If
\[
\left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x
\]
and
\[
y(0)=2,
\]
then find the value of
\[
y\left(\frac{\pi}{2}\right).
\]
Show Hint
Whenever an integral contains a fraction of the form
\[
\frac{f'(x)}{f(x)},
\]
directly use:
\[
\int \frac{f'(x)}{f(x)}dx=\ln|f(x)|+C
\]
This is one of the most frequently used formulas in differential equations.
Step 1: Understanding the Concept:
This problem is a variation of the first question, emphasizing the same technique of separation of variables.
Differential equations like this are fundamental in engineering to describe systems where the rate of change is proportional to a function of state and time.
The process involves separating differentials and finding the specific function that satisfies a boundary condition. Step 2: Key Formula or Approach:
The equation is:
\[ \frac{dy}{1 + y} = \frac{-\cos x}{2 + \sin x} dx \]
Integrated form:
\[ \ln|1+y| + \ln|2+\sin x| = K \] Step 3: Detailed Explanation:
Rearranging:
\[ \int \frac{dy}{1 + y} = -\int \frac{\cos x dx}{2 + \sin x} \]
Both sides follow the \( \int \frac{f'}{f} = \ln f \) rule.
\[ \ln|1 + y| = -\ln|2 + \sin x| + C \]
\[ \ln|(1 + y)(2 + \sin x)| = C \]
\[ (1 + y)(2 + \sin x) = e^C = K \]
Given \( y(0) = 2 \):
\[ (1 + 2)(2 + \sin 0) = K \]
\[ 3 \times 2 = K \implies K = 6 \]
Equation: \( (1 + y)(2 + \sin x) = 6 \).
Evaluate at \( x = \pi/2 \):
\[ (1 + y)(2 + \sin(\pi/2)) = 6 \]
\[ (1 + y)(3) = 6 \implies 1 + y = 2 \implies y = 1 \] Step 4: Final Answer:
The final calculation yields \( y = 1 \).
This is Option (D).