If \( \frac{dy}{dx} = \frac{2^x y + 2^y \cdot 2^x}{2^x + 2^{x+y} \log_e 2}, y(0) = 0 \), then for \( y = 1 \), the value of \( x \) lies in the interval :
Show Hint
Recognizing the derivative of the denominator in the numerator is a standard trick in differential equations.
Always check if the RHS can be simplified by factoring out common terms from numerator and denominator.
Step 1: Understanding the Concept:
The given differential equation can be simplified by dividing terms or regrouping to make it variable separable or exact. Step 2: Key Formula or Approach:
Simplify the equation:
\[ \frac{dy}{dx} = \frac{2^x (y + 2^y)}{2^x (1 + 2^y \log_e 2)} = \frac{y + 2^y}{1 + 2^y \log_e 2} \] Step 3: Detailed Explanation:
The equation becomes:
\[ \frac{1 + 2^y \log_e 2}{y + 2^y} dy = dx \]
Integrate both sides:
\[ \int \frac{1 + 2^y \ln 2}{y + 2^y} dy = \int dx \]
Notice that the numerator is the derivative of the denominator (\( \frac{d}{dy}(y + 2^y) = 1 + 2^y \ln 2 \)).
\[ \ln |y + 2^y| = x + C \]
Use initial condition \( y(0) = 0 \):
\[ \ln |0 + 2^0| = 0 + C \implies \ln 1 = C \implies C = 0 \]
The solution is \( \ln(y + 2^y) = x \).
For \( y = 1 \):
\[ x = \ln(1 + 2^1) = \ln 3 \]
Since \( e \approx 2.718 \), \( \ln e = 1 \). Since \( 3>e \), \( \ln 3>1 \).
Also \( e^2 \approx 7.38 \), so \( \ln 3<2 \).
Thus, \( x \in (1, 2) \). Step 4: Final Answer:
The value of \( x \) lies in \( (1, 2) \).
