
The general solution of the differential equation is \(y\sin x = x + c\).