If $y = f(x)$ is the solution of the differential equation $(1 + \sin x) \frac{dy}{dx} + \cos x = 0$, such that $f(0) = 0$, then $f\left(\frac{\pi}{2}\right)$ is:
Step 1: Understanding the Concept:
The given equation is a first-order ordinary differential equation.
By observing the terms, we can see that it is a variable separable differential equation.
We need to isolate \( y \) terms on one side and \( x \) terms on the other to integrate. Step 2: Key Formula or Approach:
The standard method for variable separable equations is:
\[ \int g(y) dy = \int h(x) dx \]
In this case, we rearrange the equation as:
\[ (1 + \sin x) \frac{dy}{dx} = -\cos x \implies dy = -\frac{\cos x}{1 + \sin x} dx \] Step 3: Detailed Explanation:
Integrating both sides:
\[ \int dy = -\int \frac{\cos x}{1 + \sin x} dx \]
For the RHS integral, let \( 1 + \sin x = t \), then \( \cos x \, dx = dt \).
\[ y = -\int \frac{1}{t} dt = -\ln |t| + C \]
\[ y = -\ln (1 + \sin x) + C \]
Now, apply the initial condition \( f(0) = 0 \):
\[ 0 = -\ln (1 + \sin 0) + C \implies 0 = -\ln(1) + C \implies C = 0 \]
The specific solution is \( f(x) = -\ln (1 + \sin x) \).
To find \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = -\ln \left(1 + \sin \frac{\pi}{2}\right) = -\ln(1 + 1) = -\ln 2 \] Step 4: Final Answer:
The value of \( f\left(\frac{\pi}{2}\right) \) is \( -\ln 2 \).
